add a string prefix to each value in a string column using Pandas


I would like to append a string to the start of each value in a said column of a pandas dataframe (elegantly).
I already figured out how to kind-of do this and I am currently using:

df.ix[(df['col'] != False), 'col'] = 'str'+df[(df['col'] != False), 'col']

This seems one hell of an inelegant thing to do – do you know any other way (which maybe also adds the character to rows where that column is 0 or NaN)?

In case this is yet unclear, I would like to turn:

1     a
2     0


1     stra
2     str0
Asked By: TheChymera



df['col'] = 'str' + df['col'].astype(str)


>>> df = pd.DataFrame({'col':['a',0]})
>>> df
0   a
1   0
>>> df['col'] = 'str' + df['col'].astype(str)
>>> df
0  stra
1  str0
Answered By: Roman Pekar

As an alternative, you can also use an apply combined with format (or better with f-strings) which I find slightly more readable if one e.g. also wants to add a suffix or manipulate the element itself:

df = pd.DataFrame({'col':['a', 0]})

df['col'] = df['col'].apply(lambda x: "{}{}".format('str', x))

which also yields the desired output:

0  stra
1  str0

If you are using Python 3.6+, you can also use f-strings:

df['col'] = df['col'].apply(lambda x: f"str{x}")

yielding the same output.

The f-string version is almost as fast as @RomanPekar’s solution (python 3.6.4):

df = pd.DataFrame({'col':['a', 0]*200000})

%timeit df['col'].apply(lambda x: f"str{x}")
117 ms ± 451 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit 'str' + df['col'].astype(str)
112 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Using format, however, is indeed far slower:

%timeit df['col'].apply(lambda x: "{}{}".format('str', x))
185 ms ± 1.07 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
Answered By: Cleb

If you load you table file with dtype=str

or convert column type to string df['a'] = df['a'].astype(str)
then you can use such approach:

df['a']= 'col' + df['a'].str[:]

This approach allows prepend, append, and subset string of df.
Works on Pandas v0.23.4, v0.24.1. Don’t know about earlier versions.

Answered By: Vasyl Vaskivskyi

Another solution with .loc:

df = pd.DataFrame({'col': ['a', 0]})
df.loc[df.index, 'col'] = 'string' + df['col'].astype(str)

This is not as quick as solutions above (>1ms per loop slower) but may be useful in case you need conditional change, like:

mask = (df['col'] == 0)
df.loc[mask, 'col'] = 'string' + df['col'].astype(str)
Answered By: Lukas

You can use :


In this example, it will apply the word str before all your values.

Answered By: Boxtell

Contributing to prefixing columns while controlling NaNs for things like human readable values on csv export.

"_" + df['col1'].replace(np.nan,'').astype(str)


import sys
import platform
import pandas as pd
import numpy as np

print("python {}".format(platform.python_version(), sys.executable))
print("pandas {}".format(pd.__version__))
print("numpy {}".format(np.__version__))

df = pd.DataFrame({

df['col1_prefixed'] = "_" + df['col1'].replace(np.nan,'no value').astype(str)
df['col4_prefixed'] = "_" + df['col4'].replace(np.nan,'no value').astype(str)

python 3.7.3
pandas 1.2.3
numpy 1.18.5
  col1 col2  col3  col4 col1_prefixed col4_prefixed
0   1a   2a    31   NaN           _1a     _no value
1   1b   2b    32  42.0           _1b         _42.0
2   1c  NaN    33  43.0           _1c         _43.0
3  NaN   2d    34   NaN     _no value     _no value

(Sorry for the verbosity, I found this Q while working on an unrelated column type issue and this is my reproduction code)

Answered By: Rich Andrews
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