我有一个清单:
a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50,
35, 41, 49, 37, 19, 40, 41, 31]
最大元素是55(两个元素在位置9和12)
我需要找到在哪个位置(s)的最大值是位于。请帮助。
当前回答
还有一个解决方案，只给出第一个外观，可以通过使用numpy实现:
>>> import numpy as np
>>> a_np = np.array(a)
>>> np.argmax(a_np)
9
其他回答
类似的想法与列表理解，但没有枚举
m = max(a)
[i for i in range(len(a)) if a[i] == m]
还有一个解决方案，只给出第一个外观，可以通过使用numpy实现:
>>> import numpy as np
>>> a_np = np.array(a)
>>> np.argmax(a_np)
9
>>> max(enumerate([1,2,3,32,1,5,7,9]),key=lambda x: x[1])
>>> (3, 32)
@shash在其他地方回答了这个问题
找到最大列表元素的索引的python方法是 Position = max(enumerate(a)， key=lambda x: x[1])[0]
一个通过。然而，它比@Silent_Ghost的解决方案慢，甚至比@nmichaels的解决方案更慢:
for i in s m j n; do echo $i; python -mtimeit -s"import maxelements as me" "me.maxelements_${i}(me.a)"; done
s
100000 loops, best of 3: 3.13 usec per loop
m
100000 loops, best of 3: 4.99 usec per loop
j
100000 loops, best of 3: 3.71 usec per loop
n
1000000 loops, best of 3: 1.31 usec per loop
这里是最大值和它出现的索引:
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> for i, x in enumerate(a):
... d[x].append(i)
...
>>> k = max(d.keys())
>>> print k, d[k]
55 [9, 12]
后来:为了满足@SilentGhost
>>> from itertools import takewhile
>>> import heapq
>>>
>>> def popper(heap):
... while heap:
... yield heapq.heappop(heap)
...
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> h = [(-x, i) for i, x in enumerate(a)]
>>> heapq.heapify(h)
>>>
>>> largest = heapq.heappop(h)
>>> indexes = [largest[1]] + [x[1] for x in takewhile(lambda large: large[0] == largest[0], popper(h))]
>>> print -largest[0], indexes
55 [9, 12]
