4 
import numpy as np
mainList = []
numpyArray0 = np.array([1,2,3])
numpyArray1 = np.array([4,5,6])
mainList.append(numpyArray0)
mainList.append(numpyArray1)

print("numpyArray0 in mainList:")
try:
  print(numpyArray0 in mainList)
except ValueError:
  print("ValueError")

print("numpyArray1 in mainList:")
try:
  print(numpyArray1 in mainList)
except ValueError:
  print("ValueError")

print("mainList in numpyArray0:")
try:
  print(mainList in numpyArray0)
except ValueError:
  print("ValueError")

print("mainList in numpyArray1:")
try:
  print(mainList in numpyArray1)
except ValueError:
  print("ValueError")

print(numpyArray1 in mainList)

So I have the above code basically it creates 2 numpy arrays inside a normal python list (mainList) and then it checks to see if those 2 arrays are inside the list. The code should output:

numpyArray0 in mainList:
True
numpyArray1 in mainList:
**True**
mainList in numpyArray0:
True
mainList in numpyArray1:
True
**True**

But instead of outputing the above it outputs the following:

numpyArray0 in mainList:
True
numpyArray1 in mainList:
ValueError
mainList in numpyArray0:
True
mainList in numpyArray1:
True
Traceback (most recent call last):
  File "/home/user/Documents/pythonCode/temp.py", line 31, in <module>
    print(numpyArray1 in mainList)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Am I doing anything incorrectly? Note that I tried updating python, numpy, and my os (debian) before running the code.

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5
  • Possible duplicate: stackoverflow.com/questions/5488307/numpy-array-in-python-list Commented Aug 22, 2018 at 22:33
  • I am confused about why the above code works with numpyArray0, but not with numpyArray1, but thank you for the link. Commented Aug 22, 2018 at 22:35
  • I understand; it is not really a dupe, rather a reference. I suggest you remove everything from your code from print("numpyArray0:") all the way down and keep numpyArray0 in mainList and numpyArray1 in mainList - to make the question clear and simple. Commented Aug 22, 2018 at 22:35
  • yep, done and also fixed an incorrect print Commented Aug 22, 2018 at 22:44
  • In general an in test is not reliable unless you are clearly comparing object ids (e.g. an object class without the compare method), or the compare method itself is reliable (for your purposes). Commented Aug 23, 2018 at 3:42

4 Answers 4

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3

numpyArray0 in mainList calls list.__contains__. A list's __contains__ method calls PyObject_RichCompareBool for each element of the list to check if the elements are equal. As it happens, PyObject_RichCompareBool checks for identity equality first, and then does a full comparison.

numpyArray0 is mainList[0] returns True, so full comparison is never done. If full comparison was done, numpy would raise ValueError since a numpy array cannot be interepreted as a boolean.

numpyArray1 in mainList shows that as well (since identity comparison fails for numpyArray1 vs mainList[0].

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4 Comments

Why is mainList in numpyArray0 then? And mainList in numpyArray1?
As a further test, numpyArray0[:] in mainList raises the ValueError. The object id's no longer match, so it goes on to doing the element test.
@DYZ stackoverflow.com/questions/18320624/…
Yeah, pretty surprising :). From my answer to that question: mail-archive.com/[email protected]/msg31578.html and github.com/numpy/numpy/issues/3016 are relevant.
1

Looks like it is a well-known feature related to the way the == operator is overloaded for Numpy arrays.

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2 Comments

This is the right answer, not the accepted one.
It's a combination of testing for id first, as described in @Alok's answer, followed by an array equality test - it's that equality test that produces the value error.
0

here the gist:

>>> numpyArray1 in mainList
....    
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

this will work:

>>> any([list(numpyArray1) == list(litem)  for litem in mainList])

True

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2 Comments

Your answer does not explain why numpyArray0 in mainList works.
as @alok mentions it is essentially short circuited logic at work. since numpyArray0 is the first item in the mainList it is Ture because it is a single value. i.e. >>> numpyArray0 in [numpyArray0] True >>> numpyArray1 in [numpyArray1] True
0

Compare the same operations with lists instead of numpy arrays:

In[171]: mainList = []
In[172]: list0 = [1,2,3]
In[173]: list1 = [4,5,6]
In[174]: mainList.append(list0)
In[175]: mainList.append(list1)
In[176]: list0 in mainList
Out[176]: True
In[177]: list1 in mainList
Out[177]: True

What am I trying to show with this?

Two things.

  1. That the right answer of OP question gave DYZ, not Alok Singhal.
  2. That the OP question is suspiciously similar to the reported bug. Compare with this part of reported issue in Python member operation does not work on lists because of overloaded ndarray equality #10400:
cache = []
cache.append(numpy.ndarray([1,2,3]))
cache.append(numpy.ndarray([4,5,6]))
numpy.ndarray([4,5,6]) in cache
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2 Comments

why 'suspiciously' ?
"Bug" questions almost always get upvoted if they can be replicated. I imagine trawling bug reports on github and then making questions of them would get a new user some quick rep. But considering that even in that case, it still produces a useful question and answer I don't find it too disruptive.

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