Finding the Min for positive and negative in python 3.3 list

88888 Dihedral dihedral88888 at googlemail.com
Thu Mar 14 12:45:30 CET 2013


Wolfgang Maier於 2013年3月13日星期三UTC+8下午6時43分38秒寫道:
> Steven D'Aprano <steve+comp.lang.python <at> pearwood.info> writes:
> 
> 
> 
> > 
> 
> > On Tue, 12 Mar 2013 17:03:08 +0000, Norah Jones wrote:
> 
> > 
> 
> > > For example:
> 
> > > a=[-15,-30,-10,1,3,5]
> 
> > > 
> 
> > > I want to find a negative and a positive minimum.
> 
> > > 
> 
> > > example: negative
> 
> > > print(min(a)) = -30
> 
> > >  
> 
> > > positive
> 
> > > print(min(a)) = 1
> 
> > 
> 
> > Thank you for providing examples, but they don't really cover all the 
> 
> > possibilities. For example, if you had:
> 
> > 
> 
> > a = [-1, -2, -3, 100, 200, 300]
> 
> > 
> 
> > I can see that you consider -3 to be the "negative minimum". Do you 
> 
> > consider the "positive minimum" to be 100, or 1?
> 
> > 
> 
> > If you expect it to be 100, then the solution is:
> 
> > 
> 
> >     min([item for item in a if item > 0])
> 
> > 
> 
> > If you expect it to be 1, then the solution is:
> 
> > 
> 
> >     min([abs(item) for item in a])
> 
> > 
> 
> > which could also be written as:
> 
> > 
> 
> >     min(map(abs, a))
> 
> > 
> 
> > A third alternative is in Python 3.3:
> 
> > 
> 
> >     min(a, key=abs)
> 
> > 
> 
> > which will return -1.
> 
> > 
> 
> 
> 
> thinking again about the question, then the min() solutions suggested so far
> 
> certainly do the job and they are easy to understand.
> 
> However, if you need to run the function repeatedly on larger lists, using min()
> 
> is suboptimal because its performance is an O(n) one.
> 
> It's faster, though less intuitive, to sort your list first, then use bisect on
> 
> it to find the zero position in it. Two manipulations running at O(log(n)).
> 
> 
> 
> compare these two functions:
> 
> 
> 
> def with_min(x):
> 
>     return (min(n for n in a if n<0), min(n for n in a if n>=0))
> 
> 
> 
> def with_bisect(x):
> 
>     b=sorted(x)
> 
>     return (b[0] if b[0]<0 else None, b[bisect.bisect_left(b,0)])
> 
> 
> 
> then either time them for small lists or try:
> 
> 
> 
> a=range(-10000000,10000000)
> 
> with_min(a)
> 
> with_bisect(a)
> 
> 
> 
> of course, the disadvantage is that you create a huge sorted list in memory and
> 
> that it's less readable.
> 
> 
> 
> Best,
> 
> Wolfgang

Sorting numbers of such range M in a list of length N by radix sort 
is faster but requires more memory.



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