如何使用正则表达式替换字符串中的数字

Question

I am quite new in regular expression so I am having confusion in replacing numbers inside an string.

a="12ab34cde56" 

I want to replace it by 12abXXcde56

b="abc1235ef"

I want to replace it by abcXXXXef

c="1ab12cd"

I want to replace it by 1abXXcd

I am trying those in python and in php, but with no luck. This is what I had in my mind:

^([0-9]+)([a-z]+)(.*)([a-z]+)([0-9]+)$

Answer 1

You can use this regex to capture all digits that is not leading or trailing:

(?<!^|\d)\d+(?!$|\d)

Then in Python, you can supply a function that replace the match with corresponding number of X .

For PHP, you can enable PREG_OFFSET_CAPTURE to know the position of the match, and loop through the list of matches and process them.

Note that with the regex above " 5 ddds" will be changed into " X ddds"

Answer 2

我们用X替换字符串s中用非数字（ \\D ）包围的每组数字（ \\d+ ）。

re.sub(r'(?<=\D)\d+(?=\D)',lambda match : 'X' * len(match.group(0)) , s)

Answer 3

The following pattern catch the string to remove in group 1:

^.*[a-z]+(\d+)[a-z]+.*$

Demo .

Answer 4

import re
re1 = re.compile("([\d]*[a-zA-Z])([\d\w]+)([a-zA-Z][\d]*)")
re2 = re.compile("([\d])")

s = "4f6g6h7"
def x(matchobj):
    return ''.join([matchobj.groups()[0],
        re2.sub('X', matchobj.groups()[1]), matchobj.groups()[2]])

print re1.sub(x, s)

---

Update: The original method won't work for case "4f6g6h7" or any string only has one alphabet char between digit.

If using two regular expression instead of one is acceptable. The following code should work for u.

import re
re1 = re.compile("([\d]*[a-zA-Z])([\d\w]+)([a-zA-Z][\d]*)")
re2 = re.compile("([\d])")

s = ['12ab34cde56', "abc1235ef","1ab12cd", "4f6g6h7"]

def x(matchobj):
    return ''.join([matchobj.groups()[0],
        re2.sub('X', matchobj.groups()[1]), matchobj.groups()[2]])

for ss in s:
    print ss, '->', re1.sub(x, ss)

>>>
12ab34cde56 -> 12abXXcde56
abc1235ef -> abcXXXXef
1ab12cd -> 1abXXcd
4f6g6h7 -> 4fXgXh7
>>> 

Answer 5

The only possibility with the stock re module appears to be a replace function, for example:

xs = ["12ab34cde56", "abc1235ef", "1ab12cd"]

import re
for x in xs:
    print x, re.sub(r'(\D)(\d+)(\D)', lambda m: m.group(1) + 'X' * len(m.group(2)) + m.group(3), x)

With the more advanced regex module you can use variable-width lookaround assertions:

import regex
for x in xs:
    print x, regex.sub(r'(?<=\D\d*)\d(?=\d*\D)', 'X', x)