# Finding the Min for positive and negative in python 3.3 list

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