python 从左到右替换字符串中的字符

Question

Ok so I found a similar question to this but it focused on splitting the string into pairs of two characters,

The thing is thought That I want to be able to factor in multiple possibilities for replacements strings that are 2 characters long and 4 characters long. so without splitting the string and keeping it intact I would like to be able to scan the string from left to right and upon finding any "matches" replace this and then continue on scanning. along with prioratising the longer replacement sets first. "0000" becomes "e" and not "aa" or "00 00"

the usual.replace() function re-scans the string for each different value, I want to avoid this.

this is my script:

s = "0000000110110110100111111111"

x = s.replace("00","a").replace("11","b").replace("01","c").replace("10","d").replace("0000","e").replace("1111","f").replace("0101","g").replace("1010","h")

print(x)

My script so far produces: aaa0b0b0bcabbbb1

But I would like to get the result: eacdbchcff

based on the replacement possibilities of: 0000 00 01 10 11 01 1010 01 1111 1111

Answer 1

You could put the translations into a dict, and also combine the search-patterns into a single regular expression, which gives priority to the longer patterns. Then use the callback argument that re.sub accepts to make the replacement using the dict.

import re

trans = {
    "00": "a",
    "11": "b",
    "01": "c",
    "10": "d",
    "0000": "e",
    "1111": "f",
    "0101": "g",
    "1010": "h"
}

regex = "|".join(sorted(trans.keys(), key=len, reverse=True))

# demo
s =  "0000000110110110100111111111"
result = re.sub(regex, lambda x: trans[x.group(0)], s)
print(result)  # eacdbchcff

Answer 2

Non-regex approach would be to assess each section as a set of 4 characters, see if theres a match for those, or split into two halves of the 4 and get a match for them...

replacements = {'0000': 'e', '1111': 'f', '1010': 'h', '0101': 'g', '10': 'd', '01': 'c', '11': 'b', '00': 'a'}
s = "0000000110110110100111111111"
r_d = replacement_dict  # only here to shorten comprehension

for i in range(0, len(s), 4):
     print(r_d.get(s[i:i+4], r_d.get(s[i:i+2], "") +r_d.get(s[i+2:i+4],"")), end="")

or with loop as a list comprehension

"".join(r_d.get(s[i:i+4], r_d.get(s[i:i+2], "") +r_d.get(s[i+2:i+4], "")) for i in range(0, len(s), 4))
'eacdbcddcff'