在Python中查找数组中最低5个数字的索引

Question

How to find out which indices belong to the lowest x (say, 5) numbers of an array?

[10.18398473, 9.95722384, 9.41220631, 9.42846614, 9.7300549 , 9.69949144, 9.86997862, 10.28299122, 9.97274071, 10.08966867, 9.7]

Also, how to directly find the sorted (from low to high) lowest x numbers?

Answer 1

The existing answers are nice, but here's the solution if you're using numpy :

mylist = np.array([10.18398473, 9.95722384, 9.41220631, 9.42846614, 9.7300549 , 9.69949144, 9.86997862, 10.28299122, 9.97274071, 10.08966867, 9.7])
x = 5
lowestx = np.argsort(mylist)[:x]
#array([ 2,  3,  5, 10,  4])

Answer 2

You could do something like this:

>>> l = [5, 1, 2, 4, 6]
>>> sorted(range(len(l)), key=lambda i: l[i])
[1, 2, 3, 0, 4]

Answer 3

mylist = [10.18398473, 9.95722384, 9.41220631, 9.42846614, 9.7300549 , 9.69949144, 9.86997862, 10.28299122, 9.97274071, 10.08966867, 9.7]

# lowest 5
lowest = sorted(mylist)[:5]

# indices of lowest 5
lowest_ind = [i for i, v in enumerate(mylist) if v in lowest]

# 5 indices of lowest 5
import operator
lowest_5ind = [i for i, v in sorted(enumerate(mylist), key=operator.itemgetter(1))[:5]]

Answer 4

[a.index(b) for b in sorted(a)[:5]]
sorted(a)[.x]