What would be the most pythonesque way to find the first index in a list that is greater than x?
For example, with
list = [0.5, 0.3, 0.9, 0.8]
The function
f(list, 0.7)
would return
2.
views:
188answers:
7
A:
>>> f=lambda seq, m: [ii for ii in xrange(0, len(seq)) if seq[ii] > m][0]
>>> f([.5, .3, .9, .8], 0.7)
2
flybywire
2010-02-10 13:03:56
That looks pretty slick. But theoretically it will traverse the whole list and then return the first result (greater than x), right? Is there any way to make one that stops straight after finding the first result?
c00kiemonster
2010-02-10 13:08:52
yes, it traverses all the list
flybywire
2010-02-10 13:12:01
what's wrong with traversing whole list? if the first value greater than 0.7 is near the end of the list, it doesn't make a difference.
ghostdog74
2010-02-10 13:12:08
True. But in this particular case the lists I intend to use the function on are pretty long, so I'd prefer it to quit traversing as soon as a match is found...
c00kiemonster
2010-02-10 13:15:03
whether its long or not, if the first value is the last second item of the list, you will still have to traverse the whole list to get there!
ghostdog74
2010-02-10 13:18:23
@ghostdog74: Yes, but this is not a reason to want all cases to be worst cases.
UncleBens
2010-02-10 16:26:16
+1: Although I would prefer to avoid the magic numbers: next(idx for idx, value in enumerate(L) if value > 0.7)
truppo
2010-02-10 13:12:28
+1 for simplicity and `next()`, but maybe this for readability: `next(i for i,v in enumerate(L) if v > 0.7)`
Will Hardy
2010-02-10 13:14:26
While this is nice looking, the case where there's no result will raise a confusing StopIteration.
Virgil Dupras
2010-02-10 13:15:15
Correct me if I'm wrong here, but the next() method will basically just keep on asking for the next item until it has found a match, right? So all I got to do is to catch a possible StopIteration as Virgil points out.
c00kiemonster
2010-02-10 13:18:38
@c00kiemonster no, next() just gets the first item of the iterator. But the iterator is filtered, so the first item is the right item.
Virgil Dupras
2010-02-10 13:19:42
@flybywire: `next()` is in 2.6+. Call the `next()` method of the genex in earlier versions.
Ignacio Vazquez-Abrams
2010-02-10 13:38:18
To avoid the StopIteration problem and return -1 instead, you can use ([i for i,v in enumerate(L) if v > 0.7] + [-1])[0]
Wim
2010-02-10 20:16:07
@Wim: But then you regress back to evaluating the entire sequence. Use `itertools.chain()` instead of adding lists like this.
Ignacio Vazquez-Abrams
2010-02-10 20:18:26
Oh right, I didn't get the quick exit part yet... `next(itertools.chain(iter(i for i,v in enumerate(L) if v > 0.7), [-1]))` then?
Wim
2010-02-10 23:29:38
+3
A:
for index, elem in enumerate(elements):
if elem > reference:
return index
raise ValueError("Nothing Found")
Virgil Dupras
2010-02-10 13:09:28
+4
A:
>>> alist= [0.5, 0.3, 0.9, 0.8]
>>> [ n for n,i in enumerate(alist) if i>0.7 ][0]
2
ghostdog74
2010-02-10 13:09:31
@mshsayem: Problem is ill-defined for this case. Failure may be the right thing to do.
S.Lott
2010-02-10 14:43:09
-1: While technically correct, dont use filter where a list comprehension is both more readable and more performant
truppo
2010-02-10 13:18:20
A:
if list is sorted then bisect_left(alist, value) is faster for a large list than next(i for i, x in enumerate(alist) if x >= value).
J.F. Sebastian
2010-02-10 20:13:18