Python中的正则表达式替换列表中的项目

Question

This is for an exercise we are doing in uni. I am trying to get all k1-9 and p1-9 strings in the txt file and change them so that each k(n) = 1*n and each p(n) = 0*n (ie p5= 00000, k3= 111, p2= 00). I have managed to gather the k1-9 and p1-9 in a list called codes but I dont know how to proceed.

import re

with open("suspicious_knitting.txt") as file:
    string = file.read()
    codes = re.findall("k[1-9]|p[1-9]" ,string)

Printing codes is like this.:

['k1', 'p1', 'k1', 'p1', 'k1', 'p2', 'k1', 'p2', 'k1', 'p3', 'k1', 'p3', 'k1', 'p1', 'k2', 'p1', 'k2', 'p3', 'k1', 'p2', 'k2', 'p1', 'k2', 'p1', 'k1', 'p1', 'k1', 'p1', 'k2', 'p2', 'k3', 'p1', 'k1', 'p2', 'k1', 'p2', 'k2', 'p1', 'k1', 'p1', 'k1', 'p2', 'k1', 'p2', 'k1', 'p2', 'k2', 'p2', 'k5', 'p2', 'k3', 'p1', 'k1', 'p1', 'k1', 'p2', 'k3', 'p1', 'k2', 'p3']

Answer 1

You could use sub :

import re

text = ' '.join(
    ['k1', 'p1', 'k1', 'p1', 'k1', 'p2', 'k1', 'p2', 'k1', 'p3', 'k1', 'p3', 'k1', 'p1', 'k2', 'p1', 'k2', 'p3',
     'k1', 'p2', 'k2', 'p1', 'k2', 'p1', 'k1', 'p1', 'k1', 'p1', 'k2', 'p2', 'k3', 'p1', 'k1', 'p2', 'k1', 'p2',
     'k2', 'p1', 'k1', 'p1', 'k1', 'p2', 'k1', 'p2', 'k1', 'p2', 'k2', 'p2', 'k5', 'p2', 'k3', 'p1', 'k1', 'p1',
     'k1', 'p2', 'k3', 'p1', 'k2', 'p3'])


def repl(match):
    return int(match.group(2)) * match.group(1)


result = re.sub('([kp])([1-9])', repl, text)
print(result)

Output

k p k p k pp k pp k ppp k ppp k p kk p kk ppp k pp kk p kk p k p k p kk pp kkk p k pp k pp kk p k p k pp k pp k pp kk pp kkkkk pp kkk p k p k pp kkk p kk ppp

Explanation

The pattern ([kp])([1-9]) matches a k or a p followed by any digit between 1 and 9 . For the sub part let's look at the documentation:

Return the string obtained by replacing the leftmost non-overlapping occurrences of pattern in string by the replacement repl.

It turns out that repl can be a function, that receives a match object . In this case the repl takes the second matching group (number of repetitions) cast it to int an multiplies for the first matching group, the letter k or p.

Note that I used as input the example in your question joined by space.

Answer 2

Here is a more classic approach. I continued based on your code, and just replaced values as described in code comments.

    import re

    with open("suspicious_knitting.txt") as file:
        string = file.read()
        codes = re.findall("k[1-9]|p[1-9]" ,string)

        for i in range(len(codes) - 1):
            letter = codes[i][0] # this will be k or p
            number = codes[i][1] # this is number after k/p
            if letter == 'k':
                codes[i] = letter + ('1' * int(number)) # for example, if variable number is 5, 1 will be repeated 5 times
            else:
                codes[i] = letter + ('0' * int(number))

        # now array codes contains desireable values :)