python replace regex match without spaces-CodePudding

I basically want to 'join' numbers that should clearly go together. I want to replace the regex match with itself but without any spaces.

I have:

df
               a
'Fraxiparine 9 500 IU (anti-Xa)/1 ml'
'Colobreathe 1 662 500 IU inhalačný prášok v tvrdej kapsule'

I want to have:

df
               a
'Fraxiparine 9500 IU (anti-Xa)/1 ml'
'Colobreathe 1662500 IU inhalačný prášok v tvrdej kapsule'

I'm using r'\d \s \d \s*\d ' to match the numbers, and I've created the following function to remove the spaces within the string:

def spaces(x):
    match = re.findall(r'\d \s \d \s*\d ', x)
    return match.replace(" ","")

Now I'm having trouble applying that function to the full dataframe, but I also don't know exactly how to replace the original match with the string without any spaces.

CodePudding user response：

Try using the following code:

def spaces(s):
    return re.sub('(?<=\d) (?=\d)', '', s)

df['a'] = df['a'].apply(spaces)

The regex will match:

any space
preceeded by a digit (?<=\d)
and followed by a digit (?=\d).

Then, the pandas.Series.apply function will apply your function to all rows of your dataframe.

Output:

0   Fraxiparine 9500 IU (anti-Xa)/1 ml
1   Colobreathe 1662500 IU inhalačný prášok v tvrd...

CodePudding user response：

I believe that your problem can be solved by tweaking a bit your function in order to be applied on the whole string 'match' as follows :

import pandas as pd
import re

df = pd.DataFrame({'a' : ['Fraxiparine 9 500 IU (anti-Xa)/1 ml','Colobreathe 1 662 500 IU inhalačný prášok v tvrdej kapsule']})

# your function
def spaces(x):
    match = re.findall(r'\d \s \d \s*\d ', x)
    replace_with = match[0].replace(" ","")
    return x.replace(match[0], replace_with)

# now apply it on the whole dataframe, row per row
df['a'] = df['a'].apply(lambda x: spaces(x))

CodePudding user response：

Use

df['a'] = df['a'].str.replace(r'(?<=\d)\s (?=\d)', '', regex=True)

EXPLANATION

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  (?<=                     look behind to see if there is:
--------------------------------------------------------------------------------
    \d                       digits (0-9)
--------------------------------------------------------------------------------
  )                        end of look-behind
--------------------------------------------------------------------------------
  \s                       whitespace (\n, \r, \t, \f, and &quot; &quot;) (1 or
                           more times (matching the most amount
                           possible))
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    \d                       digits (0-9)
--------------------------------------------------------------------------------
  )                        end of look-ahead

If your plan is to remove spaces only in \d \s \d \s*\d :

df['a'] = df['a'].str.replace(r'\d \s \d \s*\d ', lambda m: re.sub(r'\s ', '', m.group()), regex=True)

See str.replace:

repl : str or callable
Replacement string or a callable. The callable is passed the regex match object and must return a replacement string to be used. See re.sub().