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