解析字符串以获取特定字符前后的数字

Question

I am trying to parse digits before and after X from this string, but unable to get all the digits. Can someone help me pointing out what I am missing here?

>>> import re
>>> f = "abc_xyz1024X137M4B4abc_xyz"
>>> re.findall(".*\w+(\d+)X(\d+).*", f)
[('4', '137')]

Answer 1

Note that .*\\w+(\\d+)X(\\d+).* first grabs all the 0+ chars as many as possible (the whole string) and then backracks trying to match the subsequent patterns. \\w+ backtracks up to the point where the next char is a digit before X , so the first capturing group only contains the single digit before X , and the second one contains all the digits after X . Check this .*\\w+(\\d+)X(\\d+).* debugger page .

You should only match and capture the digits, then match the X and then again match and capture the digits.

You may use

import re
f = "abc_xyz1024X137M4B4abc_xyz"
print(re.findall(r"(\d+)X(\d+)", f))
# => [('1024', '137')]

Or, if you are only interested in a single match:

m = re.search(r"(?P<x>\d+)X(?P<y>\d+)", f)
if m:
    print(m.groupdict()) # => {'y': '137', 'x': '1024'}

See the Python demo and the regex demo .

Answer 2

In this particular example, another option is to split the string on the character "X" . Then find the last set of consecutive digits in the left half of the split and the first set of consecutive digits in the right half of the split.

For example:

import re
f = "abc_xyz1024X137M4B4abc_xyz"

left, right = f.split("X")
print(right)
#137M4B4abc_xyz

print(left)
#abc_xyz1024

print((re.findall('\d+', left)[-1], re.findall('\d+', right)[0]))
#('1024', '137')