Efficiently replace part of value from one column with value from another column in pandas using regex?

问题

I have a pandas dataframe df with dates as strings:

Date1        Date2
2017-08-31   1970-01-01 17:35:00
2017-10-31   1970-01-01 15:00:00
2017-11-30   1970-01-01 16:30:00
2017-10-31   1970-01-01 16:00:00
2017-10-31   1970-01-01 16:12:00

What I want to do is replace each date part in the Date2 column with the corresponding date in Date1 but leave the time untouched, so the output is:

Date1        Date2
2017-08-31   2017-08-31 17:35:00
2017-10-31   2017-10-31 15:00:00
2017-11-30   2017-11-30 16:30:00
2017-10-31   2017-10-31 16:00:00
2017-10-31   2017-10-31 16:12:00

I have achieved this using pandas replace and regex's as such

import re
date_reg = re.compile(r"([0-9]{4}\-[0-9]{2}\-[0-9]{2})")
df['Market Close Time'].replace(to_replace=date_reg, value=df['Date1'], inplace=True)

But this method is very slow (>10 minutes) for a dataframe with only 150k rows.

The solution from this post implements numpy np.where which is much faster - how can I use the np.where in this example, or is there another more efficient way to perform this operation?

回答1:

One idea is:

df['Date3'] =  ['{} {}'.format(a, b.split()[1]) for a, b in zip(df['Date1'], df['Date2'])]

Or:

df['Date3'] = df['Date1'] + ' ' + df['Date2'].str.split().str[1]
print (df)
        Date1                Date2                Date3
0  2017-08-31  1970-01-01 17:35:00  2017-08-31 17:35:00
1  2017-10-31  1970-01-01 15:00:00  2017-10-31 15:00:00
2  2017-11-30  1970-01-01 16:30:00  2017-11-30 16:30:00
3  2017-10-31  1970-01-01 16:00:00  2017-10-31 16:00:00
4  2017-10-31  1970-01-01 16:12:00  2017-10-31 16:12:00

Or:

df['Date3'] = pd.to_datetime(df['Date1']) + pd.to_timedelta(df['Date2'].str.split().str[1])
print (df)
        Date1                Date2               Date3
0  2017-08-31  1970-01-01 17:35:00 2017-08-31 17:35:00
1  2017-10-31  1970-01-01 15:00:00 2017-10-31 15:00:00
2  2017-11-30  1970-01-01 16:30:00 2017-11-30 16:30:00
3  2017-10-31  1970-01-01 16:00:00 2017-10-31 16:00:00
4  2017-10-31  1970-01-01 16:12:00 2017-10-31 16:12:00

Timings:

In [302]: %timeit df['Date3'] =  ['{} {}'.format(a, b.split()[1]) for a, b in zip(df['Date1'], df['Date2'])]
30.2 ms ± 137 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [303]: %timeit df['Date3'] = df['Date1'] + ' ' + df['Date2'].str.split().str[1]
66.4 ms ± 3.18 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

回答2:

Another way is to

df.Date2 = df.Date1.str[:].values + df.Date2.str[10:].values

df.Date1.str[:].values will get the Date1 field as a numpy array and likewise with Date2 field.

str[10:] is done to extract the time part of Date2 which is appended to the date from Date1.

Timings: 2.26 ms ± 82.2 µs

%timeit df.d2 = df.d1.str[:].values + df.d2.str[10:].values
2.26 ms ± 82.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

来源：https://stackoverflow.com/questions/50583265/efficiently-replace-part-of-value-from-one-column-with-value-from-another-column

标签

python

pandas

vectorization