Calculate Time Difference Between Two Pandas Columns in Hours and Minutes

I have two columns, fromdate and todate, in a dataframe.

import pandas as pd

data = {'todate': [pd.Timestamp('2014-01-24 13:03:12.050000'), pd.Timestamp('2014-01-27 11:57:18.240000'), pd.Timestamp('2014-01-23 10:07:47.660000')],
        'fromdate': [pd.Timestamp('2014-01-26 23:41:21.870000'), pd.Timestamp('2014-01-27 15:38:22.540000'), pd.Timestamp('2014-01-23 18:50:41.420000')]}

df = pd.DataFrame(data)

I add a new column, diff, to find the difference between the two dates using

df['diff'] = df['fromdate'] - df['todate']

I get the diff column, but it contains days, when there’s more than 24 hours.

                   todate                fromdate                   diff
0 2014-01-24 13:03:12.050 2014-01-26 23:41:21.870 2 days 10:38:09.820000
1 2014-01-27 11:57:18.240 2014-01-27 15:38:22.540 0 days 03:41:04.300000
2 2014-01-23 10:07:47.660 2014-01-23 18:50:41.420 0 days 08:42:53.760000

How do I convert my results to only hours and minutes (i.e. days are converted to hours)?

Answers:

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Method 1

Pandas timestamp differences returns a datetime.timedelta object. This can easily be converted into hours by using the *as_type* method, like so

import pandas
df = pandas.DataFrame(columns=['to','fr','ans'])
df.to = [pandas.Timestamp('2014-01-24 13:03:12.050000'), pandas.Timestamp('2014-01-27 11:57:18.240000'), pandas.Timestamp('2014-01-23 10:07:47.660000')]
df.fr = [pandas.Timestamp('2014-01-26 23:41:21.870000'), pandas.Timestamp('2014-01-27 15:38:22.540000'), pandas.Timestamp('2014-01-23 18:50:41.420000')]
(df.fr-df.to).astype('timedelta64[h]')

to yield,

0    58
1     3
2     8
dtype: float64

Method 2

This was driving me bonkers as the .astype() solution above didn’t work for me. But I found another way. Haven’t timed it or anything, but might work for others out there:

t1 = pd.to_datetime('1/1/2015 01:00')
t2 = pd.to_datetime('1/1/2015 03:30')

print pd.Timedelta(t2 - t1).seconds / 3600.0

…if you want hours. Or:

print pd.Timedelta(t2 - t1).seconds / 60.0

…if you want minutes.

UPDATE: There used to be a helpful comment here that mentioned using .total_seconds() for time periods spanning multiple days. Since it’s gone, I’ve updated the answer.

Method 3

• How do I convert my results to only hours and minutes
  • The accepted answer only returns days + hours. Minutes are not included.
• To provide a column that has hours and minutes as hh:mm or x hours y minutes, would require additional calculations and string formatting.
• This answer shows how to get either total hours or total minutes as a float, using timedelta math, and is faster than using .astype('timedelta64[h]').
• Pandas Time Deltas User Guide
• Pandas Time series / date functionality User Guide
• python timedelta objects: See supported operations.
• The following sample data is already a datetime64[ns] dtype. It is required that all relevant columns are converted using pandas.to_datetime().

import pandas as pd

# test data from OP, with values already in a datetime format
data = {'to_date': [pd.Timestamp('2014-01-24 13:03:12.050000'), pd.Timestamp('2014-01-27 11:57:18.240000'), pd.Timestamp('2014-01-23 10:07:47.660000')],
        'from_date': [pd.Timestamp('2014-01-26 23:41:21.870000'), pd.Timestamp('2014-01-27 15:38:22.540000'), pd.Timestamp('2014-01-23 18:50:41.420000')]}

# test dataframe; the columns must be in a datetime format; use pandas.to_datetime if needed
df = pd.DataFrame(data)

# add a timedelta column if wanted. It's added here for information only
# df['time_delta_with_sub'] = df.from_date.sub(df.to_date)  # also works
df['time_delta'] = (df.from_date - df.to_date)

# create a column with timedelta as total hours, as a float type
df['tot_hour_diff'] = (df.from_date - df.to_date) / pd.Timedelta(hours=1)

# create a colume with timedelta as total minutes, as a float type
df['tot_mins_diff'] = (df.from_date - df.to_date) / pd.Timedelta(minutes=1)

# display(df)
                  to_date               from_date             time_delta  tot_hour_diff  tot_mins_diff
0 2014-01-24 13:03:12.050 2014-01-26 23:41:21.870 2 days 10:38:09.820000      58.636061    3518.163667
1 2014-01-27 11:57:18.240 2014-01-27 15:38:22.540 0 days 03:41:04.300000       3.684528     221.071667
2 2014-01-23 10:07:47.660 2014-01-23 18:50:41.420 0 days 08:42:53.760000       8.714933     522.896000

Other methods

• An item of note from the podcast in Other Resources, .total_seconds() was added and merged when the core developer was on vacation, and would not have been approved.
  • This is also why there aren’t other .total_xx methods.

# convert the entire timedelta to seconds
# this is the same as td / timedelta(seconds=1)
(df.from_date - df.to_date).dt.total_seconds()
[out]:
0    211089.82
1     13264.30
2     31373.76
dtype: float64

# get the number of days
(df.from_date - df.to_date).dt.days
[out]:
0    2
1    0
2    0
dtype: int64

# get the seconds for hours + minutes + seconds, but not days
# note the difference from total_seconds
(df.from_date - df.to_date).dt.seconds
[out]:
0    38289
1    13264
2    31373
dtype: int64

Other Resources

• Talk Python to Me: Episode #271: Unlock the mysteries of time, Python’s datetime that is!
  • Timedelta begins at 31 minutes
  • As per Python core developer Paul Ganssle and python dateutil maintainer:
    • Use (df.from_date - df.to_date) / pd.Timedelta(hours=1)
    • Don’t use (df.from_date - df.to_date).dt.total_seconds() / 3600
      • pandas.Series.dt.total_seconds
      • .dt accessor
• Real Python: Using Python datetime to Work With Dates and Times
• The dateutil module provides powerful extensions to the standard datetime module.

%%timeit test

import pandas as pd

# dataframe with 2M rows
data = {'to_date': [pd.Timestamp('2014-01-24 13:03:12.050000'), pd.Timestamp('2014-01-27 11:57:18.240000')], 'from_date': [pd.Timestamp('2014-01-26 23:41:21.870000'), pd.Timestamp('2014-01-27 15:38:22.540000')]}
df = pd.DataFrame(data)
df = pd.concat([df] * 1000000).reset_index(drop=True)

%%timeit
(df.from_date - df.to_date) / pd.Timedelta(hours=1)
[out]:
43.1 ms ± 1.05 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%%timeit
(df.from_date - df.to_date).astype('timedelta64[h]')
[out]:
59.8 ms ± 1.29 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0