import pandas as pd
date_stngs = ('2008-12-20','2008-12-21','2008-12-22','2008-12-23')
a = pd.Series(range(4),index = (range(4)))
for idx, date in enumerate(date_stngs):
a[idx]= pd.to_datetime(date)
This code bit produces error:
TypeError:" 'int' object is not iterable"
Can anyone tell me how to get this series of date time strings into a DataFrame as DateTime objects?
>>> import pandas as pd
>>> date_stngs = ('2008-12-20','2008-12-21','2008-12-22','2008-12-23')
>>> a = pd.Series([pd.to_datetime(date) for date in date_stngs])
>>> a
0 2008-12-20 00:00:00
1 2008-12-21 00:00:00
2 2008-12-22 00:00:00
3 2008-12-23 00:00:00
UPDATE
Use pandas.to_datetime(pd.Series(..)). It's concise and much faster than above code.
>>> pd.to_datetime(pd.Series(date_stngs))
0 2008-12-20 00:00:00
1 2008-12-21 00:00:00
2 2008-12-22 00:00:00
3 2008-12-23 00:00:00
pd.to_datetime(pd.Series(date_stngs)). Strongly not recommend this method.
timeit.timeit mine, yours, shows very little difference.In [46]: pd.to_datetime(pd.Series(date_stngs))
Out[46]:
0 2008-12-20 00:00:00
1 2008-12-21 00:00:00
2 2008-12-22 00:00:00
3 2008-12-23 00:00:00
dtype: datetime64[ns]
In [43]: dates = [(dt.datetime(1960, 1, 1)+dt.timedelta(days=i)).date().isoformat() for i in range(20000)]
In [44]: timeit pd.Series([pd.to_datetime(date) for date in dates])
1 loops, best of 3: 1.71 s per loop
In [45]: timeit pd.to_datetime(pd.Series(dates))
100 loops, best of 3: 5.71 ms per loop
A simple solution involves the Series constructor. You can simply pass the data type to the dtype parameter. Also, the to_datetime function can take a sequence of strings now.
Create Data
date_strings = ('2008-12-20','2008-12-21','2008-12-22','2008-12-23')
pd.Series(date_strings, dtype='datetime64[ns]')
pd.Series(pd.to_datetime(date_strings))
pd.to_datetime(pd.Series(date_strings))
The benchmarks provided by @waitingkuo are wrong. The first method is a bit slower than the other two, which have the same performance.
import datetime as dt
dates = [(dt.datetime(1960, 1, 1)+dt.timedelta(days=i)).date().isoformat()
for i in range(20000)] * 100
%timeit pd.Series(dates, dtype='datetime64[ns]')
730 ms ± 9.06 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit pd.Series(pd.to_datetime(dates))
426 ms ± 3.45 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit pd.to_datetime(pd.Series(dates))
430 ms ± 5.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
