Python正则表达式提取包含数字和字母的子字符串

Question

I am attempting to extract a substring that contains numbers and letters:

string = "LINE     : 11m56.95s CPU    13m31.14s TODAY"

I only want 11m56.95s and 13m31.14s

I have tried doing this:

re.findall('\d+', string)

that doesn't give me what I want, I also tried this:

re.findall('\d{2}[m]+\d[.]+\d|\+)

that did not work either, any other suggestions?

Answer 1

Your current regular expression does not match what you expect it to.

You could use the following regular expression to extract those substrings.

re.findall(r'\d+m\d+\.\d+s', string)

Live Demo

Example :

>>> import re
>>> s = 'LINE     : 11m56.95s CPU    13m31.14s TODAY'
>>> for x in re.findall(r'\d+m\d+\.\d+s', s):
...     print x

11m56.95s
13m31.14s

Answer 2

Try this:

re.findall("[0-9]{2}[m][0-9]{2}\.[0-9]{2}[s]", string)

Output:

['11m56.95s', '13m31.14s']

Answer 3

Your Regex pattern is not formed correctly. It is currently matching:

\d{2}  # Two digits
[m]+   # One or more m characters
\d     # A digit
[.]+   # One or more . characters
\d|\+  # A digit or +

Instead, you should use:

>>> import re
>>> string = "LINE     : 11m56.95s CPU    13m31.14s TODAY"
>>> re.findall('\d+m\d+\.\d+s', string)
['11m56.95s', '13m31.14s']
>>>

Below is an explanation of what the new pattern matches:

\d+  # One or more digits
m    # m
\d+  # One or more digits
\.   # .
\d+  # One or more digits
s    # s

Answer 4

\b   #word boundary
\d+  #starts with digit
.*?   #anything (non-greedy so its the smallest possible match)
s    #ends with s
\b   #word boundary

Answer 5

If your lines are all like your example split will work:

s = "LINE     : 11m56.95s CPU    13m31.14s TODAY"

spl = s.split()

a,b = spl[2],spl[4]
print(a,b)
('11m56.95s', '13m31.14s')