CODEWITHSUNDEEP
pythonnumpy
Moe Steen
Moe Steen

Reputation: 829

Use a.any() or a.all()

x = np.arange(0,2,0.5)
valeur = 2*x

if valeur <= 0.6:
    print ("this works")
else:   
    print ("valeur is too high")

here is the error I get:

if valeur <= 0.6:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I have read several posts about a.any() or a.all() but still can't find a way that really clearly explain how to fix the problem. I see why Python does not like what I wrote but I am not sure how to fix it.

Upvotes: 82

Views: 565130

Answers (7)

BrokenBenchmark
BrokenBenchmark

Reputation: 19253

For this specific pattern (mapping elements to one value if they meet a certain condition, else mapping them to another value), you can use np.where():

import numpy as np

x = np.arange(0,2,0.5)
valeur = 2*x

print(*np.where(2 * x <= 0.6, "this works", "valeur is too high"), sep='\n')

This outputs:

this works
valeur is too high
valeur is too high
valeur is too high

Upvotes: 0

AnandOdayil
AnandOdayil

Reputation: 11

import numpy as np
x = np.arange(0,2,0.5)
valeur = 2*x
print(valeur <= 0.6)
'''# [ True False False False] due to this "if statement" cannot decide if the 
value is true or false
 #This is where you got ValueError of ambiguity
 #This error recomended you to use a.any() or a.all() built-in functions
 #a.any() :- any True is True i.e. ([ True False False False].any) will be 
True/1
 #a.all() :- all Ture is Ture i.e. ([ True False False False].all) will be 
False/0'''

if valeur.all() <= 0.6:
'''# Now in "if statement" valeur.all() is Flase/0
# so is 0 <= 0.6? 
# Ture right? '''
    print ("this works")
'''# This is the reason "this works"'''
else:   
    print ("valeur is too high")

Upvotes: 1

AsukaMinato
AsukaMinato

Reputation: 1432

if you want do something for each element:

use numpy's function

import numpy as np
def temp(x):
    if x <= 0.6:
        print ("this works")
    else:   
        print ("valeur is too high")
arr = np.arange(0,1,0.1)
np.vectorize(temp)(arr)

Upvotes: 0

Prajot Kuvalekar
Prajot Kuvalekar

Reputation: 6658

There is one more way you can get this

import numpy as np

a = np.array([1,2,3,4])
b = np.array([5,6,7,8])
c = np.array([1,2,3,4])

print((a == b ).all())  #False
print((a == c ).all())   # True
print((a == b ).any())   #False
print((a == c ).any())   #True
print((a > 3 ).all())    #False

Upvotes: 14

Gursel Karacor
Gursel Karacor

Reputation: 1167

This should also work and is a closer answer to what is asked in the question:

for i in range(len(x)):
    if valeur.item(i) <= 0.6:
        print ("this works")
    else:   
        print ("valeur is too high")

Upvotes: 0

hpaulj
hpaulj

Reputation: 231615

You comment:

valeur is a vector equal to [ 0. 1. 2. 3.] I am interested in each single term. For the part below 0.6, then return "this works"....

If you are interested in each term, then write the code so it deals with each. For example.

for b in valeur<=0.6:
    if b:
        print ("this works")
    else:   
        print ("valeur is too high")

This will write 2 lines.

The error is produced by numpy code when you try to use it a context that expects a single, scalar, value. if b:... can only do one thing. It does not, by itself, iterate through the elements of b doing a different thing for each.

You could also cast that iteration as list comprehension, e.g.

['yes' if b else 'no' for b in np.array([True, False, True])]

Upvotes: 4

poke
poke

Reputation: 388153

If you take a look at the result of valeur <= 0.6, you can see what’s causing this ambiguity:

>>> valeur <= 0.6
array([ True, False, False, False], dtype=bool)

So the result is another array that has in this case 4 boolean values. Now what should the result be? Should the condition be true when one value is true? Should the condition be true only when all values are true?

That’s exactly what numpy.any and numpy.all do. The former requires at least one true value, the latter requires that all values are true:

>>> np.any(valeur <= 0.6)
True
>>> np.all(valeur <= 0.6)
False

Upvotes: 106

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