To identify repeated words and get the distance between two words as a list in python

Question

A = "AEJXKWKJSSSJKZJLJLEJSSLKXMPPLSSKKDNEMSMLDMMEPPLETFMM"
print Repeat_Letter : [PPL:16, JSS:10]   --> What I want

String A is a string that lists random characters that I suggested. Some of them are repetitive string. Repeated string within long strings is "PPL" and "JSS" in string A, respectively. 16 is a distance between the letter "PPL", and 10 is the distance between "JSS". And finally, the goal is to automatically determine repeated words and to express the distance between these words as a list in python.

trigrams = [A[i:i+3] for i in range(len(A)-2)]
counts = collections.Counter(trigrams)
repeated = [trigram for trigram, count in counts.items() if count > 1]

Through this, I checked which word is repeated. However, I'm wondering how to get the distance of these discriminated words. For example, we don't know how to get the distance between "PPL" and other "PPL".

What is the role of the "16" and "10" in your example output? — user1717828, Mar 14 '21 at 13:10
@thethiny We have a better shot of not putting off new SO contributors if we nicely ask for improvements in the comments. — user1717828, Mar 14 '21 at 13:13
@user1717828 Thank you for the comment. I edited my question. 16 is a distance between the letter "PPL", and 10 is the distance between "JSS". — Emma Lim, Mar 14 '21 at 13:13
@NathanXabedi I defined the repetitive part of that long string as "word." I intended the word to be the "PPL" and "JSS" parts. — Emma Lim, Mar 14 '21 at 13:17
@thethiny I edited the content with adding what I tried. Only thing left is to discriminate the distance between the letters. — Emma Lim, Mar 14 '21 at 13:23

sharath k · Answer 1 · 2021-03-14T14:36:11.440

1

This uses find and string slicing method

def find_distance_in_dups(string,length):
  dict_words={}
  for i in range(len(string)-length-1):
    word = string[i:length+i]
    distance=string[string.find(word)+length:].find(word)+1
    #print(distance)
    if distance > 0:
      dict_words[word]=distance
  #print(dict_words)
  return dict_words
print(find_distance_in_dups("AEJXKWKJSSSJKZJLJLEJSSLKXMPPLSSKKDNEMSMLDMMEPPLETFMM",3))

edited Mar 14 '21 at 14:36

answered Mar 14 '21 at 14:05

sharath k

56
2

The error occurs this. "NameError: name 'dict_words' is not defined" What did you mean? – Emma Lim Mar 14 '21 at 14:17
1

I added the empty initialisation for dict_words. Please try again – sharath k Mar 14 '21 at 14:36

score 0 · Answer 2 · edited Mar 14 '21 at 13:30

0

Just wrote this one,

To solve your problem you however have to indicate the length of the word you are looking for I think as a parameter.

import numpy as np 

def find_duplicates(string,length=3):
    all_words = np.array([string[0+i:length+i] for i in range(len(string)-length)])
    unique_words = np.unique(all_words)
    dico = {}
    for unique in unique_words:
        loc = np.where(all_words==unique)[0]
        if len(loc)>1:
            dico[unique] = np.diff(loc) - (length - 1) #distance between consecutive duplicates
    return dico

A = "AEJXKWKJSSSJKZJLJLEJSSLKXMPPLSSKKDNEMSMLDMMEPPLETFMM"

print(find_duplicates(A,length=3)) # {'JSS': 10, 'PPL': 16}

edited Mar 14 '21 at 13:30

Dharman

30,962
25
85
135

answered Mar 14 '21 at 13:25

Michaël Cretignier

61
6

What's the role of np.where(three_words==unique)[0]? It seems "np.where(three_words==unique)" prints array[x] and dtype. So adding [0] means array[x], but I can't understand what this means. – Emma Lim Mar 14 '21 at 13:50
Yes, and you want the array[x] from the tuple output of np.where – Michaël Cretignier Mar 14 '21 at 14:00

makes · Accepted Answer · 2021-03-14T14:21:09.100

You can use regular expressions as described in this answer:

import collections, re

Code copied from question:

A = "AEJXKWKJSSSJKZJLJLEJSSLKXMPPLSSKKDNEMSMLDMMEPPLETFMM"                                                                                                                                                                            

trigrams = [A[i:i+3] for i in range(len(A)-2)]
counts = collections.Counter(trigrams)
repeated = [trigram for trigram, count in counts.items() if count > 1]

Find distances using regex:

dists = {}
for r in repeated:
    matches = [m for m in re.finditer(r, A)]
    dists[r] = matches[1].start() - (matches[0].end()-1)                                                                                                                                                                                

print(dists)

Output:

{'JSS': 10, 'PPL': 16}

This of course finds only the distance between the first two matches. You didn't specify if further occurrences should count.

To identify repeated words and get the distance between two words as a list in python

3 Answers3