efficient way to manipulating numpy array

Question

So let's say I have a numpy array a= np.array([1,2,3,4,5]) and a value x=4, then I want to create a numpy array of values -1 and 1 where there is a 1 in the 4th position and -1 everywhere else.

Here is what I tried:

for i in range(a):
    if i == x:
        a[i]=1
    else:
        a[i]=-1

Is this good?

range(a) doesn't work in your example. It would need to be range(len(a)) — David Buck
– David Buck, Commented Oct 23, 2019 at 15:26

Lior Cohen · Accepted Answer · 2019-10-23 15:26:32Z

4

No, this is not numpy'ish

b=-np.ones(a.shape)
b[x] = 1

Edit: added example

import numpy as np

x=3
a= np.array([1, 2, 3, 4, 5])
b=-np.ones(a.shape)
b[x] = 1
print(b)

> [-1. -1. -1.  1. -1.]

edited Oct 23, 2019 at 15:26

answered Oct 23, 2019 at 15:22

Lior Cohen

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2 Comments

David Buck Over a year ago

Or b=-np.ones(a.shape, dtype=int) if ints are required, as otherwise they're floats

tstanisl Over a year ago

Construction of b could be slightly optimized with np.full_like(a, -1). Less operations, less mem consumed.

Georgina Skibinski · Accepted Answer · 2019-10-23 15:28:20Z

2

Try:

import numpy as np

a= np.array([1,2,3,4,5]) 

x=np.where(a==4, 1, -1)

print(x)

Output:

[-1 -1 -1  1 -1]

[Program finished]

answered Oct 23, 2019 at 15:28

Georgina Skibinski

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Comments

Milad Barazandeh · Accepted Answer · 2019-10-23 15:27:18Z

1

try this:

b = np.array([1 if i == 4 else -1 for i in range(a.shape)])

answered Oct 23, 2019 at 15:27

Milad Barazandeh

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Comments

ilmarinen · Accepted Answer · 2019-10-23 16:07:04Z

0

Another alternative. Utilizes casting from bool to int.

b=2*(a==x)-1

answered Oct 23, 2019 at 16:07

ilmarinen

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