# 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.

You could use the `total_seconds` method:

``````time_d_float = time_d.total_seconds()
``````

In Python 3.2 or higher, you can divide two `timedelta`s 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)
``````

I had the same problem before and I used timedelta.total_seconds to get Estimated duration into seconds with float and it works.
I hope this works for you.

``````from datetime import timedelta,datetime

time_d = datetime_1 - datetime_2

time_d.total_seconds()
``````

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)
``````

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)
``````

``````from datetime import timedelta,datetime
x1= timedelta(seconds=40, minutes=40, hours=5)
x2= timedelta( seconds=50, minutes=20, hours=4)
x3=x1-x2
x5 = x3.total_seconds()
print(x5)
print(type(x5))
print(type(x1))
print(x1)

# if you are working with Dataframe then use loop (* for-loop).
``````
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