Converting a Panda DF List into a string
You should certainly not convert to string before you transform the list. Try:
df['col'].apply(', '.join)
Also note that apply
applies the function to the elements of the series, so using df['col']
in the lambda function is probably not what you want.
Edit: thanks Yakym for pointing out that there is no need for a lambda function.
Edit: as noted by Anton Protopopov, there is a native .str.join
method, but it is (surprisingly) a bit slower than apply
.
When you cast col
to str
with astype
, you get a string representation of a python list, brackets and all. You do not need to do that, just apply
join
directly:
import pandas as pd
df = pd.DataFrame({
'A': [['a', 'b', 'c'], ['A', 'B', 'C']]
})
# Out[8]:
# A
# 0 [a, b, c]
# 1 [A, B, C]
df['Joined'] = df.A.apply(', '.join)
# A Joined
# 0 [a, b, c] a, b, c
# 1 [A, B, C] A, B, C
You could convert your list to str with astype(str)
and then remove '
, [
, ]
characters. Using @Yakim example:
In [114]: df
Out[114]:
A
0 [a, b, c]
1 [A, B, C]
In [115]: df.A.astype(str).str.replace('\[|\]|\'', '')
Out[115]:
0 a, b, c
1 A, B, C
Name: A, dtype: object
Timing
import pandas as pd
df = pd.DataFrame({'A': [['a', 'b', 'c'], ['A', 'B', 'C']]})
df = pd.concat([df]*1000)
In [2]: timeit df['A'].apply(', '.join)
292 µs ± 10.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [3]: timeit df['A'].str.join(', ')
368 µs ± 24.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [4]: timeit df['A'].apply(lambda x: ', '.join(x))
505 µs ± 5.74 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [5]: timeit df['A'].str.replace('\[|\]|\'', '')
2.43 ms ± 62.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)