[Tutor] Testing if a number occurs more than once

[Yann Le Du]

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On Mon, 2 Dec 2002, Michael Williams wrote:

> as not only convoluted, but slow (this is the dominant loop of the 
> program). There must be a better way! Any suggestions?

try :

def morethanone(l):
        for n in l:
                if l.count(n)>1:
                        return 1
        return 0


then this gives :

In [372]: l=[1,2,3,4,5]

In [373]: morethanone(l)
Out[373]: 0

In [374]: l=[1,2,3,4,5,23,2,48,23,90]

In [375]: morethanone(l)
Out[375]: 1


Yann
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