[Baypiggies] What is happening here with true/false comparisons
Max Slimmer
max at theslimmers.net
Tue Jan 26 01:06:01 CET 2010
Great explanation, this should be a good newbie nugget.
Thanks,
Max Slimmer
On Mon, Jan 25, 2010 at 3:03 PM, Damon McCormick <damonmc at gmail.com> wrote:
> While the suggested *fixes* are all correct (when in doubt, explicilty
> parenthesize!), none of the *explanations* for the unexpected output are
> quite right. Since this involves a subtle issue, I thought I'd send a full
> explanation.
>
> It's tempting to assume that
>
> >>> a in alist == b in alist
> is equivalent to
> >>> ((a in alist) == b) in alist
>
> However, this is not correct!
>
> For a simpler (but perhaps more confusing) example of the hazards of using
> "in" and "==", non-parenthesized, in an expression like this, consider the
> following:
>
> >>> a = 1 # as before
> >>> alist = [5,6] # as before
> >>> a in alist == False
> False
> >>> (a in alist) == False
> True
>
> Weird, right? And no, putting parens around (alist == False) won't
> work--that would be an exception because the right side of the 'in' operator
> wouldn't be iterable.
>
> Here's one last example:
>
> >>> blist = [1, [5,6]]
> >>> 5 in alist == [5,6] in blist
> True
>
> You might enjoy the exercise of figuring out why the above output is
> correct. But to cut to the chase, what's going on is the following. Python
> allows comparisons to be chained, as in the following:
>
> >>> a == 1 == 2/2
> True
> >>> 1 < 5 < 7
> True
>
> The way the chaining works (see 5.9 in http://tinyurl.com/3vsb6m) is that
>
> >>> a == 1 == 2/2
> is equivalent to
> >>> (a == 1) and (1 == 2/2)
>
> and
>
> >>> 1 < 5 < 7
> is equivalent to
> >>> (1 < 5) and (5 < 7)
>
> Since 'in' is just another comparison operator, it works the same way.
> Thus, the the original expression
>
> >>> a in alist == b in alist
> is equivalent to
> >>>(a in alist) and (alist == b) and (b in alist)
>
> which is False because all three comparisons are False. You'll see that
> the two other examples I came up with make sense in this context as well.
>
> -Damon
>
>
>
> On Mon, Jan 25, 2010 at 1:27 PM, Asher Langton <langton2 at llnl.gov> wrote:
>
>> On Jan 25, 2010, at 1:12 PM, Max Slimmer wrote:
>>
>>> Can anyone explain the following:
>>>
>>> >>> a = 1
>>> >>> b = 2
>>> >>> alist = [5,6]
>>> >>> print a in alist
>>> False
>>>
>>> >>> a in alist == b in alist
>>> False
>>> >>> a in alist == a in alist
>>> False
>>> >>> bool(a in alist) == bool(b in alist) # this does what we expect
>>> True
>>> >>> c = 5
>>> >>> c in alist == c in alist
>>> False
>>> >>>
>>>
>>
>> The '==' and 'in' operators have the same precedence, so the expression 'a
>> in alist == b in alist' is evaluated left-to-right as:
>>
>> >>> ( (a in alist) == b) in alist
>>
>> Since 'a in alist' is False, this is the same as
>>
>> >>> ( False == b) in alist
>>
>> which can be simplified to
>>
>> >>> False in alist
>>
>> which is False.
>>
>>
>> -Asher
>>
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>
>
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