Odd truth result with in and ==
John Pote
johnpote at jptechnical.co.uk
Fri Nov 23 19:02:11 EST 2018
On 21/11/2018 19:18, Python wrote:
> $ python3
> Python 3.5.2 (default, Nov 23 2017, 16:37:01)
> [GCC 5.4.0 20160609] on linux
> Type "help", "copyright", "credits" or "license" for more information.
>>>> 1 in [1,2,3] == True
> False
>>>> 1 in ([1,2,3] == True)
> Traceback (most recent call last):
> File "<stdin>", line 1, in <module>
> TypeError: argument of type 'bool' is not iterable
>>>> (1 in [1,2,3]) == True
> True
>
> How is the first not equivalent to either one of the second or third?
> My expectation is it should produce the same result as the second. It
> *seems* like Python is ignoring the '1 in' part and just giving the
> result for '[1,2,3] == True'... Is this just a bug?
>
I've followed this thread with interest, as I do with threads like this,
and learnt a useful detail about Python.
But the following I found unexpected. (Python 3.6 on a Windows 7 64 bit box)
>>> if []: print("Truthy")
...
>>> if [1,2,3]: print("Truthy")
...
Truthy
>>>
from which I concluded [] is Falsey and [1,2,3] is Truthy and the above
if statements work as expected.
but,
>>> [1,2,3] == True
False
>>>
is unexpected as to my mind as [1,2,3] is 'Truthy' and True has ultimate
'Truthiness'.
Any ideas? Is there an implicit 'casting' taking place and if so is this
documented somewhere?
I interpret the above comparison as
>>> bool([1,2,3]) == bool(True)
True
>>>
Thanks everyone.
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