Question

我有两个pandas数据框：一个（df1）有三列（StartDate，EndDate和ID）和第二个（df2）与日期。我想基于df1和df2之间的df2.Date合并df1.StartDate和df2.EndDate。

df1中的每个日期范围都是唯一的，并且不会与数据框中的任何其他行重叠。

日期格式为YYYY-MM-DD。

Answer 1

仅提供使用np.piecewise的替代方式。性能甚至比np.searchedsort更快。

import pandas as pd
import numpy as np

# data
# ====================================
df1 = pd.DataFrame({'StartDate': pd.date_range('2010-01-01', periods=9, freq='5D'), 'EndDate': pd.date_range('2010-01-04', periods=9, freq='5D'), 'ID': np.arange(1, 10, 1)})

df2 = pd.DataFrame(dict(values=np.random.randn(50), date_time=pd.date_range('2010-01-01', periods=50, freq='D')))

df1.StartDate

Out[139]: 
0   2010-01-01
1   2010-01-06
2   2010-01-11
3   2010-01-16
4   2010-01-21
5   2010-01-26
6   2010-01-31
7   2010-02-05
8   2010-02-10
Name: StartDate, dtype: datetime64[ns]

df2.date_time

Out[140]: 
0    2010-01-01
1    2010-01-02
2    2010-01-03
3    2010-01-04
4    2010-01-05
5    2010-01-06
6    2010-01-07
7    2010-01-08
8    2010-01-09
9    2010-01-10
        ...    
40   2010-02-10
41   2010-02-11
42   2010-02-12
43   2010-02-13
44   2010-02-14
45   2010-02-15
46   2010-02-16
47   2010-02-17
48   2010-02-18
49   2010-02-19
Name: date_time, dtype: datetime64[ns]


df2['ID_matched'] = np.piecewise(np.zeros(len(df2)), [(df2.date_time.values >= start_date)&(df2.date_time.values <= end_date) for start_date, end_date in zip(df1.StartDate.values, df1.EndDate.values)], df1.ID.values)


Out[143]: 
    date_time  values  ID_matched
0  2010-01-01 -0.2240           1
1  2010-01-02 -0.4202           1
2  2010-01-03  0.9998           1
3  2010-01-04  0.4310           1
4  2010-01-05 -0.6509           0
5  2010-01-06 -1.4987           2
6  2010-01-07 -1.2306           2
7  2010-01-08  0.1940           2
8  2010-01-09 -0.9984           2
9  2010-01-10 -0.3676           0
..        ...     ...         ...
40 2010-02-10  0.5242           9
41 2010-02-11  0.3451           9
42 2010-02-12  0.7244           9
43 2010-02-13 -2.0404           9
44 2010-02-14 -1.0798           0
45 2010-02-15 -0.6934           0
46 2010-02-16 -2.3380           0
47 2010-02-17  1.6623           0
48 2010-02-18 -0.2754           0
49 2010-02-19 -0.7466           0

[50 rows x 3 columns]

%timeit df2['ID_matched'] = np.piecewise(np.zeros(len(df2)), [(df2.date_time.values >= start_date)&(df2.date_time.values <= end_date) for start_date, end_date in zip(df1.StartDate.values, df1.EndDate.values)], df1.ID.values)
1000 loops, best of 3: 466 µs per loop

Answer 2

对@JianxunLi答案的较小修正。有点参与评论。

这使用len(funclist) == len(condlist) + 1的{{1}}属性为不匹配时分配默认值。否则，默认的不匹配值为零，这可能会导致问题...

piecewise

PS。还纠正了### Data / inits import pandas as pd import numpy as np df1 = pd.DataFrame({'StartDate': pd.date_range('2010-01-01', periods=9, freq='5D'), 'EndDate': pd.date_range('2010-01-04', periods=9, freq='5D'), 'ID': np.arange(1, 10, 1)}) df2 = pd.DataFrame(dict(values=np.random.randn(50), date_time=pd.date_range('2010-01-01', periods=50, freq='D'))) ### Processing valIfNoMatch = np.nan df2['ID_matched'] = np.piecewise(np.zeros(len(df2)),\ [(df2.date_time.values >= start_date)&(df2.date_time.values < end_date) for start_date, end_date in zip(df1.StartDate.values, df1.EndDate.values)],\ np.append(df1.ID.values, valIfNoMatch))和>=的错字测试；在两个间隔之间精确间隔上的时间戳记对于两个不同的间隔将返回true，这打破了该方法的关键假设。</ p>

根据日期范围合并数据框

2 个答案: