The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() in python

Question

suppose I have this function:

a = np.array([-2,1,-5,2])
if a <0:
    a += 10
print (a)

why I am getting this error.

If I pass only one value then it is okey. How to solve this problem?

thank you in advance.

You're comparing list with an integer, which will give you a type error. This question is really unclear, but if you want to use `any` or `all`, as in the title, use `if any(a):` or `if all(a):`. And please elaborate on what you're trying to do. — FlipTack, Dec 29 '16 at 14:56
You are trying to compare ONE value with MANY. It is not possible. — Psytho, Dec 29 '16 at 14:57
There are many duplicates on SO that discuss this error message. — Alex Riley, Dec 29 '16 at 14:59
@ajcr In this case you've linked to the wrong duplicate (not saying it isn't a duplicate, it almost certainly is). It's **not** about evaluating the boolean value of the array. — MSeifert, Dec 29 '16 at 15:03
There may well be a better duplicate to link to, but I don't agree that the linked duplicate is necessarily wrong. Regardless of what the correct approach to incrementing certain values of an array may be, the OP *is* evaluating the boolean value of the array `a` in the code (although probably unintentionally). The OP asks why the error arises here and the linked duplicate explains the reason why. — Alex Riley, Dec 29 '16 at 15:29

Psidom · Answer 1 · 2016-12-29T15:00:44.383

4

numpy does element wise comparison and addition and is vectorized. A direct translation of if-else in numpy is np.where():

import numpy as np
a = np.where(a < 0, a + 10, a)
# array([8, 1, 5, 2])

edited Dec 29 '16 at 15:00

answered Dec 29 '16 at 14:57

Psidom

score 2 · Accepted Answer · answered Dec 29 '16 at 14:58

2

You can simply increment by indexing with a boolean array (a < 0):

a[a < 0] += 10

answered Dec 29 '16 at 14:58

MSeifert

2 Answers2