Weird behavior in search in a list

Su Y suyuancn at gmail.com
Thu Mar 29 14:34:14 CEST 2007


On 3月29日, 下午8时22分, Michael Bentley <mich... at jedimindworks.com> wrote:
> On Mar 29, 2007, at 6:51 AM, Su Y wrote:
>
>
>
>
>
> > I want find the first number in extend[] which is larger than num, so
> > I wrote:
> > def find(num):
> >     count=0
> >     for elem in extend:
> >         if elem<num:
> >             count+=1
> >     return count
>
> > I found that if extend[] is monotonous, like [1.1, 2.3, 3.2, 4.5,
> > 5.6],
> > it works fine: find(4) returns 3, extend[3] is 4.5.
> > But, if extend[] is not monotonous, like [1.1, 2.3, 3.2, 4.5, 5.6,
> > 4.6, 3.4, 2.1, 0.3],
> > find(4) returns 6, extend[6] is 3.4!
>
> > what's going on here? I really can't understand....
>
> find() loops through the list, and every time it finds a value less
> than num it increments count.  So in your second example the values
> 1.1, 2.3, 3.2,  3.4, 2.1, and 0.3 are all less than 4, which means
> count will be 6.
>
> As you learned, a sorted list behaves as you expect.  So one approach
> is to sort your list first.  But another perhaps better approach is
> to do something like:
>
> def find(num):
>         # check to make sure there *is* a value greater than num
>         if max(extend) > num:
>                 # then return the smallest value that is greater than num
>                 return min([x for x in extend if x > num])
>         else:
>                 return None
>
> Hope this helps,
> Michael

oh yes, it's the "break" I forgot... Thank you, it helps me a lot!




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