grammar.py 2.3 KB

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  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. import re
  30. def plural(noun):
  31. # This is a dumb plural() implementation that is just enough for our uses.
  32. if re.search("h$", noun):
  33. return noun + "es"
  34. else:
  35. return noun + "s"
  36. def pluralize(noun, count):
  37. if count != 1:
  38. noun = plural(noun)
  39. return "%d %s" % (count, noun)
  40. def join_with_separators(list_of_strings, separator=', ', only_two_separator=" and ", last_separator=', and '):
  41. if not list_of_strings:
  42. return ""
  43. if len(list_of_strings) == 1:
  44. return list_of_strings[0]
  45. if len(list_of_strings) == 2:
  46. return only_two_separator.join(list_of_strings)
  47. return "%s%s%s" % (separator.join(list_of_strings[:-1]), last_separator, list_of_strings[-1])