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89 lines
2.7 KiB
Python
89 lines
2.7 KiB
Python
# Copyright (C) 2008 Canonical Ltd
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#
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# This program is free software; you can redistribute it and/or modify
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# it under the terms of the GNU General Public License as published by
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# the Free Software Foundation; either version 2 of the License, or
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# (at your option) any later version.
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#
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# This program is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU General Public License for more details.
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#
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# You should have received a copy of the GNU General Public License
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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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"""Miscellaneous useful stuff."""
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import os
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def single_plural(n, single, plural):
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"""Return a single or plural form of a noun based on number."""
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if n == 1:
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return single
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else:
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return plural
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def invert_dict(d):
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"""Invert a dictionary with keys matching each value turned into a list."""
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# Based on recipe from ASPN
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result = {}
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for k, v in d.iteritems():
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keys = result.setdefault(v, [])
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keys.append(k)
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return result
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def invert_dictset(d):
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"""Invert a dictionary with keys matching a set of values, turned into lists."""
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# Based on recipe from ASPN
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result = {}
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for k, c in d.iteritems():
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for v in c:
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keys = result.setdefault(v, [])
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keys.append(k)
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return result
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def _common_path_and_rest(l1, l2, common=[]):
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# From http://code.activestate.com/recipes/208993/
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if len(l1) < 1: return (common, l1, l2)
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if len(l2) < 1: return (common, l1, l2)
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if l1[0] != l2[0]: return (common, l1, l2)
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return _common_path_and_rest(l1[1:], l2[1:], common+[l1[0]])
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def common_path(path1, path2):
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"""Find the common bit of 2 paths."""
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return ''.join(_common_path_and_rest(path1, path2)[0])
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def common_directory(paths):
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"""Find the deepest common directory of a list of paths.
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:return: if no paths are provided, None is returned;
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if there is no common directory, '' is returned;
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otherwise the common directory with a trailing / is returned.
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"""
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def get_dir_with_slash(path):
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if path == '' or path.endswith('/'):
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return path
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else:
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dirname, basename = os.path.split(path)
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if dirname == '':
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return dirname
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else:
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return dirname + '/'
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if not paths:
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return None
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elif len(paths) == 1:
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return get_dir_with_slash(paths[0])
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else:
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common = common_path(paths[0], paths[1])
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for path in paths[2:]:
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common = common_path(common, path)
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return get_dir_with_slash(common)
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