Count space character before a letter/number in a python pandas column

Question

I have a pandas dataframe, which looks like this:

columnA columnB
A        10
 B       12
  C      13
  D      14
 010     17

How can i count the space characters before the first string/number/letter in the column A in a new column? So for example:

columnA columnB counter
A        10      0
 B       12      1
  C      13      2
  D      14      2
 010     17      1

Asked By: PV8

||

Source

Answer 1

You can combine str.extract and str.len:

df['counter'] = df['columnA'].str.extract('^( *)', expand=False).str.len()

Output (I added quotes around the string for visualization):

  columnA  columnB  counter
0     "A"       10        0
1    " B"       12        1
2   "  C"       13        2
3   "  D"       14        2
4  " 010"       17        1

Reproducible input:

df = pd.DataFrame({'columnA': ['A', ' B', '  C', '  D', ' 010'],
                   'columnB': [10, 12, 13, 14, 17],
                   'counter': [0, 1, 2, 2, 1]})

Answered By: mozway

Answer 2

You can use str.findall then extract the len of the first item:

df['counter'] = df['columnA'].str.findall('^ *').str[0].str.len()
print(df)

# Output
  columnA  columnB  counter
0       A       10        0
1       B       12        1
2       C       13        2
3       D       14        2
4     010       17        1

Answered By: Corralien

Answer 3

You can use RegExp with apply(), why not:

import pandas as pd
import re

df = pd.DataFrame({'columnA': ['A', ' B', '  C', '  D', ' 010'], 'columnB': [10, 12, 13, 14, 17]})

pattern = r'^s*'

def count_spaces(s):
    return len(re.match(pattern, s).group())

df['counter'] = df['columnA'].apply(count_spaces)

print(df)

Output:

  columnA  columnB  counter
0       A       10        0
1       B       12        1
2       C       13        2
3       D       14        2
4     010       17        1

Answered By: angwrk

Count space character before a letter/number in a python pandas column

Question:

Answers: