python - 絕對到相對時間

Question

我的熊貓數據幀具有以下當前結構：

{
'Temperature': [1,2,3,4,5,6,7,8,9],
'machining': [1,1,1,2,2,2,3,3,3],
'timestamp': [1560770645,1560770646,1560770647,1560770648,1560770649,1560770650,1560770651,1560770652,1560770653]
}

我想添加一個包含每個加工過程的相對時間的列，以便每次“加工”列更改其值時都會刷新。
因此，所需的結構是：

{
'Temperature': [1,2,3,4,5,6,7,8,9],
'machining': [1,1,1,2,2,2,3,3,3],
'timestamp': [1560770645,1560770646,1560770647,1560770648,1560770649,1560770650,1560770651,1560770652,1560770653]
'timestamp_machining': [1,2,3,1,2,3,1,2,3]
}

我正在努力以一種干凈的方式做到這一點：如果需要，任何幫助也將不勝感激。

Answer 1

減去GroupBy.transform創建的每個組的第一個值：

#if values are not sorted
df = df.sort_values(['machining','timestamp'])

print (df.groupby('machining')['timestamp'].transform('first'))
0    1560770645
1    1560770645
2    1560770645
3    1560770648
4    1560770648
5    1560770648
6    1560770651
7    1560770651
8    1560770651
Name: timestamp, dtype: int64

df['new'] = df['timestamp'].sub(df.groupby('machining')['timestamp'].transform('first')) + 1
print (df)

   Temperature  machining   timestamp  timestamp_machining  new
0            1          1  1560770645                    1    1
1            2          1  1560770646                    2    2
2            3          1  1560770647                    3    3
3            4          2  1560770648                    1    1
4            5          2  1560770649                    2    2
5            6          2  1560770650                    3    3
6            7          3  1560770651                    1    1
7            8          3  1560770652                    2    2
8            9          3  1560770653                    3    3

如果只需要計數器，那么GroupBy.cumcount是您的朋友：

df['new'] = df.groupby('machining').cumcount() + 1