python-2.7 - 如何快速选择日期之间的行pandas dataframe

Question

我想知道在索引内两个日期之间选择行的速度方面最有效的方法是什么。例如

>>> import pandas as pd
>>> index = pd.date_range('2018-01-01', '2030-01-02', freq='BM')
>>> df = pd.DataFrame(np.zeros((len(index), 1)), index=index)
>>> df.head()
              0
2018-01-31  0.0
2018-02-28  0.0
2018-03-30  0.0
2018-04-30  0.0
2018-05-31  0.0

然后一种选择之间所有行的方法，例如2018-05-30 2027-07-03 是

>>> df.loc[(df.index >= '2018-05-30') & (df.index <= '2027-07-03')]

在我的应用程序中，我不预先知道值 2018-05-30 2027-07-03。实现所需选择的最快方法是什么？

Answer 1

您可以使用 truncate :

print (df.truncate(before='2018-05-30', after='2027-07-03'))

print (df.loc['2018-05-30':'2027-07-03'])

print (df.loc[(df.index >= '2018-05-30') & (df.index <= '2027-07-03')])

时间:

In [366]: %timeit (df.loc['2018-05-30':'2027-07-03'])
The slowest run took 5.08 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.43 ms per loop

In [367]: %timeit (df.loc[(df.index >= '2018-05-30') & (df.index <= '2027-07-03')])
The slowest run took 4.97 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 502 µs per loop

In [368]: %timeit (df.truncate(before='2018-05-30', after='2027-07-03'))
The slowest run took 4.98 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 450 µs per loop

如果您稍微更改条件但不包含最后一个值(如果存在)- <=更改为< :

In [372]: %timeit (df.loc[(df.index >= '2018-05-31') & (df.index < '2027-05-31')])
The slowest run took 4.81 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 520 µs per loop

In [373]: %timeit (df.iloc[df.index.searchsorted('2018-05-31'): df.index.searchsorted('2027-05-31')])
10000 loops, best of 3: 136 µs per loop