On Mon, Oct 8, 2012 at 10:36 AM, Guido van Rossum wrote:

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It's not about equality. If you ask whether two NaNs are *unequal* the answer is *also* False.

Does this mean that the following behaviour of lists is a bug?

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x=float('NAN') [x]==[x], [x]<=[x], [x]>=[x] (True, True, True)

No. That's a special case in the comparisons for sequences.

[Now that I'm back at a real keyboard I can elaborate...] This applies to all container comparisons: without the rule that if two contained items reference the same object they are to be considered equal without calling their __eq__, containers couldn't take the shortcut that a container is always equal to itself (i.e. c1 is c2 => c1 == c2). Without this shortcut, container comparisons would be much more expensive: any time a large container was compared to itself, it would be forced to recursively compare all the contained items. You might say that it has to do this anyway when comparing to a container that is not itself, but if the anser is "unequal" the comparison can stop as soon as two unequal items are found, whereas if the answer is "equal" you end up comparing all items. For two different containers there is no possible shortcut, but comparing a container to itself is quite common and really does deserve the shortcut. We discussed this in the past and always came to the same conclusion: despite the rules for NaN, the shortcut for containers is required. A similar shortcut exists for 'x in [x]' BTW. -- --Guido van Rossum (python.org/~guido)