Timedelta without Date-CodePudding

I have a column with times that are not timestamps and would like to know the timedelta to 00:30:00 o'clock. However, I can only find methods for timestamps.

df['Time'] = ['22:30:00', '23:30:00', '00:15:00']

The intended result should look something like this:

df['Output'] = ['02:00:00', '01:00:00', '00:15:00']

CodePudding user response：

This code convert a type of Time value from str to datetime (date is automatically set as 1900-01-01). Then, calculated timedelta by setting standardTime as 1900-01-02-00:30:00.

import pandas as pd
from datetime import datetime, timedelta

df = pd.DataFrame()
df['Time'] = ['22:30:00', '23:30:00', '00:15:00']

standardTime = datetime(1900, 1, 2, 0, 30, 0)
df['Time'] = pd.to_datetime(df['Time'], format='%H:%M:%S')
df['Output'] = df['Time'].apply(lambda x: standardTime-x).astype(str).str[7:] # without astype(str).str[7:], the Output value include a day such as "0 days 01:00:00"

print(df)

#                 Time    Output
#0 1900-01-01 22:30:00  02:00:00
#1 1900-01-01 23:30:00  01:00:00
#2 1900-01-01 00:15:00  00:15:00

CodePudding user response：

One could want to use datetime.time as data structures, but these cannot be subtracted, so you can't conveniently get a timedelta from them.

On the other hand, datetime.datetime objects can be subtracted, so if you're always interested in positive deltas, you could construct a datetime object from your time representation using 1970-01-01 as date, and compare that to 1970-01-02T00:30.

For instance, if your times are stored as strings (as per your snippet):

import datetime as dt

def timedelta_to_0_30(time_string: str) -> dt.timedelta:
    time_string_as_datetime = dt.datetime.fromisoformat(f"1970-01-01T{time_string}")
    return dt.datetime(1970, 1, 2, 0, 30) - time_string_as_datetime

my_time_string = "22:30:00"
timedelta_to_0_30(my_time_string)  # 2:00:00