Getting the index of datetime function within Numpy Python

Question

The code down below finds the index of newdates within Setups with the init_location function. Then I increment the index by one to get the next date. But I cannot add to the init_location function. How would I be able to do that and get the expected value?

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]")
    

newdates = np.array(['2017-09-15T07:11:00.000000000','2017-12-22T03:26:00.000000000'],dtype="datetime64[ns]")
init_location= np.where(np.in1d(Setups,newdates[1]))
print("init location: ", Setups[init_location],"Address right after: ",Setups[init_location+1])

Output Error

TypeError: can only concatenate tuple (not "int") to tuple

Expected Output

init location: 2017-12-22T03:26:00. Address right after: 2017-12-22T03:31:00.00

you'll want np.nonzero; something like init_location = np.nonzero(np.in1d(Setups, newdates[1]))[0][0] — FObersteiner
– FObersteiner, Commented Jul 28, 2021 at 6:35

Siva-Sg · Accepted Answer · 2021-07-28 14:51:38Z

0

becuase init_location is a tuple, you cannot add "1" directly . Instead use the following format.

print("init location: ", Setups[init_location],"Address right after: ",Setups[init_location[0]+1])

answered Jul 28, 2021 at 14:51

Siva-Sg

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2 Comments

georgehere Over a year ago

Thank you it works but it prints like this ['2018-11-14T16:21:00.000000000'] - ['2018-12-09T08:43:00.000000000'] is there anychance I could make it print like '2018-11-14 16:21:00 - 2018-12-09 08:43:00

Siva-Sg Over a year ago

you can change the code to as follows

init_location= np.where(np.in1d(Setups,newdates[1]))[0][0] print("init location: ", Setups[init_location],"Address right after: ",Setups[init_location+1])

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