Compare Pandas DatetimeIndex to a smaller DatetimeIndex with condition

Question

I am trying to compare two Pandas DatetineIndexs with different lengths. I want to simply add a new column to my larger Dataframe (df) and place 1 where it matches the smaller DatetineIndex (events).

I tried

df['filter'] = np.where(df.index == tEvents, 1, np.nan)

but I get "ValueError: Lengths must match to compare"

I've been stuck here longer than I like to admit

tEvents and df.index are different lengths. I believe you want df.index in tEvents not df.index == tEvents — W Stokvis
– W Stokvis, Commented Jul 20, 2018 at 17:34
@W Stokvis, thanks I believe this is what I am looking for. Can you post it as an answer so I can mark it as solved! — BillyRay
– BillyRay, Commented Jul 20, 2018 at 17:40

W Stokvis · Accepted Answer · 2018-07-20 17:59:31Z

3

tEvents and df.index are different lengths. df.index == tEvents looks to compare the two lists.

You want to check if an element in df.index is in tEvents. Thus replace df.index == tEvents with df.index.isin(tEvents)

To see add a True or false value if date matches, use DataFrame.isin()

edited Jul 20, 2018 at 17:59

answered Jul 20, 2018 at 17:42

W Stokvis

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2 Comments

DSM Over a year ago

df.index in tEvents won't work, because indexes aren't hashable. That's trying to check if the df.index object itself, not its constituent elements, is in tEvents. You probably want df.index.isin(tEvents).

W Stokvis Over a year ago

@DSM adjusted the answer. Thanks for the feedback.

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