[Tutor] Compound if statement question.

[Corey Richardson]

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On 05/01/2011 02:28 PM, Greg Christian wrote:
> Is there a way to write an if statement that will pick up duplicates (two ‘1’s):
> 
> L = ['1', '4', '1']
> if (L[0]) != (L[1]) != (L[2]):
>     print "THEY ARE NOT EQUAL"
> else:
>     print "THEY ARE EQUAL"
> 
> When I run this code, it prints “THEY ARE NOT EQUAL” when it should print the else “THEY ARE EQUAL”.
> 

Well, think about what that if-statement is doing. It's making sure that
the first three elements of L aren't equal to each other. Due to
python's short-circuit evaluation, it breaks the if after the first one,
because '1' != '4'. The parentheses there are useless.

You might look at [1] to see if there is anything that will count how
many times something appears in a list...

[1] -- http://docs.python.org/tutorial/datastructures.html#more-on-lists

- -- 
Corey Richardson
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