In a dataframe
df = pd.DataFrame({'c1': ['c10:b', 'c11', 'c12:k'], 'c2': ['c20', 'c21', 'c22']})
c1 c2
0 c10:b c20
1 c11 c21
2 c12:k c22
I'd like to modify the string values of column c1 so that everything after (and including) the colon removes, so it ends up like this:
c1 c2
0 c10 c20
1 c11 c21
2 c12 c22
I've tried slicing
df[’c1’].str[:df[’c1’].str.find(’:’)]
but it doesn't work. How do I accomplish this?
Using replace with regex=True:
df.replace(r'\:.*', '', regex=True)
c1 c2
0 c10 c20
1 c11 c21
2 c12 c22
To only replace this pattern in a single column, use the str accessor:
df.c1.str.replace(r'\:.*', '')
If performance is a concern, use a list comprehension and partition instead of pandas string methods:
[i.partition(':')[0] for i in df.c1]
# ['c10', 'c11', 'c12']
Timings
df = pd.concat([df]*10000)
%timeit df.replace(r'\:.*', '', regex=True)
30.8 ms ± 340 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit df.c1.str.replace(r'\:.*', '')
31.2 ms ± 449 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit df['c1'].str.partition(':')[0]
56.7 ms ± 269 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit [i.partition(':')[0] for i in df.c1]
4.2 ms ± 22.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
