Numpy apply function to array

Question

For example, I have function:

f1 = lambda x: x % 2

If I want to modify array = np.linspace(0, 5, 6) I can do f1(array). Everything works as expected:

[0. 1. 0. 1. 0. 1.]

If I change function to:

f2 = lambda x: 0
print(f2(array))

gives me 0 while I expected [0. 0. 0. 0. 0. 0.]. How to achieve consistency?

The return value of f2 is not an array. Nothing unexpected happening here. — meowgoesthedog
– meowgoesthedog, Commented Mar 4, 2019 at 11:26
I expected that f2 will iterate over array and evaluate every element then store it to a resulting array. — Kenenbek Arzymatov
– Kenenbek Arzymatov, Commented Mar 4, 2019 at 11:27
Python functions don't automatically iterate over their arguments; the expression x % 2 does (when x is an np.array). — meowgoesthedog
– meowgoesthedog, Commented Mar 4, 2019 at 11:28
you want the mod... np.mod(array, 2) => ,,, array([0., 1., 0., 1., 0., 1.]) — NaN
– NaN, Commented Mar 4, 2019 at 11:43

Mayur · Accepted Answer · 2019-03-04 11:28:38Z

6

You can use below code to achieve desirable output

import numpy as np
array = np.linspace(0, 5, 6)
f2 = lambda x: x-x
print(f2(array))

answered Mar 4, 2019 at 11:28

Mayur

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2 Comments

Areza Over a year ago

how about if the array was array of array ?

Mayur Over a year ago

Thanks, @user702846 for your concern, in that case, we need some changes. The above code works for the mentioned case

PiRK · Accepted Answer · 2019-03-04 15:18:35Z

3

Slightly more explicit than previous answer :

import numpy as np
array = np.linspace(0, 5, 6)
f2 = lambda x: np.zeros_like(x)
print(f2(array))

Documentation for numpy.zeros_like: Return an array of zeros with the same shape and type as a given array.

answered Mar 4, 2019 at 15:18

PiRK

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Comments

Benjamin Loison · Accepted Answer · 2024-04-15 12:31:58Z

1

To iterate over an array, evaluate the function for every element, then store it to a resulting array, a list iterator works consistently:

import numpy as np
array = np.linspace(0, 5, 6)
f1 = lambda x: x % 2
f2 = lambda x: 0

print ([f1(x) for x in array])

[0.0, 1.0, 0.0, 1.0, 0.0, 1.0]

print ([f2(x) for x in array])

[0, 0, 0, 0, 0, 0]

edited Apr 15, 2024 at 12:31

Benjamin Loison

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answered Sep 13, 2020 at 22:46

Jim Bander

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