I have a dataframe like so:
import pandas as pd
import numpy as np
date_rng = pd.date_range(start="2023-11-18", periods=3, freq="10S")
values = [4, 2, 3]
df = pd.DataFrame(data={"values": values}, index=date_rng)
df["dt"] = df.index.to_series().diff().dt.seconds
df["dt"] = df.index.to_series().diff(periods=2).dt.seconds
df["dt_neg"] = df.index.to_series().diff(periods=-1).dt.seconds
print(df)
gives
values dt dt_neg
2023-11-18 00:00:00 4 NaN 86390.0
2023-11-18 00:00:10 2 NaN 86390.0
2023-11-18 00:00:20 3 20.0 NaN
Shouldn’t negative values work, too?
I read the answers here and here.
>Solution :
seconds return the number of seconds independently of the hours/minutes. You need to use total_seconds:
df["dt_neg"] = df.index.to_series().diff(periods=-1).dt.total_seconds()
Output:
values dt dt_neg
2023-11-18 00:00:00 4 NaN -10.0
2023-11-18 00:00:10 2 NaN -10.0
2023-11-18 00:00:20 3 20.0 NaN
Indeed, negative timedelta have a peculiar format:
df.index.to_series().diff(periods=-1)
2023-11-18 00:00:00 -1 days +23:59:50
2023-11-18 00:00:10 -1 days +23:59:50
2023-11-18 00:00:20 NaT
Freq: 10S, dtype: timedelta64[ns]