Convert timedelta to floating-point

Question

I got a timedelta object from the subtraction of two datetimes. I need this value as floating point for further calculations. All that I've found enables the calculation with floating-points, but the result is still a timedelta object.

time_d = datetime_1 - datetime_2
time_d_float = float(time_d)

does not work.

Solution 1

You could use the total_seconds method:

time_d_float = time_d.total_seconds()

Solution 2

In Python 3.2 or higher, you can divide two timedeltas to give a float. This is useful if you need the value to be in units other than seconds.

time_d_min = time_d / datetime.timedelta(minutes=1)
time_d_ms  = time_d / datetime.timedelta(milliseconds=1)

Solution 3

You could use numpy to solve that:

import pandas as pd
import numpy as np

time_d = datetime_1 - datetime_2

#for a single value
number_of_days =pd.DataFrame([time_d]).apply(np.float32)

#for a Dataframe
number_of_days = time_d.apply(np.float32)

Hope it is helpful!

Solution 4

If you needed the number of days as a floating number you can use timedelta's days attribute

time_d = datetime_1 - datetime_2
number_of_days = float(time_d.days)

View more solutions

Comments

[umbreonben]

I get AttributeError: 'datetime.timedelta' object has no attribute 'total_seconds'.

[Will Hardy]

timedelta.total_seconds() was introduced in Python 2.7/3.2

[Ironcache]

How does this add anything over the currently-accepted answer?

[lancew]

I get this error with your suggestion: AttributeError: 'datetime.timedelta' object has no attribute 'apply'

[Monique Marins]

Hi @lancew, thanks for catching that! In that case, I needed to convert a dataframe to float so it worked out for me. If you want to convert a unique value, you could do it: pd.DataFrame([time_d]).apply(np.float32) . In this case, you need to import both 'pandas as pd' and 'numpy as np'. Although I'll be testing other ways. Please, let me know if it works.

[Wolf]

I see this as the most natural way to do it. Besides that, it's also great for adapting time implementations of other languages/systems.

[Mache]

@aaa90210 - it's a method and should be used with ()

[Skippy le Grand Gourou]

Not to be confused with timedelta.seconds (which returns an int, BTW).

[supercooler8]

Now, I get this error: "TypeError: float() argument must be a string or a number, not 'Timedelta'"

[lpnorm]

This gives the number of days only. You lose whatever amount of hours minutes or seconds you have in time_d. This is not accurate.

[Just_Alex]

I had a similar issue. I solved it using to_numpy() e.g. df["column"].to_numpy().astype(float). Also see solution here stackoverflow.com/questions/49370756/….

[Joe Sadoski]

@SkippyleGrandGourou furthermore, it doesn't round up for you - it's the literal seconds value without the milliseconds.