CODEWITHSUNDEEP
pythonpandasdataframetime
Vincent Tep
Vincent Tep

Reputation: 124

Calculating difference in seconds

I'm calculating time differences, in seconds, between busses' expected and actual stop times.

My problem looks like this:

# creating data
d = {
    'time_A': ['2022-08-30 06:21:00', '2022-08-30 16:41:00'], 
    'time_B': ['2022-08-30 06:21:09', '2022-08-30 16:40:16'], 
}

# creating DataFrame
my_df = pd.DataFrame(d)
my_df['time_A'] = pd.to_datetime(my_df['time_A'])
my_df['time_B'] = pd.to_datetime(my_df['time_B'])

# subtracting times
my_df['difference'] = my_df['time_B'] - my_df['time_A']

my_df

result:

    time_A  time_B  difference
0   2022-08-30 06:21:00 2022-08-30 06:21:09 0 days 00:00:09
1   2022-08-30 16:41:00 2022-08-30 16:40:16 -1 days +23:59:16

I don't understand why the difference between today 16:40:16 and today 16:41:00 is -1 days +23:59:16.

if I do this

my_df['difference'] = (my_df['time_B'] - my_df['time_A']).dt.seconds

Then I get

    time_A  time_B  difference
0   2022-08-30 06:21:00 2022-08-30 06:21:09 9
1   2022-08-30 16:41:00 2022-08-30 16:40:16 86356

I would like the "difference" cell on row O to display something like "+9", and the one below to display "-44". How do I do this? Thanks!

Upvotes: 0

Views: 36

Answers (1)

Daweo
Daweo

Reputation: 36450

Subtracting datetime.datetimes gives datetime.timedeltas which are represented that way, use .total_seconds() to get numeric value of seconds, consider following simple example

import datetime
import pandas as pd
df = pd.DataFrame({"schedule":pd.to_datetime(["2000-01-01 12:00:00"]),"actual":pd.to_datetime(["2000-01-01 12:00:05"])})
df['difference_sec'] = (df['schedule'] - df['actual']).apply(datetime.timedelta.total_seconds)
print(df)

output

             schedule              actual  difference_sec
0 2000-01-01 12:00:00 2000-01-01 12:00:05            -5.0

Note that this is feature of datetime.timedelta, it is not specific to pandas.

Upvotes: 2

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