optimal_string_alignment.py 2.1 KB

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  1. # Copyright (c) 2018 luozhouyang
  2. #
  3. # Permission is hereby granted, free of charge, to any person obtaining a copy
  4. # of this software and associated documentation files (the "Software"), to deal
  5. # in the Software without restriction, including without limitation the rights
  6. # to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
  7. # copies of the Software, and to permit persons to whom the Software is
  8. # furnished to do so, subject to the following conditions:
  9. #
  10. # The above copyright notice and this permission notice shall be included in all
  11. # copies or substantial portions of the Software.
  12. #
  13. # THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
  14. # IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
  15. # FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
  16. # AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
  17. # LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
  18. # OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
  19. # SOFTWARE.
  20. from .string_distance import StringDistance
  21. class OptimalStringAlignment(StringDistance):
  22. def distance(self, s0, s1):
  23. if s0 is None:
  24. raise TypeError("Argument s0 is NoneType.")
  25. if s1 is None:
  26. raise TypeError("Argument s1 is NoneType.")
  27. if s0 == s1:
  28. return 0.0
  29. n, m = len(s0), len(s1)
  30. if n == 0:
  31. return 1.0 * n
  32. if m == 0:
  33. return 1.0 * m
  34. d = [[0] * (m + 2) for _ in range(n + 2)]
  35. for i in range(n + 1):
  36. d[i][0] = i
  37. for j in range(m + 1):
  38. d[0][j] = j
  39. for i in range(1, n + 1):
  40. for j in range(1, m + 1):
  41. cost = 1
  42. if s0[i - 1] == s1[j - 1]:
  43. cost = 0
  44. d[i][j] = min(d[i - 1][j - 1] + cost, d[i][j - 1] + 1, d[i - 1][j] + 1)
  45. if i > 1 and j > 1 and s0[i - 1] == s1[j - 2] and s0[i - 2] == s1[j - 1]:
  46. d[i][j] = min(d[i][j], d[i - 2][j - 2] + cost)
  47. return d[n][m]