index of min element of sequence
Neal Becker
sequence? (I'm fairly sure sorting the sequence is not the fastest
approach)
Peter Otten
>>> items = "defbhkamnz"
>>> min(xrange(len(items)), key=items.__getitem__)
6
>>> items[6]
'a'
Peter
Paul Rubin
Python 2.5 (untested):
from operator import itemgetter
minindex = min(enumerate(seq), key=itemgetter(1))[1]
Roger Miller
> Neal Becker wrote:
> > What's a good/fast way to find the index of the minimum element of a
> > sequence?
...
> >>> min(xrange(len(items)), key=items.__getitem__)
...
Or just
items.index(min(items))
I found this to be significantly faster in a simple test (searching a
list of 1000 ints with the minimum in the middle), despite the fact
that it requires two passes. I'm sure that one could find cased where
Peter's approach is faster, so you if you are concerned about speed
you should measure with your own data.
John Machin
Bzzzt!
>>> seq = [1000, 9, 8, 7, 2000, 3000]
>>> from operator import itemgetter
>>> minindex = min(enumerate(seq), key=itemgetter(1))[1]
>>> minindex
7
>>> min(enumerate(seq), key=itemgetter(1))
(3, 7)
s/[1]/[0]/ or more generally:
minindex, minvalue = min(enumerate(seq), key=itemgetter(1))
Peter Otten
I suppose those cases are rare (slow equality check), and even then I
might prefer your solution because it's so much clearer.
Peter
Paul Rubin
> s/[1]/[0]/ or more generally:
Oops, got two spellings confused. Originally was going to use
from itertools import count, izip
min(izip(seq, count()))[1]
but did it with enumerate instead. I don't know which is actually
faster.
> minindex, minvalue = min(enumerate(seq), key=itemgetter(1))
Cool, I like this best of all. Or alternatively,
minindex, minvalue = min(izip(seq, count()))
John Machin
Bzzzt #2!
>>> from itertools import count, izip
>>> min(izip(seq, count()))
(7, 3)
Paul Rubin
> >>> from itertools import count, izip
> >>> min(izip(seq, count()))
> (7, 3)
Serves me right for cutting and pasting.
minvalue, minindex = min(izip(seq, count()))