另一种方法是使用pandas.DataFrame.query()方法。让我给你们看一个关于下面数据帧的例子，叫做df。

>>> df = pd.DataFrame(np.random.random((5, 1)), columns=['col_1'])
>>> df['date'] = pd.date_range('2020-1-1', periods=5, freq='D')
>>> print(df)
      col_1       date
0  0.015198 2020-01-01
1  0.638600 2020-01-02
2  0.348485 2020-01-03
3  0.247583 2020-01-04
4  0.581835 2020-01-05

作为参数，使用条件进行过滤，如下所示:

>>> start_date, end_date = '2020-01-02', '2020-01-04'
>>> print(df.query('date >= @start_date and date <= @end_date'))
      col_1       date
1  0.244104 2020-01-02
2  0.374775 2020-01-03
3  0.510053 2020-01-04

如果你不想包含边界，只需要像下面这样改变条件:

>>> print(df.query('date > @start_date and date < @end_date'))
      col_1       date
2  0.374775 2020-01-03

你也可以用between:

df[df.some_date.between(start_date, end_date)]

您可以像这样在日期列上使用isin方法 df (df .isin (pd(“日期”)。date_range (start_date end_date)))

注意:这只适用于日期(正如问题所要求的)，而不适用于时间戳。

例子:

import numpy as np   
import pandas as pd

# Make a DataFrame with dates and random numbers
df = pd.DataFrame(np.random.random((30, 3)))
df['date'] = pd.date_range('2017-1-1', periods=30, freq='D')

# Select the rows between two dates
in_range_df = df[df["date"].isin(pd.date_range("2017-01-15", "2017-01-20"))]

print(in_range_df)  # print result

这给了

           0         1         2       date
14  0.960974  0.144271  0.839593 2017-01-15
15  0.814376  0.723757  0.047840 2017-01-16
16  0.911854  0.123130  0.120995 2017-01-17
17  0.505804  0.416935  0.928514 2017-01-18
18  0.204869  0.708258  0.170792 2017-01-19
19  0.014389  0.214510  0.045201 2017-01-20

import pandas as pd

technologies = ({
    'Courses':["Spark","PySpark","Hadoop","Python","Pandas","Hadoop","Spark"],
    'Fee' :[22000,25000,23000,24000,26000,25000,25000],
    'Duration':['30days','50days','55days','40days','60days','35days','55days'],
    'Discount':[1000,2300,1000,1200,2500,1300,1400],
    'InsertedDates':["2021-11-14","2021-11-15","2021-11-16","2021-11-17","2021-11-18","2021-11-19","2021-11-20"]
               })
df = pd.DataFrame(technologies)
print(df)

使用pandas.DataFrame.loc按日期过滤行

方法1:

    mask = (df['InsertedDates'] > start_date) & (df['InsertedDates'] <= end_date)

    df2 = df.loc[mask]
    print(df2)

方法2:

    start_date = '2021-11-15'
    end_date = '2021-11-19'
    after_start_date = df["InsertedDates"] >= start_date
    before_end_date = df["InsertedDates"] <= end_date
    between_two_dates = after_start_date & before_end_date


    df2 = df.loc[between_two_dates]
    print(df2)

使用pandas.DataFrame.query()选择数据帧行

start_date = '2021-11-15'
end_date   = '2021-11-18'
df2 = df.query('InsertedDates >= @start_date and InsertedDates <= @end_date')
print(df2)

使用datafframe .query()选择两个日期之间的行

start_date = '2021-11-15'
end_date = '2021-11-18'
df2 = df.query('InsertedDates > @start_date and InsertedDates < @end_date')
print(df2)

pandas.Series.between()函数使用两个日期

df2 = df.loc[df["InsertedDates"].between("2021-11-16", "2021-11-18")]
print(df2)

使用DataFrame.isin()在两个日期之间选择数据帧行

df2 = df[df["InsertedDates"].isin(pd.date_range("2021-11-15", "2021-11-17"))]
print(df2)

Pandas 0.22有一个between()函数。 使回答这个问题更容易，代码更可读。

# create a single column DataFrame with dates going from Jan 1st 2018 to Jan 1st 2019
df = pd.DataFrame({'dates':pd.date_range('2018-01-01','2019-01-01')})

假设你想获取2018年11月27日至2019年1月15日之间的日期:

# use the between statement to get a boolean mask
df['dates'].between('2018-11-27','2019-01-15', inclusive=False)

0    False
1    False
2    False
3    False
4    False

# you can pass this boolean mask straight to loc
df.loc[df['dates'].between('2018-11-27','2019-01-15', inclusive=False)]

    dates
331 2018-11-28
332 2018-11-29
333 2018-11-30
334 2018-12-01
335 2018-12-02

注意包含的参数。当你想要明确你的范围时，这非常有用。注意当设置为True时，我们也会返回2018年11月27日:

df.loc[df['dates'].between('2018-11-27','2019-01-15', inclusive=True)]

    dates
330 2018-11-27
331 2018-11-28
332 2018-11-29
333 2018-11-30
334 2018-12-01

这种方法也比前面提到的isin方法快:

%%timeit -n 5
df.loc[df['dates'].between('2018-11-27','2019-01-15', inclusive=True)]
868 µs ± 164 µs per loop (mean ± std. dev. of 7 runs, 5 loops each)


%%timeit -n 5

df.loc[df['dates'].isin(pd.date_range('2018-01-01','2019-01-01'))]
1.53 ms ± 305 µs per loop (mean ± std. dev. of 7 runs, 5 loops each)

但是，它并不比当前接受的unutbu提供的答案快，只有在掩码已经创建的情况下。但如果掩码是动态的，需要一遍又一遍地重新分配，我的方法可能会更有效:

# already create the mask THEN time the function

start_date = dt.datetime(2018,11,27)
end_date = dt.datetime(2019,1,15)
mask = (df['dates'] > start_date) & (df['dates'] <= end_date)

%%timeit -n 5
df.loc[mask]
191 µs ± 28.5 µs per loop (mean ± std. dev. of 7 runs, 5 loops each)

另一种方法是使用pandas.DataFrame.query()方法。让我给你们看一个关于下面数据帧的例子，叫做df。

>>> df = pd.DataFrame(np.random.random((5, 1)), columns=['col_1'])
>>> df['date'] = pd.date_range('2020-1-1', periods=5, freq='D')
>>> print(df)
      col_1       date
0  0.015198 2020-01-01
1  0.638600 2020-01-02
2  0.348485 2020-01-03
3  0.247583 2020-01-04
4  0.581835 2020-01-05

作为参数，使用条件进行过滤，如下所示:

>>> start_date, end_date = '2020-01-02', '2020-01-04'
>>> print(df.query('date >= @start_date and date <= @end_date'))
      col_1       date
1  0.244104 2020-01-02
2  0.374775 2020-01-03
3  0.510053 2020-01-04

如果你不想包含边界，只需要像下面这样改变条件:

>>> print(df.query('date > @start_date and date < @end_date'))
      col_1       date
2  0.374775 2020-01-03