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February 10, 2018 03:38
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Pretty inefficient computation of minimum edit distance between two strings
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| # Generate all substrings of string recursively | |
| def substrings(string): | |
| #base case | |
| if len(string) == 0: return '' | |
| else: | |
| ret = [string] | |
| #Slide from left | |
| for x in range(0, len(string)): | |
| sub = string[x:-1] | |
| ret.append(sub) | |
| subs = substrings(sub) | |
| for substr in subs: | |
| if substr != '' and substr not in ret: ret.append(substr) | |
| #Slide from right | |
| for x in range(1, len(string)-1): | |
| ret.append(string[0:-x]) | |
| sub = string[x:] | |
| ret.append(sub) | |
| subs = substrings(sub) | |
| for substr in subs: | |
| if substr != '' and substr not in ret: ret.append(substr) | |
| return ret | |
| def edit_distance(n1, n2): | |
| minD = len(n2) | |
| minSub = '' | |
| for sub in substrings(n1): | |
| if sub in n2: | |
| dist = abs(len(n2) - len(sub)) | |
| if dist <= minD: | |
| minD = dist | |
| minSub = sub | |
| if minSub != '': return minD | |
| else: return len(n2) + abs(len(n2)-len(n1)) | |
| test_cases = [ | |
| {'n1': 'rob', 'n2': 'bob', 'dist': 1}, | |
| {'n1': 'abcdxyz', 'n2': 'bbcdxywa', 'dist': 3}, | |
| {'n1': 'randy', 'n2': 'robert', 'dist': 5}, | |
| {'n1': 'randy', 'n2': 'conrados', 'dist': 6}, | |
| {'n1': 'randy', 'n2': 'josh', 'dist': 5} | |
| ] | |
| for test in test_cases: | |
| n1 = test['n1'] | |
| n2 = test['n2'] | |
| assert(edit_distance(n1, n2) == test['dist']) | |
| print('Count of character edits to convert ' + n1 + ' to ' + n2 + ' is ' + str(edit_distance(n1, n2))) |
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