df['col'] = 'str' + df['col'].astype(str)
Example:
>>> df = pd.DataFrame({'col':['a',0]})
>>> df
col
0 a
1 0
>>> df['col'] = 'str' + df['col'].astype(str)
>>> df
col
0 stra
1 str0
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:
col
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)
You can use pandas.Series.map :
df['col'].map('str{}'.format)
It will apply the word "str" before all your values.
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.
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