正则表达式在不同长度的空格后捕获数字

Question

I try to use a non-capturing group to detect the spaces (before the numbers I needed) and not to bring spaces into my result, so I use

(?: 1+)\\d*.?\\d*

to process my text:

 input: kMPCV/epS4SgFoNdLo3LOuClO/URXS/5         134.686356921  2018-06-14 21:50:35.494
 input: pRVh7kPpFbtmuwS1NILiCzwHUVwJ4NcK         839.680408921  2018-06-14 22:13:39.996
 input: Ga7MIXmXAsrbaEc1Yj60qYYblcRQpnpz         4859.688276920  2018-06-14 23:02:11.125
 input: 4mqdb5njytfDOFpgeG3XS0Iv1OXFPEnb        1400.684675920  2018-06-14 23:33:42.031

and try to get the numbers.

But line 2 and 3 returns None result and line 1 and 4 returns numbers with 1 space before it: " 134.686356921"

Why I get different results? Code is below:

import re
def calcprice(filename):

    try:
        print ('ok')
        f = open(filename, 'r')
        data = f.read()
        rows = data.split('\n')

        for row in rows:
            print (re.search("[(?: 1+)\d*\.?\d*][1]",row))


    except Exception as e:
        print(e)


if __name__ == "__main__": ## If we are not importing this:
    calcprice('dfk balance.txt')

Result:

<_sre.SRE_Match object; span=(52, 66), match=' 134.686356921'>

None

None

<_sre.SRE_Match object; span=(51, 66), match=' 1400.684675920'>

Answer 1

Your current regex is basically one big character set :

[(?: 1+)\d*\.?\d*]

which doesn't make much sense, looks like a misunderstanding of how regex works. If you want to match the numbers, it would probably make more sense to lookbehind for a couple spaces, match digits and periods, and lookahead for another couple spaces:

(?<=  )[\d.]+(?=  )

https://regex101.com/r/NRnXWb/1

for row in rows:
    print (re.search(r"(?<=  )[\d.]+(?=  )",row))

Answer 2

Your regex [(?: 1+)\\d*\\.?\\d*][1] consists or 2 times a character class .

If the number you want to match always contains a dot, you could use a word boundary and a positive lookahead to assert that what followes is a whitespace:

\\b\\d+\\.\\d+(?= )

If it could also be without a dot you could check for a leading and a trailing whitespace using lookrounds and make the part which will match a dot and one or more times a digit optional (?:\\.\\d+)? .

(?<= )\\d+(?:\\.\\d+)?(?= )

Demo

Answer 3

Try the regex \\b(\\d+[\\d\\.]*)\\b

Your regex doesn't align to what you're trying to do.. It's pretty erroneous.

Answer 4

Try this pattern: +(\\d+(\\.\\d+)?) + .

Explanation: pattern will match number preceeded and followed by one or more spaces ( + ). It will match numbers with optional decimal part ( (\\.\\d+)? ), which will become second capturing group in a match (but you won't need it anyway).

In every match, first capturing group \\1 will be your number.

Demo