Split and replace all strings in a pandas dataframe

Question

I have a large dataframe where each row contains a string. I want to split each string into several columns, and also replace two character types.

The code below does the job, but it is slow on a large dataframe. Is there a faster way than using a for loop?

import re
import pandas as pd

df = pd.DataFrame(['[3.4, 3.4, 2.5]', '[3.4, 3.4, 2.5]'])

df_new = pd.DataFrame({'col1': [0,0], 'col2': [0,0], 'col3': [0,0]})

for i in range(df.shape[0]):
    df_new.iloc[i, :] = re.split(',', df.iloc[i, 0].replace('[', '').replace(']', ''))

Answer 1

Your solution should be changed with Series.str.strip and Series.str.split :

df1 = df[0].str.strip('[]').str.split(', ', expand=True).add_prefix('col')
print(df1)
  col0 col1 col2
0  3.4  3.4  2.5
1  3.4  3.4  2.5

If performance is important use list comprehension instead pandas functions:

df1 = pd.DataFrame([x.strip('[]').split(', ') for x in df[0]]).add_prefix('col')

Timings :

#20k rows
df = pd.concat([df] * 10000, ignore_index=True)

In [208]: %timeit df[0].str.strip('[]').str.split(', ', expand=True).add_prefix('col')
61.5 ms ± 1.68 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [209]: %timeit pd.DataFrame([x.strip('[]').split(', ') for x in df[0]]).add_prefix('col')
29.8 ms ± 1.85 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 2

You can do it with:

import pandas as pd
df = pd.DataFrame(['[3.4, 3.4, 2.5]', '[3.4, 3.4, 2.5]'])

df_new = df[0].str[1:-1].str.split(",", expand=True)
df_new.columns = ["col1", "col2", "col3"]

The idea is to first get rid of the [ and ] and then split by , and expand the dataframe. The last step would be to rename the columns.