1
# -*- coding: UTF-8 -*-
2
"""Directed graph production.
4
This module contains the code to produce an ordered directed graph of a
5
bzr branch, such as we display in the tree view at the top of the bzrk
9
__copyright__ = "Copyright © 2005 Canonical Ltd."
10
__author__ = "Scott James Remnant <scott@ubuntu.com>"
12
from bzrlib.revision import NULL_REVISION
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from bzrlib.tsort import merge_sort
16
def linegraph(graph, start_revs, maxnum=None, broken_line_length=None,
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graph_data=True, mainline_only=False, root_progress=None):
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"""Produce a directed graph of a bzr repository.
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Returns a tuple of (line_graph, revid_index, columns_len) where
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* line_graph is a list of tuples of (revid,
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* revid_index is a dict of each revision with the key being the revid, and
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the value the row index, and
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* columns_len is the number of columns need to draw the line graph.
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Node is a tuple of (column, colour) with column being a zero-indexed
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column number of the graph that this revision represents and colour
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being a zero-indexed colour (which doesn't specify any actual colour
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in particular) to draw the node in.
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Lines is a list of tuples which represent lines you should draw away
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from the revision, if you also need to draw lines into the revision
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you should use the lines list from the previous iteration. Each
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typle in the list is in the form (start, end, colour) with start and
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end being zero-indexed column numbers and colour as in node.
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It's up to you how to actually draw the nodes and lines (straight,
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curved, kinked, etc.) and to pick the actual colours for each index.
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assert isinstance(start_revs, list)
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def update_root_progress(step_number):
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"""IFF our container received a root progress bar, then update it."""
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if root_progress is not None:
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root_progress.update(current_cnt=step_number)
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update_root_progress(1)
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progress_bar = ui.ui_factory.nested_progress_bar()
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progress_bar.update(msg="Arranging tree fragments")
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for i, (revid, parent_revids) in enumerate(graph.iter_ancestry(start_revs)):
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if parent_revids is None:
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if parent_revids == (NULL_REVISION,):
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graph_parents[revid] = ()
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graph_parents[revid] = parent_revids
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for parent in parent_revids:
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graph_children.setdefault(parent, []).append(revid)
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graph_children.setdefault(revid, [])
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progress_bar.finished()
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update_root_progress(2)
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progress_bar = ui.ui_factory.nested_progress_bar()
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progress_bar.update(msg="Removing ghosts", total_cnt=len(ghosts))
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for i, ghost in enumerate(ghosts):
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progress_bar.update(None, current_cnt=i)
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for ghost_child in graph_children[ghost]:
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graph_parents[ghost_child] = [p for p in graph_parents[ghost_child]
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progress_bar.finished()
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graph_parents["top:"] = start_revs
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if len(graph_parents)>0:
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merge_sorted_revisions = merge_sort(
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merge_sorted_revisions = ()
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merge_sorted_revisions = [elem for elem in merge_sorted_revisions \
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assert merge_sorted_revisions[0][1] == "top:"
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merge_sorted_revisions = merge_sorted_revisions[1:]
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# This will hold an item for each "branch". For a revisions, the revsion
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# number less the least significant digit is the branch_id, and used as the
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# key for the dict. Hence revision with the same revsion number less the
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# least significant digit are considered to be in the same branch line.
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# e.g.: for revisions 290.12.1 and 290.12.2, the branch_id would be 290.12,
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# and these two revisions will be in the same branch line. Each value is
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# a list of rev_indexes in the branch.
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update_root_progress(3)
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progress_bar = ui.ui_factory.nested_progress_bar()
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progress_bar.update(msg="Finding nodes", total_cnt=len(merge_sorted_revisions))
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for (rev_index, (sequence_number,
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end_of_merge)) in enumerate(merge_sorted_revisions):
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if rev_index % 25 == 0:
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progress_bar.update(None, current_cnt=rev_index)
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if maxnum and rev_index >= maxnum:
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revid_index[revid] = rev_index
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parents = graph_parents[revid]
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linegraph.append([revid,
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revno_index[revno_sequence] = rev_index
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branch_id = revno_sequence[0:-1]
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if branch_id not in branch_lines:
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branch_lines[branch_id] = branch_line
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branch_line = branch_lines[branch_id]
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branch_line.append(rev_index)
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progress_bar.finished()
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branch_ids = branch_lines.keys()
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def branch_id_cmp(x, y):
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"""Compaire branch_id's first by the number of digits, then reversed
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return cmp(len_x, len_y)
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branch_ids.sort(branch_id_cmp)
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# This will hold a tuple of (child_index, parent_index, col_index) for each
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# line that needs to be drawn. If col_index is not none, then the line is
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# drawn along that column, else the the line can be drawn directly between
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# the child and parent because either the child and parent are in the same
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# branch line, or the child and parent are 1 row apart.
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empty_column = [False for i in range(len(graph_parents))]
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# This will hold a bit map for each cell. If the cell is true, then the
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# cell allready contains a node or line. This use when deciding what column
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# to place a branch line or line in, without it overlaping something else.
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columns = [list(empty_column)]
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update_root_progress(4)
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progress_bar = ui.ui_factory.nested_progress_bar()
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progress_bar.update(msg="Organizing edges", total_cnt=len(branch_ids))
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for i, branch_id in enumerate(branch_ids):
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progress_bar.update(None, current_cnt=i)
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branch_line = branch_lines[branch_id]
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# Find the col_index for the direct parent branch. This will be the
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# starting point when looking for a free column.
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if len(branch_id) > 1:
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parent_revno = branch_id[0:-1]
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if parent_revno in revno_index:
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parent_index = revno_index[parent_revno]
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parent_node = linegraph[parent_index][1]
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parent_col_index = parent_node[0]
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col_search_order = _branch_line_col_search_order(columns,
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color = reduce(lambda x, y: x+y, branch_id, 0)
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last_rev_index = None
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for rev_index in branch_line:
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if broken_line_length and \
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rev_index - last_rev_index > broken_line_length:
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line_range.append(last_rev_index+1)
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line_range.append(rev_index-1)
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line_range.extend(range(last_rev_index+1, rev_index))
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line_range.append(rev_index)
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last_rev_index = rev_index
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if broken_line_length and \
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parent_index - last_rev_index > broken_line_length:
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line_range.append(last_rev_index+1)
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line_range.extend(range(last_rev_index+1, parent_index))
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col_index = _find_free_column(columns,
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node = (col_index, color)
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for rev_index in branch_line:
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linegraph[rev_index][1] = node
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columns[col_index][rev_index] = True
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for rev_index in branch_line:
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end_of_merge) = merge_sorted_revisions[rev_index]
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linegraph[rev_index][4] = graph_children[revid]
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col_index = linegraph[rev_index][1][0]
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for parent_revid in graph_parents[revid]:
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if parent_revid in revid_index:
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parent_index = revid_index[parent_revid]
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parent_node = linegraph[parent_index][1]
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parent_col_index = parent_node[0]
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parent_col_index = None
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_line_col_search_order(columns,
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# If this line is really long, break it.
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if len(branch_id) > 0 and \
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broken_line_length and \
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parent_index - rev_index > broken_line_length:
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child_line_col_index = \
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_find_free_column(columns,
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_mark_column_as_used(columns,
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child_line_col_index,
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# Recall _line_col_search_order to reset it back to
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_line_col_search_order(columns,
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parent_col_line_index = \
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_find_free_column(columns,
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_mark_column_as_used(columns,
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parent_col_line_index,
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lines.append((rev_index,
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(child_line_col_index,
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parent_col_line_index)))
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line_col_index = col_index
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if parent_index - rev_index >1:
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line_range = range(rev_index + 1, parent_index)
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_find_free_column(columns,
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_mark_column_as_used(columns,
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lines.append((rev_index,
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progress_bar.finished()
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update_root_progress(5)
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progress_bar = ui.ui_factory.nested_progress_bar()
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progress_bar.update(msg="Pretifying graph", total_cnt=len(lines))
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for i, (child_index, parent_index, line_col_indexes) in enumerate(lines):
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progress_bar.update(None, current_cnt=i)
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(child_col_index, child_color) = linegraph[child_index][1]
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(parent_col_index, parent_color) = linegraph[parent_index][1]
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if len(line_col_indexes) == 1:
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if parent_index - child_index == 1:
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linegraph[child_index][2].append(
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# line from the child's column to the lines column
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linegraph[child_index][2].append(
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# lines down the line's column
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for line_part_index in range(child_index+1, parent_index-1):
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linegraph[line_part_index][2].append(
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(line_col_indexes[0],
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# line from the line's column to the parent's column
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linegraph[parent_index-1][2].append(
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(line_col_indexes[0],
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# line from the child's column to the lines column
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linegraph[child_index][2].append(
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linegraph[child_index+1][2].append(
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(line_col_indexes[0],
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linegraph[parent_index-2][2].append(
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# line from the line's column to the parent's column
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linegraph[parent_index-1][2].append(
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(line_col_indexes[1],
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progress_bar.finished()
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return (linegraph, revid_index, len(columns))
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return (linegraph, revid_index, 0)
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def _branch_line_col_search_order(columns, parent_col_index):
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for col_index in range(parent_col_index, len(columns)):
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for col_index in range(parent_col_index-1, -1, -1):
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def _line_col_search_order(columns, parent_col_index, child_col_index):
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if parent_col_index is not None:
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max_index = max(parent_col_index, child_col_index)
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min_index = min(parent_col_index, child_col_index)
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for col_index in range(max_index, min_index -1, -1):
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max_index = child_col_index
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min_index = child_col_index
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yield child_col_index
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while max_index + i < len(columns) or \
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if max_index + i < len(columns):
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if min_index - i > -1:
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def _find_free_column(columns, empty_column, col_search_order, line_range):
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for col_index in col_search_order:
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column = columns[col_index]
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has_overlaping_line = False
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for row_index in line_range:
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if column[row_index]:
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has_overlaping_line = True
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if not has_overlaping_line:
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col_index = len(columns)
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column = list(empty_column)
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columns.append(column)
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def _mark_column_as_used(columns, col_index, line_range):
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column = columns[col_index]
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for row_index in line_range:
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column[row_index] = True
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def same_branch(a, b):
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"""Return whether we think revisions a and b are on the same branch."""
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if len(a.parent_ids) == 1:
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# Defacto same branch if only parent
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elif a.committer == b.committer:
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# Same committer so may as well be
1
# -*- coding: UTF-8 -*-
2
"""Directed graph production.
4
This module contains the code to produce an ordered directed graph of a
5
bzr branch, such as we display in the tree view at the top of the bzrk
9
__copyright__ = "Copyright © 2005 Canonical Ltd."
10
__author__ = "Scott James Remnant <scott@ubuntu.com>"
13
from bzrlib.revision import Revision
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from bzrlib.tsort import merge_sort
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class DummyRevision(Revision):
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"""Dummy bzr revision.
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Sometimes, especially in older bzr branches, a revision is referenced
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as the parent of another but not actually present in the branch's store.
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When this happens we use an instance of this class instead of the real
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Revision object (which we can't get).
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def __init__(self, revid):
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super(DummyRevision, self).__init__(revid)
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class RevisionProxy(object):
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"""A revision proxy object.
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This will demand load the revision it represents when the committer or
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message attributes are accessed in order to populate them. It is
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constructed with the revision id and parent_ids list and a repository
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object to request the revision from when needed.
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def __init__(self, revid, parent_ids, repository):
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self.revision_id = revid
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self.parent_ids = parent_ids
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self._repository = repository
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def _get_attribute_getter(attr):
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def get_attribute(self):
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if self._revision is None:
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return getattr(self._revision, attr)
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committer = property(_get_attribute_getter('committer'))
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message = property(_get_attribute_getter('message'))
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properties = property(_get_attribute_getter('properties'))
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timestamp = property(_get_attribute_getter('timestamp'))
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timezone = property(_get_attribute_getter('timezone'))
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"""Load the revision object."""
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self._revision = self._repository.get_revision(self.revision_id)
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class DistanceMethod(object):
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def __init__(self, branch, start):
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self.children_of_id = {start: set()}
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self.parent_ids_of = {}
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self.colours = { start: 0 }
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self.direct_parent_of = {}
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def fill_caches(self):
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graph = self.branch.repository.get_revision_graph_with_ghosts([self.start])
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for revid in graph.ghosts:
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self.cache_revision(DummyRevision(revid))
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for revid, parents in graph.get_ancestors().items():
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self.cache_revision(RevisionProxy(revid, parents, self.branch.repository))
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def cache_revision(self, revision):
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"Set the caches for a newly retrieved revision."""
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revid = revision.revision_id
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# Build a revision cache
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self.revisions[revid] = revision
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# Build a children dictionary
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for parent_id in revision.parent_ids:
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self.children_of_id.setdefault(parent_id, set()).add(revision)
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# Build a parents dictionnary, where redundant parents will be removed,
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# and that will be passed along tothe rest of program.
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if len(revision.parent_ids) != len(set(revision.parent_ids)):
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# fix the parent_ids list.
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parent_ids_set = set()
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for parent_id in revision.parent_ids:
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if parent_id in parent_ids_set:
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parent_ids.append(parent_id)
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parent_ids_set.add(parent_id)
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revision.parent_ids = parent_ids
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self.parent_ids_of[revision] = list(revision.parent_ids)
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self.graph[revid] = revision.parent_ids
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def make_children_map(self):
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revisions = self.revisions
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return dict((revisions[revid], c)
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for (revid, c) in self.children_of_id.iteritems())
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def sort_revisions(self, sorted_revids, maxnum):
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revisions = self.revisions
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parent_ids_of = self.parent_ids_of
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children_of_id = self.children_of_id
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# Try to compact sequences of revisions on the same branch.
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expected_id = sorted_revids[0]
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revid = sorted_revids.pop(0)
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if revid != expected_id:
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skipped_revids.append(revid)
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revision = revisions[revid]
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for child in children_of_id[revid]:
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# postpone if any child is missing
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if child.revision_id not in distances:
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if expected_id not in pending_ids:
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pending_ids.append(expected_id)
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expected_id = pending_ids.pop(0)
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skipped_revids.append(revid)
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sorted_revids[:0] = skipped_revids
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del skipped_revids[:]
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# all children are here, push!
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distances[revid] = len(distances)
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if maxnum is not None and len(distances) > maxnum:
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# bail out early if a limit was specified
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sorted_revids[:0] = skipped_revids
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for revid in sorted_revids:
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distances[revid] = len(distances)
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# all parents will need to be pushed as soon as possible
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for parent in parent_ids_of[revision]:
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if parent not in pending_ids:
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pending_ids.insert(0, parent)
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expected_id = pending_ids.pop(0)
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# if the next expected revid has already been skipped, requeue
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# the skipped ids, except those that would go right back to the
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if expected_id in skipped_revids:
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pos = skipped_revids.index(expected_id)
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sorted_revids[:0] = skipped_revids[pos:]
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del skipped_revids[pos:]
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self.distances = distances
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return sorted(distances, key=distances.get)
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def choose_colour(self, revid):
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revision = self.revisions[revid]
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children_of_id = self.children_of_id
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parent_ids_of = self.parent_ids_of
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colours = self.colours
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the_children = children_of_id[revid]
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if len(the_children) == 1:
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[child] = the_children
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if len(parent_ids_of[child]) == 1:
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# one-one relationship between parent and child, same
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colours[revid] = colours[child.revision_id]
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self.choose_colour_one_child(revision, child)
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self.choose_colour_many_children(revision, the_children)
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def choose_colour_one_child(self, revision, child):
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revid = revision.revision_id
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direct_parent_of = self.direct_parent_of
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revisions = self.revisions
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# one child with multiple parents, the first parent with
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# the same committer gets the colour
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direct_parent = direct_parent_of.get(child)
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if direct_parent is None:
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# if it has not been found yet, find it now and remember
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for parent_id in self.parent_ids_of[child]:
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parent_revision = revisions[parent_id]
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if parent_revision.committer == child.committer:
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# found the first parent with the same committer
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direct_parent = parent_revision
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direct_parent_of[child] = direct_parent
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if direct_parent == revision:
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self.colours[revid] = self.colours[child.revision_id]
200
self.colours[revid] = self.last_colour = self.last_colour + 1
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def choose_colour_many_children(self, revision, the_children):
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"""Colour revision revision."""
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revid = revision.revision_id
205
direct_parent_of = self.direct_parent_of
206
# multiple children, get the colour of the last displayed child
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# with the same committer which does not already have its colour
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for child in the_children:
211
if child.committer != revision.committer:
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direct_parent = direct_parent_of.get(child)
214
if direct_parent == revision:
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self.colours[revid] = self.colours[child.revision_id]
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# FIXME: Colouring based on whats been displayed MUST be done with
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# knowledge of the revisions being output.
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# until the refactoring to fold graph() into this more compactly is
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# done, I've disabled this reuse. RBC 20060403
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# if direct_parent is None:
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# available[child] = distances[child.revision_id]
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# .. it will be something like available[child] = \
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# revs[child.revision_id][0] - which is the sequence number
227
sorted_children = sorted(available, key=available.get)
228
child = sorted_children[-1]
229
direct_parent_of[child] = revision
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self.colours[revid] = self.colours[child.revision_id]
232
# no candidate children is available, pick the next
234
self.colours[revid] = self.last_colour = self.last_colour + 1
237
def distances(branch, start):
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"""Sort the revisions.
240
Traverses the branch revision tree starting at start and produces an
241
ordered list of revisions such that a revision always comes after
242
any revision it is the parent of.
244
Returns a tuple of (revids, revisions, colours, children)
246
distance = DistanceMethod(branch, start)
247
distance.fill_caches()
248
distance.merge_sorted = merge_sort(distance.graph, distance.start)
249
children = distance.make_children_map()
251
for seq, revid, merge_depth, end_of_merge in distance.merge_sorted:
252
distance.choose_colour(revid)
254
revisions = distance.revisions
255
colours = distance.colours
256
parent_ids_of = distance.parent_ids_of
257
return (revisions, colours, children, parent_ids_of, distance.merge_sorted)
260
def graph(revisions, colours, merge_sorted):
261
"""Produce a directed graph of a bzr branch.
263
For each revision it then yields a tuple of (revision, node, lines).
264
If the revision is only referenced in the branch and not present in the
265
store, revision will be a DummyRevision object, otherwise it is the bzr
266
Revision object with the meta-data for the revision.
268
Node is a tuple of (column, colour) with column being a zero-indexed
269
column number of the graph that this revision represents and colour
270
being a zero-indexed colour (which doesn't specify any actual colour
271
in particular) to draw the node in.
273
Lines is a list of tuples which represent lines you should draw away
274
from the revision, if you also need to draw lines into the revision
275
you should use the lines list from the previous iteration. Each
276
typle in the list is in the form (start, end, colour) with start and
277
end being zero-indexed column numbers and colour as in node.
279
It's up to you how to actually draw the nodes and lines (straight,
280
curved, kinked, etc.) and to pick the actual colours for each index.
282
if not len(merge_sorted):
284
# split merge_sorted into a map:
286
# FIXME: get a hint on this from the merge_sorted data rather than
287
# calculating it ourselves
288
# mapping from rev_id to the sequence number of the next lowest rev
290
# mapping from rev_id to next-in-branch-revid - may be None for end
292
next_branch_revid = {}
293
# the stack we are in in the sorted data for determining which
294
# next_lower_rev to set. It is a stack which has one list at each
295
# depth - the ids at that depth that need the same id allocated.
297
for seq, revid, indent, end_merge in merge_sorted:
298
revs[revid] = (seq, indent, end_merge)
299
if indent == len(current_stack):
300
# new merge group starts
301
current_stack.append([revid])
302
elif indent == len(current_stack) - 1:
303
# part of the current merge group
304
current_stack[-1].append(revid)
306
# end of a merge group
307
while current_stack[-1]:
308
stack_rev_id = current_stack[-1].pop()
309
# record the next lower rev for this rev:
310
next_lower_rev[stack_rev_id] = seq
311
# if this followed a non-end-merge rev in this group note that
312
if len(current_stack[-1]):
313
if not revs[current_stack[-1][-1]][2]:
314
next_branch_revid[current_stack[-1][-1]] = stack_rev_id
316
# append to the now-current merge group
317
current_stack[-1].append(revid)
318
# assign a value to all the depth 0 revisions
319
while current_stack[-1]:
320
stack_rev_id = current_stack[-1].pop()
321
# record the next lower rev for this rev:
322
next_lower_rev[stack_rev_id] = len(merge_sorted)
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# if this followed a non-end-merge rev in this group note that
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if len(current_stack[-1]):
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if not revs[current_stack[-1][-1]][2]:
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next_branch_revid[current_stack[-1][-1]] = stack_rev_id
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# a list of the current revisions we are drawing lines TO indicating
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# the sequence of their lines on the screen.
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# i.e. [A, B, C] means that the line to A, to B, and to C are in
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# (respectively), 0, 1, 2 on the screen.
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hanging = [merge_sorted[0][1]]
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for seq, revid, indent, end_merge in merge_sorted:
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# a list of the lines to draw: their position in the
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# previous row, their position in this row, and the colour
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# (which is the colour they are routing to).
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for h_idx, hang in enumerate(hanging):
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# one of these will be the current lines node:
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# we are drawing a line. h_idx
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# we have found the current lines node
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node = (h_idx, colours[revid])
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# note that we might have done the main parent
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drawn_parents = set()
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def draw_line(from_idx, to_idx, revision_id):
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n_idx = new_hanging.index(revision_id)
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# force this to be vertical at the place this rev was
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new_hanging.insert(to_idx, revision_id)
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lines.append((from_idx, n_idx, colours[revision_id]))
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# we want to draw a line to the next commit on 'this' branch
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# drop this line first.
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parent_id = next_branch_revid[revid]
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draw_line(h_idx, h_idx, parent_id)
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# we have drawn this parent
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drawn_parents.add(parent_id)
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# this is the last revision in a 'merge', show where it came from
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if len(revisions[revid].parent_ids) > 1:
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# parents means this commit was a merge, and being
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# the end point of a merge group means that all
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# the parent revisions were merged into branches
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# to the left of this before this was committed
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# - so we want to show this as a new branch from
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# to do this, we show the parent with the lowest
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# sequence number, which is the one that this
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# branch 'spawned from', and no others.
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# If this sounds like a problem, remember that:
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# if the parent was not already in our mainline
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# it would show up as a merge into this making
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# this not the end of a merge-line.
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lowest = len(merge_sorted)
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for parent_id in revisions[revid].parent_ids:
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if revs[parent_id][0] < lowest:
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lowest = revs[parent_id][0]
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assert lowest != len(merge_sorted)
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draw_line(h_idx, len(new_hanging), merge_sorted[lowest][1])
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drawn_parents.add(merge_sorted[lowest][1])
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elif len(revisions[revid].parent_ids) == 1:
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# only one parent, must show this link to be useful.
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parent_id = revisions[revid].parent_ids[0]
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draw_line(h_idx, len(new_hanging), parent_id)
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drawn_parents.add(parent_id)
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# what do we want to draw lines to from here:
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# each parent IF its relevant.
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# Now we need to hang its parents, we put them at the point
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# the old column was so anything to the right of this has
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# to move outwards to make room. We also try and collapse
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# hangs to keep the graph small.
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# RBC: we do not draw lines to parents that were already merged
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# unless its the last revision in a merge group.
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for parent_id in revisions[revid].parent_ids:
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if parent_id in drawn_parents:
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parent_seq = revs[parent_id][0]
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parent_depth = revs[parent_id][1]
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if parent_depth == indent + 1:
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# The parent was a merge into this branch determine if
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# it was already merged into the mainline via a
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# different merge: if all revisions between us and
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# parent_seq have a indent greater than there are no
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# revisions with a lower indent than us.
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# We do not use 'parent_depth < indent' because that
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# would allow un-uniqueified merges to show up, and
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# merge_sorted should take care of that for us (but
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# does not trim the values)
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if parent_seq < next_lower_rev[revid]:
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draw_line(h_idx, len(new_hanging), parent_id)
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elif parent_depth == indent and parent_seq == seq + 1:
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# part of this branch
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draw_line(h_idx, len(new_hanging), parent_id)
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# draw a line from the previous position of this line to the
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# h_idx is the old position.
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# new_indent is the new position.
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draw_line(h_idx, len(new_hanging), hang)
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# we've calculated the row, assign new_hanging to hanging to setup for
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hanging = new_hanging
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yield (revisions[revid], node, lines)
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def same_branch(a, b):
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"""Return whether we think revisions a and b are on the same branch."""
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if len(a.parent_ids) == 1:
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# Defacto same branch if only parent
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elif a.committer == b.committer:
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# Same committer so may as well be