levenshtein-distance

Find near-duplicates of comma-separated lists using Levenshtein distance [duplicate]

自闭症网瘾萝莉.ら 提交于 2019-12-04 06:43:59
问题 This question already has an answer here : Potential Duplicates Detection, with 3 Severity Level (1 answer) Closed 5 years ago . This question based on the answer of my question yesterday. To solve my problem, Jean-François Corbett suggested a Levenshtein distance approach. Then I found this code somewhere to get Levenshtein distance percentage. Public Function GetLevenshteinPercentMatch( _ ByVal string1 As String, ByVal string2 As String, _ Optional Normalised As Boolean = False) As Single

Can't install Levenshtein distance package on Windows Python 3.5

喜你入骨 提交于 2019-12-04 04:45:22
I need to install python Levenshtein distance package in order to use this library . Unfortunately, I am not able to install it succesfully. I usually install libraries with pip. However, this time I am getting error: [WinError 2] The system cannot find the file specified which had never happened to me before (when installing libraries). I have tried to install it using the python setup.py install but I get exactly the same error. This the output I get from the console. C:\Users\my_user\Anaconda3\Lib\site-packages\python-Levenshtein-0.10.2>python setup.py install running install running bdist

Reverse Levenshtein distance

﹥>﹥吖頭↗ 提交于 2019-12-04 04:09:33
In levenshtein distance you ask the question, given these two strings, what is their levenshtein distance. How would you go about taking a string and a levenshtein distance and generating all the strings within that levenshtein distance. (It would also take in a character set). So if i pass in a string x and a distance d. then it would give me all the strings within that edit distance, including d-1 and d-2....d-n; (n < d). Expected functionality: >>> getWithinDistance('apple',2,{'a','b',' '}) ['applea','appleb','appel','app le'...] Please note that the program is able to produce app le as

Damerau - Levenshtein Distance, adding a threshold

醉酒当歌 提交于 2019-12-03 16:45:08
I have the following implementation, but I want to add a threshold, so if the result is going to be greater than it, just stop calculating and return. How would I go about that? EDIT: Here is my current code, threshold is not yet used...the goal is that it is used public static int DamerauLevenshteinDistance(string string1, string string2, int threshold) { // Return trivial case - where they are equal if (string1.Equals(string2)) return 0; // Return trivial case - where one is empty if (String.IsNullOrEmpty(string1) || String.IsNullOrEmpty(string2)) return (string1 ?? "").Length + (string2 ??

How do I convert between a measure of similarity and a measure of difference (distance)?

北慕城南 提交于 2019-12-03 13:59:03
问题 Is there a general way to convert between a measure of similarity and a measure of distance? Consider a similarity measure like the number of 2-grams that two strings have in common. 2-grams('beta', 'delta') = 1 2-grams('apple', 'dappled') = 4 What if I need to feed this to an optimization algorithm that expects a measure of difference, like Levenshtein distance? This is just an example...I'm looking for a general solution, if one exists. Like how to go from Levenshtein distance to a measure

how to convert a string into a palindrome with minimum number of operations?

↘锁芯ラ 提交于 2019-12-03 11:42:49
问题 Here is the problem states to convert a string into a palindrome with minimum number of operations. I know it is similar to the Levenshtein distance but I can't solve it yet For example, for input mohammadsajjadhossain , the output is 8 . 回答1: Perform Levenshtein distance on the string and its reverse. The solution will be the minimum of the operations in the diagonal of the DP array going from bottom-left to top-right, as well as each entry just above and just below the diagonal. This works

Damerau–Levenshtein distance (Edit Distance with Transposition) c implementation

风格不统一 提交于 2019-12-03 11:42:49
I implemented the Damerau–Levenshtein distance in c++ but it does not give correct o/p for the input (pantera,aorta) the correct o/p is 4 but my code gives 5..... int editdist(string s,string t,int n,int m) { int d1,d2,d3,cost; int i,j; for(i=0;i<=n;i++) { for(j=0;j<=m;j++) { if(s[i+1]==t[j+1]) cost=0; else cost=1; d1=d[i][j+1]+1; d2=d[i+1][j]+1; d3=d[i][j]+cost; d[i+1][j+1]=minimum(d1,d2,d3); if(i>0 && j>0 && s[i+1]==t[j] && s[i]==t[j+1] ) //transposition { d[i+1][j+1]=min(d[i+1][j+1],d[i-1][j-1]+cost); } } } return d[n+1][m+1]; } I don't see any errors. Can someone find a problem with the

Is there an edit distance algorithm that takes “chunk transposition” into account?

柔情痞子 提交于 2019-12-03 09:49:59
问题 I put "chunk transposition" in quotes because I don't know whether or what the technical term should be. Just knowing if there is a technical term for the process would be very helpful. The Wikipedia article on edit distance gives some good background on the concept. By taking "chunk transposition" into account, I mean that Turing, Alan. should match Alan Turing more closely than it matches Turing Machine I.e. the distance calculation should detect when substrings of the text have simply been

How to speed up Levenshtein distance calculation

烂漫一生 提交于 2019-12-03 08:17:50
问题 I am trying to run a simulation to test the average Levenshtein distance between random binary strings. My program is in python but I am using this C extension. The function that is relevant and takes most of the time computes the Levenshtein distance between two strings and is this. lev_edit_distance(size_t len1, const lev_byte *string1, size_t len2, const lev_byte *string2, int xcost) { size_t i; size_t *row; /* we only need to keep one row of costs */ size_t *end; size_t half; /* strip

OCR: weighted Levenshtein distance

安稳与你 提交于 2019-12-03 07:40:15
I'm trying to create an optical character recognition system with the dictionary. In fact I don't have an implemented dictionary yet=) I've heard that there are simple metrics based on Levenstein distance which take in account different distance between different symbols. E.g. 'N' and 'H' are very close to each other and d("THEATRE", "TNEATRE") should be less than d("THEATRE", "TOEATRE") which is impossible using basic Levenstein distance. Could you help me locating such metric, please. This might be what you are looking for: http://en.wikipedia.org/wiki/Damerau%E2%80%93Levenshtein_distance