Unexpected output when passing array as argument of a vectorized function

Question

In the example code below

import numpy as np

def f(x):
    print(x)

x = np.array([[ 0.31432202,  7.94263361],
              [-0.5346868,   1.93901039],
              [-0.47571535,  4.17720033]])

np.vectorize(f)(x[0,:])

As output, I expected to get something like

[ 0.31432202  7.94263361]

Instead I get

0.31432202
0.31432202
7.94263361

Can anyone tell me what is wrong with it? Thank you

hpaulj · Accepted Answer · 2021-07-08 04:23:25Z

2

How much of the np.vectorize docs did you read?

In [129]: def f(x):
     ...:     print(x)
     ...: 
     ...: x = np.array([[ 0.31432202,  7.94263361],
     ...:               [-0.5346868,   1.93901039],
     ...:               [-0.47571535,  4.17720033]])
In [130]: f1=np.vectorize(f)
In [131]: f1(1)
1
1
Out[131]: array(None, dtype=object)

f gets called twice, once to determine the return dtype, and once for each element. Try it with 3 elements:

In [132]: f1([1,2,3])
1
1
2
3
Out[132]: array([None, None, None], dtype=object)

Note that the return is an array with None. That's because your f doesn't have a return statement. It just does the print.

Why are you using np.vectorize? It has a clear performance disclaimer. It also talks about the return dtype and how it determines that. It's not a high performance way of calling a function that just prints something. It may be useful for running a function of several scalar values, and you want to take advantage of numpy broadcasting.

Read the docs.

edited Jul 8, 2021 at 4:23

answered Jul 8, 2021 at 3:23

hpaulj

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2 Comments

Eduard Over a year ago

Thank you! I've read the docs but it went unnoticed. Sorry if it was a stupid question... I don't intent using the function for printing the values (as it was just an example code) but for calculating a mathematical expression with the values in a row. Is there a more efficient way to do it?

hpaulj Over a year ago

I have in the past recommended including a print to better see what vectorize is doing. In this case it reveals the initial trial call. Specifying an otypes gets rid of that.

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