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I have a list of 3x3 arrays l and I want to check if a different single 3x3 array a is in the list. I tried like this:

a in l

but it couldn't be executed beacuse of the following error:

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

I cannot grasp why such a comparison would be ambiguous (the intention is clear), but anyway, is there a way to solve my problem?

Hendrik
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3 Answers3

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Numpy is confused there on what you want to do. Do you want to know if a is equivalent to any element of l or if a is an element of l. 

a  = np.ones((3,3))
b = np.ones((3,3))
l  = [b]


b in l
>>True

a in l
>>
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-17-cf48b78477bf> in <module>()
----> 1 a in l

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

An alternative for this, if you want to know if a is in l is to use id() function.

so 

ids = map(id, l)

id(a) in ids
>> False

id(b) in ids
>>True
jkhadka
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    Sorry, what is the difference here: "if a is equivalent to any element of l or if a is an element of l"? If a is any element of l then a is an element of l right? – Hendrik Dec 05 '18 at 12:31
  • I meant, there are two specific cases to consider here : 1. if `a` is similar to an element of `l`, meaning `a` contains same thing as an element in `l`. and 2. if `a` is an element of `l`. Example: if `l` is a list of cars, `a = "red car"`, now 1. if you assign `l =["red car","yellow car"]` then `a` is equivalent (equal) to an element of `l` but if you change the value of `a`, `a = "blue car"`, now `a` is not equal to any element of `l` but if `l` was defined as `l =[a]`, then you can do what ever to `a` and `a in l` will always be true. – jkhadka Dec 05 '18 at 13:32
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a.all() and a.any() could only be done using numpy. Now numpy needs to know if you'd consider it a match if -

  • any - any of the elements match
  • all - all of the elements match

Its not about intentions. Its about providing functions that the community would find helpful. So in your case you'd probably use a.all
This SO post should clear it up for you. I essentially provided the gist above.

jar
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0

It worked well on my side when I tried it, but I found some returns that were inaccurate.

a= np.ones((3,3))
b= np.ones((3,3))

a in b
#output True

b in a 
#output True

but when I changed the length of arrays in one, and not the other some discrepancies came out.

a= np.ones((3,3))
b= np.ones((4,3))

a in b
#Output- False    #this was the same for the reverse

np.any(a ==b)
#Output False   #This also threw up a depreciation warning forelementwise == comparison

np.any(a[0] == b[0])
#OutPut True

for i in range(len(a)):
    if a[i] in b[i]:
        print('yes')
#OutPut   yes yes yes

seems it doesn't like comparing an array of array's as much as it likes doing it one by one.

Puhtooie
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