Input
New Time
11:59:57
12:42:10
12:48:45
18:44:53
18:49:06
21:49:54
21:54:48
5:28:20
Below I wrote code to create interval in min.
import pandas as pd
import numpy as np
df = pd.read_csv(r"D:\test\test1.csv")
df['Interval in min'] = (pd.to_timedelta(df['New Time'].astype(str)).diff(1).dt.floor('T').dt.total_seconds().div(60))
print(df)
Output
New Time Interval in min
11:59:57 NaN
12:42:10 42.0
12:48:45 6.0
18:44:53 356.0
18:49:06 4.0
21:49:54 180.0
21:54:48 4.0
5:28:20 -987.0
Last interval in min i.e. -987 min is not right. Please suggest.
>Solution :
Assuming you want to consider a negative difference to be a new day, you could use:
s = pd.to_timedelta(df['New Time']).diff()
df['Interval in min'] = (s
.add(pd.to_timedelta(s.lt('0').cumsum(), unit='d'))
.dt.floor('T').dt.total_seconds().div(60)
)
output:
New Time Interval in min
0 11:59:57 NaN
1 12:42:10 42.0
2 12:48:45 6.0
3 18:44:53 356.0
4 18:49:06 4.0
5 21:49:54 180.0
6 21:54:48 4.0
7 5:28:20 453.0