How to calculate total difference in milliseconds by condition?

I have the following pandas dataframe df:

timestamp	version	actual	pred
2022-01-19 11:00:00.600	1	0	0
2022-01-19 11:00:00.800	1	0	1
2022-01-19 11:00:01.200	1	1	0
2022-01-19 11:00:01.800	1	0	0
2022-01-19 11:00:02.200	2	1	1
2022-01-19 11:00:02.600	2	0	0
2022-01-19 11:00:03.200	3	0	1
2022-01-19 11:00:03.600	3	0	0
2022-01-19 11:00:03.800	3	1	0
2022-01-19 11:00:03.600	4	0	0
2022-01-19 11:00:03.800	4	0	1

I need to calculate the total distance in milliseconds between values of 1 in actual and pred, grouped by version.

The expected answer is: 1000 ms

• version 1: (2022-01-19 11:00:01.200 – 2022-01-19 11:00:00.600) = 400 ms
• version 2: (2022-01-19 11:00:02.200 – 2022-01-19 11:00:02.200) = 0 ms
• version 3: (2022-01-19 11:00:03.800 – 2022-01-19 11:00:03.200) = 600 ms

Answer

Assuming there exists a single “actual” and single “pred” values for each “version”, we could simply subtract values. In other words, if each “version” has exactly 1 “actual” and exactly 1 “pred” value, then df['actual'].eq(1) will be True for exactly one value per “version”; same for df['pred'].eq(1). Since the data is sorted by “version”, the versions match.

df['timestamp'] = pd.to_datetime(df['timestamp'])
out = ((df.loc[df['actual'].eq(1), 'timestamp'].to_numpy() - 
        df.loc[df['pred'].eq(1), 'timestamp'].to_numpy())
       .astype('timedelta64[ms]').sum())

If not all versions have a 1 in either of “actual” or “pred”, we could still filter the “actual” and “pred” values; then merge on “version” (so that each “version” will have an “actual” and a “pred” value); then find difference and sum:

df['timestamp'] = pd.to_datetime(df['timestamp'])
merged = (df.loc[df['actual'].eq(1), ['version', 'timestamp']]
          .merge(df.loc[df['pred'].eq(1), ['version', 'timestamp']], 
                 on='version', suffixes=('_actual', '_pred')))

out = (merged['timestamp_actual'] - merged['timestamp_pred']).astype('timedelta64[ms]').sum()

Output:

1000