将字符串中的字符随机替换为当前字符以外的字符

Question

Suppose that I have a string that I would like to modify at random with a defined set of options from another string. First, I created my original string and the potential replacement characters:

string1 = "abcabcabc"
replacement_chars = "abc"

Then I found this function on a forum that will randomly replace n characters:

def randomlyChangeNChar(word, value):
     length = len(word)
     word = list(word)
     # This will select the distinct index for us to replace
     k = random.sample(range(0, length), value) 
     for index in k:
         # This will replace the characters at the specified index with the generated characters
         word[index] = random.choice(replacement_chars)
# Finally print the string in the modified format.
return "".join(word)

This code does what I want with one exception -- it does not account for characters in string1 that match the random replacement character. I understand that the problem is in the function that I am trying to adapt, I predict under the for loop, but I am unsure what to add to prevent the substituting character from equaling the old character from string1. All advice appreciated, if I'm overcomplicating things please educate me!

Answer 1

In the function you retrieved, replacing:

word[index] = random.choice(replacement_chars)

with

word[index] = random.choice(replacement_chars.replace(word[index],'')

will do the job. It simply replaces word[index] (the char you want to replace) with an empty string in the replacement_chars string, effectively removing it from the replacement characters.

Another approach, that will predictably be less efficient on average, is to redraw until you get a different character from the original one:

that is, replacing:

word[index] = random.choice(replacement_chars)

with

char = word[index]
while char == word[index]:
    char = random.choice(replacement_chars)
word[index] = char

or

while True:
    char = random.choice(replacement_chars)
    if char != word[index]:
        word[index] = char
        break

WARNING: if replacement_chars only features 1 character, both methods would fail when the original character is the same as the replacement one!