Calculate Pandas df timedelta of index

Question

Would anyone know how to calculate the time delta of the time stamp of the index?

import pandas as pd
import numpy as np

# simulate some data
# ===================================
np.random.seed(0)
dt_rng = pd.date_range('2015-03-02 00:00:00', '2015-07-19 23:00:00', freq='T')
dt_idx = pd.DatetimeIndex(np.random.choice(dt_rng, size=2000, replace=False))
df = pd.DataFrame(np.random.randn(2000), index=dt_idx, columns=['col']).sort_index()

df

Am I on track using df['elapsed_time'] = pd.TimedeltaIndex(df) at all with this?

This will throw an error: ValueError: Wrong number of items passed 2000, placement implies 1

You can't pass a DataFrame. Only a Series. `df['elapsed_time'] = pd.TimedeltaIndex(df.index)` — Henry Ecker, Nov 06 '21 at 18:41
It's a bit unclear what you're trying to do though because this will just take the underlying ns datetime and _force_ it into a time delta. It won't do any timedelta comparison. — Henry Ecker, Nov 06 '21 at 18:42
It seems like you're likely looking for `df['elapsed_time'] = df.index.to_series().diff()` [Difference pandas.DateTimeIndex without a frequency](https://stackoverflow.com/q/49277932/15497888) — Henry Ecker, Nov 06 '21 at 18:43

score 0 · Answer 1 · answered Nov 07 '21 at 15:03

This answer is beautiful!

This will create another pandas column which I called time_td where then I can cast it as a timedelta64 where m stands for minutes which I am looking for.

df['time_td'] = df.index.to_series().diff().astype('timedelta64[m]')

I can then sum this time_td column with:

df.time_td.sum()

Calculate Pandas df timedelta of index

1 Answers1