Finding the index of datetime Numpy Python

Question

How can I write a code that shows me the indexes of Newdate within Setups. I want to get the index value for each of the date values, so for the first output;'2017-12-22T03:31:00.000000000' date value in Newdate comes 6th from the beginning from Setups the output will be 5.How would I be able to get the Expected Output.

Code:

import numpy as np

Setups= np.array(['2017-09-15T07:11:00.000000000','2017-09-15T11:25:00.000000000',
 '2017-09-15T12:11:00.000000000', '2017-12-22T03:14:00.000000000',
 '2017-12-22T03:26:00.000000000', '2017-12-22T03:31:00.000000000',
 '2017-12-22T03:56:00.000000000'],dtype="datetime64[ns]")

Newdate =  np.array(['2017-09-15T07:11:00.000000000', '2017-12-22T03:31:00.000000000','2017-09-15T11:25:00.000000000', '2017-12-22T03:56:00.000000000'],dtype="datetime64[ns]")

Expected Output:

[0, 5, 1, 6]

Zla Patka · Accepted Answer · 2021-08-30 08:29:55Z

2

You could convert setups to a list and use index():

setups_list = list(Setups)
indices = [setups_list.index(n) for n in Newdate]
print(indices)

# [0, 5, 1, 6]

edited Aug 30, 2021 at 8:29

answered Aug 30, 2021 at 8:24

Zla Patka

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ThePyGuy · Accepted Answer · 2021-08-30 08:25:09Z

0

You can use numpy.isin to get the truthfulness of the values i.e. either the values are common or not, then pass it to numpy.where which will give you the indices of True values

>>> np.where(np.isin(Setups, Newdate))
(array([0, 1, 5, 6], dtype=int64),)

answered Aug 30, 2021 at 8:25

ThePyGuy

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2 Comments

Oluwafemi Sule Over a year ago

This returns the position of items present in the Setups but not in order they are given in Newdate. For example if you had recurring items in Newdate , this solution doesn't provide their indices

ThePyGuy Over a year ago

@OluwafemiSule, yeah that is True that it returns the position based on the values in Setups but not in Newdate

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