# Copyright (c) 2009 Google Inc. All rights reserved. # Copyright (c) 2009 Apple Inc. All rights reserved. # # Redistribution and use in source and binary forms, with or without # modification, are permitted provided that the following conditions are # met: # # * Redistributions of source code must retain the above copyright # notice, this list of conditions and the following disclaimer. # * Redistributions in binary form must reproduce the above # copyright notice, this list of conditions and the following disclaimer # in the documentation and/or other materials provided with the # distribution. # * Neither the name of Google Inc. nor the names of its # contributors may be used to endorse or promote products derived from # this software without specific prior written permission. # # THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS # "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT # LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR # A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT # OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, # SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT # LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, # DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY # THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT # (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE # OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. import re def plural(noun): # This is a dumb plural() implementation that is just enough for our uses. if re.search("h$", noun): return noun + "es" else: return noun + "s" def pluralize(noun, count): if count != 1: noun = plural(noun) return "%d %s" % (count, noun) def join_with_separators(list_of_strings, separator=', ', only_two_separator=" and ", last_separator=', and '): if not list_of_strings: return "" if len(list_of_strings) == 1: return list_of_strings[0] if len(list_of_strings) == 2: return only_two_separator.join(list_of_strings) return "%s%s%s" % (separator.join(list_of_strings[:-1]), last_separator, list_of_strings[-1])