Rich Comparisons Gotcha

Robert Kern robert.kern at gmail.com
Mon Dec 8 08:13:26 CET 2008


James Stroud wrote:
> Robert Kern wrote:
>> James Stroud wrote:
>>> py> 112 = [1, y]
>>> py> y in 112
>>> Traceback (most recent call last):
>>>   File "<stdin>", line 1, in <module>
>>> ValueError: The truth value of an array with more than one element is...
>>>
>>> but not
>>>
>>> py> ll1 = [y,1]
>>> py> y in ll1
>>> True
>>>
>>> It's this discrepancy that seems like a bug, not that a ValueError is 
>>> raised in the former case, which is perfectly reasonable to me.
>>
>> Nothing to do with numpy. list.__contains__() checks for identity with 
>> "is" before it goes to __eq__().
> 
> ...but only for the first element of the list:
> 
> py> import numpy
> py> y = numpy.array([1,2,3])
> py> y
> array([1, 2, 3])
> py> y in [1, y]
> ------------------------------------------------------------
> Traceback (most recent call last):
>   File "<ipython console>", line 1, in <module>
> <type 'exceptions.ValueError'>: The truth value of an array with more 
> than one element is ambiguous. Use a.any() or a.all()
> py> y is [1, y][1]
> True
> 
> I think it skips straight to __eq__ if the element is not the first in 
> the list.

No, it doesn't skip straight to __eq__(). "y is 1" returns False, so (y==1) is 
checked. When y is a numpy array, this returns an array of bools. 
list.__contains__() tries to convert this array to a bool and 
ndarray.__nonzero__() raises the exception.

list.__contains__() checks "is" then __eq__() for each element before moving on 
to the next element. It does not try "is" for all elements, then try __eq__() 
for all elements.

 > That no one acknowledges this makes me feel like a conspiracy
 > is afoot.

I don't know what you think I'm not acknowledging.

-- 
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
  that is made terrible by our own mad attempt to interpret it as though it had
  an underlying truth."
   -- Umberto Eco




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