Robert Kern wrote:
James Stroud wrote:
I'm missing how a.all() solves the problem Rasmus describes, namely that the order of a python *list* affects the results of containment tests by numpy.array. E.g. "y in ll1" and "y in ll2" evaluate to different results in his example. It still seems like a bug in numpy to me, even if too much other stuff is broken if you fix it (in which case it apparently becomes an "issue").

It's an issue, if anything, not a bug. There is no consistent implementation of bool(some_array) that works in all cases. numpy's predecessor Numeric used to implement this as returning True if at least one element was non-zero. This works well for bool(x!=y) (which is equivalent to (x!=y).any()) but does not work well for bool(x==y) (which should be (x==y).all()), but many people got confused and thought that bool(x==y) worked. When we made numpy, we decided to explicitly not allow bool(some_array) so that people will not write buggy code like this again.

The deficiency is in the feature of rich comparisons, not numpy's implementation of it. __eq__() is allowed to return non-booleans; however, there are some parts of Python's implementation like list.__contains__() that still expect the return value of __eq__() to be meaningfully cast to a boolean.


You have explained

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.


All I can imagine is that something like the following lives in the bowels of the python code for list:

def __contains__(self, other):
  foundit = False
  for i, v in enumerate(self):
    if i == 0:
      # evaluates to bool numpy array
      foundit = one_kind_of_test(v, other)
    else:
      # raises exception for numpy array
      foundit = another_kind_of_test(v, other)
    if foundit:
      break
  return foundit

I'm trying to imagine some other way to get the results mentioned but I honestly can't. It's beyond me why someone would do such a thing, but perhaps it's an optimization of some sort.

James
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