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python - 为每个组创建计算行差异的列

问题描述

我想创建一个列，它是每个独立于其他组的每个单独组的['diff]相邻值的差异。StatusID

df：

        ID  Timestamp               Value   Status
103177  64  2010-09-21 23:13:21.090 21.5    1.0
252019  64  2010-09-22 00:44:14.890 21.5    1.0
271381  64  2010-09-22 00:44:15.890 21.5    0.0
268939  64  2010-09-22 00:44:17.890 23.0    0.0
259875  64  2010-09-22 00:44:18.440 23.0    1.0
18870   32  2010-09-22 00:44:19.890 24.5    1.0
205910  32  2010-09-22 00:44:23.440 24.5    1.0
103865  32  2010-09-22 01:04:33.440 23.5    0.0
152281  32  2010-09-22 01:27:01.790 22.5    1.0
138988  32  2010-09-22 02:18:52.850 21.5    0.0

可重现的例子：

from pandas import *
from numpy import nan

df = pd.DataFrame({'ID': {103177: 64,
  252019: 64,
  271381: 64,
  268939: 64,
  259875: 64,
  18870: 32,
  205910: 32,
  103865: 32,
  152281: 32,
  138988: 32},
 'Timestamp': {103177: Timestamp('2010-09-21 23:13:21.090000'),
  252019: Timestamp('2010-09-22 00:44:14.890000'),
  271381: Timestamp('2010-09-22 00:44:15.890000'),
  268939: Timestamp('2010-09-22 00:44:17.890000'),
  259875: Timestamp('2010-09-22 00:44:18.440000'),
  18870: Timestamp('2010-09-22 00:44:19.890000'),
  205910: Timestamp('2010-09-22 00:44:23.440000'),
  103865: Timestamp('2010-09-22 01:04:33.440000'),
  152281: Timestamp('2010-09-22 01:27:01.790000'),
  138988: Timestamp('2010-09-22 02:18:52.850000')},
 'Value': {103177: 21.5,
  252019: 21.5,
  271381: 21.5,
  268939: 23.0,
  259875: 23.0,
  18870: 24.5,
  205910: 24.5,
  103865: 23.5,
  152281: 22.5,
  138988: 21.5},
 'Status': {103177: 1.0,
  252019: 1.0,
  271381: 0.0,
  268939: 0.0,
  259875: 1.0,
  18870: 1.0,
  205910: 1.0,
  103865: 0.0,
  152281: 1.0,
  138988: 0.0}})

df

预期输出：


        ID  Timestamp               Value   Status  Diff
103177  64  2010-09-21 23:13:21.090 21.5    1.0     NaN
252019  64  2010-09-22 00:44:14.890 21.5    1.0     0.0
271381  64  2010-09-22 00:44:15.890 21.5    0.0    -1.0
268939  64  2010-09-22 00:44:17.890 23.0    0.0     0.0
259875  64  2010-09-22 00:44:18.440 23.0    1.0     1.0
18870   64  2010-09-22 00:44:19.890 24.5    1.0     0.0
205910  32  2010-09-22 00:44:23.440 24.5    1.0     NaN
103865  32  2010-09-22 01:04:33.440 23.5    0.0    -1.0
152281  32  2010-09-22 01:27:01.790 22.5    1.0     1.0
138988  32  2010-09-22 02:18:52.850 21.5    0.0    -1.0

我怎样才能做到这一点？

标签： pythonpandasdataframegroup-by

解决方案

您正在寻找 groupby + .diff：

df['Diff'] =  df.groupby('ID')['Status'].diff()
df

Out[1]: 
        ID               Timestamp  Value  Status  Diff
103177  64 2010-09-21 23:13:21.090   21.5     1.0   NaN
252019  64 2010-09-22 00:44:14.890   21.5     1.0   0.0
271381  64 2010-09-22 00:44:15.890   21.5     0.0  -1.0
268939  64 2010-09-22 00:44:17.890   23.0     0.0   0.0
259875  64 2010-09-22 00:44:18.440   23.0     1.0   1.0
18870   32 2010-09-22 00:44:19.890   24.5     1.0   NaN
205910  32 2010-09-22 00:44:23.440   24.5     1.0   0.0
103865  32 2010-09-22 01:04:33.440   23.5     0.0  -1.0
152281  32 2010-09-22 01:27:01.790   22.5     1.0   1.0
138988  32 2010-09-22 02:18:52.850   21.5     0.0  -1.0

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