如何在python中找到一个字符串中的6位数？

Question

An example string is "CPLR_DUK10_772989_2" . I want to pick out "772989" specifically. I would imagine using re.findall is a good way to go about it, however, I don't have a very good grasp on regular expression so I find myself stumped on this one.

Here is a sample of code that I thought would work, until I looked at the full list of strings, and saw that it definitely doesn't. I suppose I'm looking for some more robustness!

for ad in Ads:
    num = ''.join(re.findall(numbers,ad)[1:7])
    ID.append(num)
ID = pd.Series(ID)

Other sample strings: "Teb1_110765" , "PAN1_111572_5" .

Answer 1

The regex you are looking for is

p = re.findall(r'_(\d{6})', ad)

This will match a six-digit number preceded by an underscore, and give you a list of all matches ( should there be more than one )

Demo:

>>> import re
>>> stringy =  'CPLR_DUK10_772989_2'
>>> re.findall(r'_(\d{6})', stringy)
['772989']

Answer 2

This should append all sets of 6 numbers that follow an underscore

for ad in Ads:
    blocks = re.split('_', ad)
    for block in blocks[1:]:
        if len(block) == 6 and block.isdigit(): 
            ID.append(block)
ID = pd.Series(ID)

Answer 3

You can use a list comprehension:

>>> s="CPLR_DUK10_772989_2"
>>> [x for x in s.split('_') if len(x)==6 and x.isdigit()]
['772989']

If your strings are really long and you are only looking for one number, you could use intertools like so:

>>> from itertools import dropwhile
>>> next(dropwhile(lambda x: not(len(x)==6 and x.isdigit()), s.split('_')))
'772989'