Adding items to list when they already exist and when using for loop

Question

This is a pretty simple question about lists and for loop.

Assuming I have the following 2d-list:

[

['c', 'a', 't', 'c', 'a', 't']

['a', 'b', 'c', 'a', 't, 'l']

['c', 'a', 't', 'w', 'x', 'y']

]

And I want to iterate over the list using for loops, each time checking if the word 'cat' is in the list. If it is, I want to add it to the list every time it appears.

So my result should be ['cat', 'cat', 'cat, 'cat']

My function receives a word list and a given matrix (2d list containing lists of letters). My code is:

def search_for_words(word_list, matrix):
    results = []
    for word in word_list:
        for line in matrix:
            line_string = ''.join(line)
                if word in line_string:
                    results.append(word)
    return results

And it will returns me just 'cat' if cat is in thr word list.

I know I probably just need another if statement but I can figure it out.

Thanks in advance.

EDIT:

I gave a wrong example.

consider this:

matrix = [['a', 'p', 'p', 'l', 'e'], 
      ['a', 'g', 'o', 'd', 'o'],
      ['n', 'n', 'e', 'r', 't'],
      ['g', 'a', 'T', 'A', 'C'],
      ['m', 'i', 'c', 's', 'r'],
      ['P', 'o', 'P', 'o', 'P']]

word_list = ['apple', 'god', 'dog', 'CAT', 'PoP', 'poeT]

my function returns :

['apple', 'god', 'PoP']

When I expect it to return 'PoP' twice because it appears twice at the bottom list.

Welcome to SO! Can't reproduce your problem, code produces the correct output, assuming you fix your indentation error. — ggorlen, Nov 18 '18 at 20:28
For me, the returned list has 3 occurrences of `cat` (instead of 4) but that is a different problem. — slider, Nov 18 '18 at 20:34
I edited my question with the appropriate example. Sorry for the confusion. @ggorlen — Tegernako, Nov 18 '18 at 20:38
the line `if word in line_string` checks for the `word` in `line_string` only once. you will need to find the index of the first match then again look for the next match starting from the `index of the last match + 1` — Rahul Virpara, Nov 18 '18 at 20:46

score 4 · Accepted Answer · answered Nov 18 '18 at 20:46

The problem is you're not checking how many times that substring occurs in the string. You also need to account for overlapping matches:

import re

def search_for_words(word_list, matrix):
    results = []
    for word in word_list:
        for line in matrix:
            line_string = ''.join(line)
            # find all overlapping matches of word in line_string
            matches = re.findall(r'(?=(' + word + '))', line_string)
            results.extend(matches)
    return results

If we run this on your second matrix:

m = [['a', 'p', 'p', 'l', 'e'], 
     ['a', 'g', 'o', 'd', 'o'],
     ['n', 'n', 'e', 'r', 't'],
     ['g', 'a', 'T', 'A', 'C'],
     ['m', 'i', 'c', 's', 'r'],
     ['P', 'o', 'P', 'o', 'P']]

word_list = ['apple', 'god', 'dog', 'CAT', 'PoP', 'poeT']

print(search_for_words(word_list, m))

we see the following output:

['apple', 'god', 'PoP', 'PoP']

Exactly what I needed. Thanks. – Tegernako Nov 19 '18 at 11:03 — Tegernako, Nov 19 '18 at 11:03

score 1 · Answer 2 · answered Nov 18 '18 at 21:51

The answer to your first question is pretty simple. Just build a string and check it each time. Lose the first letter of the string if the string is longer than the target word.

outer = [['c', 'a', 't', 'c', 'a', 't'],
         ['a', 'b', 'c', 'a', 't', 'l'],
         ['c', 'a', 't', 'w', 'x', 'y']]

for inner in outer:
  word = ""
  target = "cat"
  t_len = len(target)
  for letter in inner:
    word += letter
    if len(word) > t_len:
      word = word[1:]
    if word in target:
      print(target)

For the answer to your second question, you're probably best to just "".join() all of the strings and parse them with regex.

Adding items to list when they already exist and when using for loop

2 Answers2

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