# 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.

## Answers:

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)
```

Hope it is helpful!

```
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).
```