Is there a straightforward way to generate two numpy arrays from one based on one logical test?

Question

import numpy as np
x = np.array([1,2,3,4,5,6,7,8])
y = x[x%2==0]
z = x[x%2==1]

I don't want to perform the second test for z. Obviously z is simply x with elements from y removed. Can I simply extract z from x with the help of y? Thanks for your help.

I would suggest to go through this [link](https://stackoverflow.com/questions/40055835/removing-elements-from-an-array-that-are-in-another-array) — Sai Sreenivas, Jul 13 '20 at 16:36
`cond = (x%2 == 0)` `y = x[cond]` `z = x[np.logical_not(cond)]` — alani, Jul 13 '20 at 16:53

score 0 · Answer 1 · answered Jul 13 '20 at 16:31

0

Try this:

import numpy as np
x = np.array([1,2,3,4,5,6,7,8])
y = x[x%2==0]
z = y-1

answered Jul 13 '20 at 16:31

doesn't always work, good one though! (atleast for this one) – Sai Sreenivas Jul 13 '20 at 16:34

score 0 · Answer 2 · answered Jul 13 '20 at 16:43

With the new dtype aware sort functions it is actually pretty fast to argsort and split the condition array:

def pp():        
    order = cond.argsort(kind="stable")
    if cond[order[0]]:                          
        return a[:0],a                          
    elif not cond[order[-1]]:
        return a,a[:0]
    split = cond.searchsorted(True,sorter=order)
    return a[order[:split]],a[order[split:]]

def OP():
    return a[~cond],a[cond]


a = np.arange(1000)
cond = a%2 == 0

timeit(OP,number=1000)
# 0.012416471960023046
timeit(pp,number=1000)
# 0.009316607960499823

Is there a straightforward way to generate two numpy arrays from one based on one logical test?

2 Answers2