从python上的列表中获取索引[重复]

你可以使用 list-comprehension 具有 enumerate() . 感谢 @MikeScotty ,性能改进将是首先计算最小值。

代码如下:

mn = min(A)
[i for i,e in enumerate(A) if e == mn]

其中给出:

[2, 5]

哪些是 1s 在里面 A -不是 [2, 8]

minx wrapper :

>>> def minx(l):
...     print("called")
...     return min(l)
... 
>>> [i for i,e in enumerate(A) if e == minx(A)]
called
called
called
called
called
called
called
called
called
[2, 5]

和一些计时使用 timeit :

>>> timeit.timeit("[i for i,e in enumerate(A) if e == min(A)]", globals=locals())
5.9568054789997404
>>> timeit.timeit("[i for i,e in enumerate(A) if e == 1]", globals=locals())
1.397674421001284

你也可以试试这个:

A = [5,2,1,5,6,1,7,9,2]
minimum = min(A)
res = list(filter(lambda m: A[m]==minimum, range(len(A))))
print(res)

[2, 5]

因为当你使用

A.index(elem)

A=[5,2,1,5,6,1,7,9,2]
minimo = min(A)
indexArray = []
for index,value in enumerate(A):
    if value == minimo:
        indexArray.append(index)
print(indexArray)