I have following dataframe in pandas
code bucket
0 08:30:00-9:00:00
1 10:00:00-11:00:00
2 12:00:00-13:00:00
I want to replace 7th character 0 with 1, my desired dataframe is
code bucket
0 08:30:01-9:00:00
1 10:00:01-11:00:00
2 12:00:01-13:00:00
How to do it in pandas?
Use indexing with str:
df['bucket'] = df['bucket'].str[:7] + '1' + df['bucket'].str[8:]
Or list comprehension:
df['bucket'] = [x[:7] + '1' + x[8:] for x in df['bucket']]
print (df)
code bucket
0 0 08:30:01-9:00:00
1 1 10:00:01-11:00:00
2 2 12:00:01-13:00:00
You lose a considerable amount of functionality by working with strings only. While this may be a one-off operation, you will find that repeated string manipulations will quickly become expensive in terms of time and memory efficiency.
pd.to_datetime insteadYou can add additional series to your dataframe with datetime objects. Below is an example which, in addition, creates an object dtype series in the format you desire.
# split by '-' into 2 series
dfs = df.pop('bucket').str.split('-', expand=True)
# convert to datetime
dfs = dfs.apply(pd.to_datetime, axis=1)
# add 1s to first series
dfs[0] = dfs[0] + pd.Timedelta(seconds=1)
# create object series from 2 times
form = '%H:%M:%S'
dfs[2] = dfs[0].dt.strftime(form) + '-' + dfs[1].dt.strftime(form)
# join to original dataframe
res = df.join(dfs)
print(res)
code 0 1 2
0 0 2018-10-02 08:30:01 2018-10-02 09:00:00 08:30:01-09:00:00
1 1 2018-10-02 10:00:01 2018-10-02 11:00:00 10:00:01-11:00:00
2 2 2018-10-02 12:00:01 2018-10-02 13:00:00 12:00:01-13:00:00
