使用pandas的TimeGrouper()以1秒的间隔查找列数 [英] Using pandas' TimeGrouper() to find column counts by intervals of 1 second

问题描述

考虑由时间戳记索引的数据帧"data"，如下所示:

Consider the dataframe "data" indexed by Timestamp as follow:

Index                                Receiver     Type         Retry
1970-01-01 00:00:00.000000000         R1          Data         1
1970-01-01 00:00:00.800000000         R1          Non-Data     1
1970-01-01 00:00:01.000287000         R2          Data         0
1970-01-01 00:00:01.600896000         R2          Data         1
1970-01-01 00:00:02.001388000         R1          Data         1
1970-01-01 00:00:02.004698000         R1          Non-Data     1
1970-01-01 00:00:02.006706000         R2          Data         1
1970-01-01 00:00:02.501351000         R2          Data         1
1970-01-01 00:00:02.810382000         R1          Data         0
1970-01-01 00:00:03.001981000         R1          Data         0
1970-01-01 00:00:03.377116000         R1          Data         0
1970-01-01 00:00:03.701811000         R2          Data         1
1970-01-01 00:00:03.910326000         R2          Data         0
1970-01-01 00:00:03.951355000         R2          Non-Data     1

我想查找Type为"Data"且Retry等于1的所有行，然后以1秒的间隔对索引进行分组，以查找每种接收器类型的计数.

I want to find all rows where Type is "Data" and Retry equals 1 and then group the index in intervals of 1 second to find counts for each Receiver type.

所需的输出为df2:

Index                        R1   R2
1970-01-01 00:00:00          1    0
1970-01-01 00:00:01          0    1
1970-01-01 00:00:02          1    2
1970-01-01 00:00:03          0    1

推荐答案

可以通过简单的pivot_table即

ndf = df[df['Type'].eq('Data')].pivot_table(columns=['Receiver'],values='Retry',
             index=pd.Grouper(key='Index',freq='s'), aggfunc='sum',fill_value=0)

Receiver             R1  R2
Index                      
1970-01-01 00:00:00   1   0
1970-01-01 00:00:01   0   1
1970-01-01 00:00:02   1   2
1970-01-01 00:00:03   0   1

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