python - 在 Pandas 中，如何将一串日期字符串转换为日期时间对象并将它们放入 DataFrame 中？

Question

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)

此代码位产生错误：

TypeError：“'int'对象不可迭代”

谁能告诉我如何将这一系列日期时间字符串作为DateTime对象放入 DataFrame 中？

Answer 1

>>> 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

更新

使用 pandas.to_datetime(pd.Series(..))。它比上面的代码简洁而且快得多。

>>> 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

Answer 2

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]

更新：基准

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

Answer 3

一个简单的解决方案涉及 Series 构造函数。您可以简单地将数据类型传递给dtype参数。此外，该to_datetime函数现在可以采用一系列字符串。

创建数据

date_strings = ('2008-12-20','2008-12-21','2008-12-22','2008-12-23')

所有三个产生相同的东西

pd.Series(date_strings, dtype='datetime64[ns]')
pd.Series(pd.to_datetime(date_strings))
pd.to_datetime(pd.Series(date_strings))

基准

@waitingkuo 提供的基准测试是错误的。第一种方法比其他两种方法慢一点，它们具有相同的性能。

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)