第一个大于 x 的 Python 列表索引？

Question

What would be the most Pythonic 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.

Answer 1

next(x[0] for x in enumerate(L) if x[1] > 0.7)

Answer 2

如果 list 已排序，则bisect.bisect_left(alist, value)对于大列表比next(i for i, x in enumerate(alist) if x >= value)更快。

Answer 3

>>> alist= [0.5, 0.3, 0.9, 0.8]
>>> [ n for n,i in enumerate(alist) if i>0.7 ][0]
2

Answer 4

filter(lambda x: x>.7, seq)[0]

Answer 5

for index, elem in enumerate(elements):
    if elem > reference:
        return index
raise ValueError("Nothing Found")

Answer 6

1) NUMPY ARGWHERE, general lists

If you are happy to use numpy (imported as np here), then the following will work on general lists (sorted or unsorted):

np.argwhere(np.array(searchlist)>x)[0]

or if you need the answer as a integer index:

np.argwhere(np.array(searchlist)>x)[0][0]

2) NUMPY SEARCHSORTED, sorted lists (very efficient for searching lists within a list)

However, if your search list [l1,l2,...] is sorted, it is much cleaner and nicer to use the function np.searchsorted :

np.searchsorted(searchlist,x)

The nice thing about using this function is that as well as searching for a single value x within the search list [l1,l2,...], you can also search for a list of values [x1,x2,x3...xn] within your search list (ie x can be a list too, and it is extremely efficient relative to a list comprehension in this case ).

Answer 7

另一个：

map(lambda x: x>.7, seq).index(True)

Answer 8

I know there are already plenty of answers, but I sometimes I feel that the word pythonic is translated into 'one-liner'.

When I think a better definition is closer to this answer :

"Exploiting the features of the Python language to produce code that is clear, concise and maintainable."

While some of the above answers are concise, I do not find them to be clear and it would take a newbie programmer a while to understand, therefore not making them extremely maintainable for a team built of many skill levels.

l = [0.5, 0.3, 0.9, 0.8]

def f(l, x):
    for i in l:
        if i >x: break
    return l.index(i)


f(l,.7)

or

l = [0.5, 0.3, 0.9, 0.8]

def f(l, x):
    for i in l:
        if i >x: return l.index(i)



f(l,.7)

I think the above is easily understood by a newbie and is still concise enough to be accepted by any veteran python programmer.

I think writing dumb code is a positive.

Answer 9

I had similar problem when my list was very long. Comprehension or filter-based solutions would go through the whole list. Instead itertools.takewhile will break the loop once the condition is false for the first time:

from itertools import takewhile

def f(l, b): return len([x for x in takewhile(lambda x: x[1] <= b, enumerate(l))])

l = [0.5, 0.3, 0.9, 0.8]
f(l, 0.7)

Answer 10

>>> 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

Answer 11

You could also do this using numpy :

import numpy as np

list(np.array(SearchList) > x).index(True)

Answer 12

Try this one:

def Renumerate(l):
    return [(len(l) - x, y) for x,y in enumerate(l)]

example code:

Renumerate(range(10))

output:

(10, 0)
(9, 1)
(8, 2)
(7, 3)
(6, 4)
(5, 5)
(4, 6)
(3, 7)
(2, 8)
(1, 9)