# 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])
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