我想得到一个列表中最小N个元素的索引如果我能把输出放到另一个列表上那就太好了。
例如：
[1, 1, 10, 5, 3, 5]
output = [0, 1]
[10, 5, 12, 5, 0, 10]
output = [4]
[9, 2, 8, 2, 3, 4, 2]
output = [1, 3, 6]
[10, 10, 10, 10, 10, 10]
output = [0, 1, 2, 3, 4, 5]
我知道
.index返回列表中最小值的第一个索引，但我不知道当最小值出现多次时如何返回所有索引。 最佳答案
>>> L = [9, 2, 8, 2, 3, 4, 2]
>>> minL = min(L)
>>> [i for i, x in enumerate(L) if x == minL]
[1, 3, 6]
目前，在迭代过程中，其他解决方案将调用
min，导致了一个很差且不必要的O（n ^ 2）复杂性。编辑KASRA：天真的解决方案的N ^ 2复杂度的证据：
>>> L1000 = [randint(0, 100) for _ in xrange(1000)]
>>> L2000 = [randint(0, 100) for _ in xrange(2000)]
>>> L3000 = [randint(0, 100) for _ in xrange(3000)]
>>> L4000 = [randint(0, 100) for _ in xrange(4000)]
>>> L5000 = [randint(0, 100) for _ in xrange(5000)]
>>> timeit [i for i, x in enumerate(L1000) if x == min(L1000)]
10 loops, best of 3: 18.8 ms per loop
>>> timeit [i for i, x in enumerate(L2000) if x == min(L2000)]
10 loops, best of 3: 73.6 ms per loop
>>> timeit [i for i, x in enumerate(L3000) if x == min(L3000)]
1 loops, best of 3: 166 ms per loop
>>> timeit [i for i, x in enumerate(L4000) if x == min(L4000)]
1 loops, best of 3: 294 ms per loop
>>> timeit [i for i, x in enumerate(L5000) if x == min(L5000)]
1 loops, best of 3: 457 ms per loop