python - 绝对到相对时间

Question

my pandas DataFrame has the following current structure :

{
'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]
}

I'd like to add a column with the relative time of each machining process, so that it refreshes every time the column 'Machining' changes its value.
Thus, the desired structure is:

{
'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]
}

I'm struggling a bit to do this in a clean way: any help would be appreciated also without pandas if needed.

Answer 1

Subtract first values per groups created by GroupBy.transform :

#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

If need counter only then GroupBy.cumcount is your friend:

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