Use:

from operator import itemgetter
from heapq import nlargest
result = nlargest(N, enumerate(your_list), itemgetter(1))

现在，结果列表将包含N个元组(index, value)，其中value是最大的。

使用argpartition的向量化2D实现:

k = 3
probas = np.array([
    [.6, .1, .15, .15],
    [.1, .6, .15, .15],
    [.3, .1, .6, 0],
])

k_indices = np.argpartition(-probas, k-1, axis=-1)[:, :k]

# adjust indices to apply in flat array
adjuster = np.arange(probas.shape[0]) * probas.shape[1]
adjuster = np.broadcast_to(adjuster[:, None], k_indices.shape)
k_indices_flat = k_indices + adjuster

k_values = probas.flatten()[k_indices_flat]

# k_indices:
# array([[0, 2, 3],
#        [1, 2, 3],
#        [2, 0, 1]])
# k_values:
# array([[0.6 , 0.15, 0.15],
#        [0.6 , 0.15, 0.15],
#       [0.6 , 0.3 , 0.1 ]])

如果你不关心第k大元素的顺序，你可以使用argpartition，它应该比通过argsort进行完整排序执行得更好。

K = 4 # We want the indices of the four largest values
a = np.array([0, 8, 0, 4, 5, 8, 8, 0, 4, 2])
np.argpartition(a,-K)[-K:]
array([4, 1, 5, 6])

这个问题值得肯定。

我运行了一些测试，看起来随着数组的大小和K值的增加，argpartition的性能优于argsort。

Use:

>>> import heapq
>>> import numpy
>>> a = numpy.array([1, 3, 2, 4, 5])
>>> heapq.nlargest(3, range(len(a)), a.take)
[4, 3, 1]

对于常规的Python列表:

>>> a = [1, 3, 2, 4, 5]
>>> heapq.nlargest(3, range(len(a)), a.__getitem__)
[4, 3, 1]

如果使用Python 2，请使用xrange而不是range。

来源:堆队列算法

您可以简单地使用字典来查找numpy数组中的前k个值和下标。 例如，如果你想找到前2个最大值和索引

import numpy as np
nums = np.array([0.2, 0.3, 0.25, 0.15, 0.1])


def TopK(x, k):
    a = dict([(i, j) for i, j in enumerate(x)])
    sorted_a = dict(sorted(a.items(), key = lambda kv:kv[1], reverse=True))
    indices = list(sorted_a.keys())[:k]
    values = list(sorted_a.values())[:k]
    return (indices, values)

print(f"Indices: {TopK(nums, k = 2)[0]}")
print(f"Values: {TopK(nums, k = 2)[1]}")


Indices: [1, 2]
Values: [0.3, 0.25]

如果你碰巧在使用一个多维数组，那么你需要平展和解开索引:

def largest_indices(ary, n):
    """Returns the n largest indices from a numpy array."""
    flat = ary.flatten()
    indices = np.argpartition(flat, -n)[-n:]
    indices = indices[np.argsort(-flat[indices])]
    return np.unravel_index(indices, ary.shape)

例如:

>>> xs = np.sin(np.arange(9)).reshape((3, 3))
>>> xs
array([[ 0.        ,  0.84147098,  0.90929743],
       [ 0.14112001, -0.7568025 , -0.95892427],
       [-0.2794155 ,  0.6569866 ,  0.98935825]])
>>> largest_indices(xs, 3)
(array([2, 0, 0]), array([2, 2, 1]))
>>> xs[largest_indices(xs, 3)]
array([ 0.98935825,  0.90929743,  0.84147098])