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 30 import re 31 32 33 def 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 41 def pluralize(noun, count): 42 if count != 1: 43 noun = plural(noun) 44 return "%d %s" % (count, noun) 45 46 47 def 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