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

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.

You can use pandas.Series.map :

df['col'].map('str{}'.format)

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

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)