1# Copyright (c) 2009 Google Inc. All rights reserved. 2# Copyright (c) 2009 Apple Inc. All rights reserved. 3# 4# Redistribution and use in source and binary forms, with or without 5# modification, are permitted provided that the following conditions are 6# met: 7# 8# * Redistributions of source code must retain the above copyright 9# notice, this list of conditions and the following disclaimer. 10# * Redistributions in binary form must reproduce the above 11# copyright notice, this list of conditions and the following disclaimer 12# in the documentation and/or other materials provided with the 13# distribution. 14# * Neither the name of Google Inc. nor the names of its 15# contributors may be used to endorse or promote products derived from 16# this software without specific prior written permission. 17# 18# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS 19# "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT 20# LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR 21# A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT 22# OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, 23# SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT 24# LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, 25# DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY 26# THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT 27# (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE 28# OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. 29 30import re 31 32 33def plural(noun): 34 # This is a dumb plural() implementation that is just enough for our uses. 35 if re.search("h$", noun): 36 return noun + "es" 37 else: 38 return noun + "s" 39 40 41def pluralize(noun, count): 42 if count != 1: 43 noun = plural(noun) 44 return "%d %s" % (count, noun) 45 46 47def join_with_separators(list_of_strings, separator=', ', only_two_separator=" and ", last_separator=', and '): 48 if not list_of_strings: 49 return "" 50 if len(list_of_strings) == 1: 51 return list_of_strings[0] 52 if len(list_of_strings) == 2: 53 return only_two_separator.join(list_of_strings) 54 return "%s%s%s" % (separator.join(list_of_strings[:-1]), last_separator, list_of_strings[-1]) 55