python replace regex match without spaces

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.

>Solution :

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...