在python中排序二维数组

Question

entropies_with_samples = []
for i in range(0,2948):
    entr = entropy(predictProbas[i])
    mixed = [proba_X_train[i],entr]
    entropies_with_samples.append(mixed)

a = np.array(entropies_with_samples)
a.flatten("F")
print(list(chain.from_iterable(entropies_with_samples)))
selection = (sorted(mixed, key=itemgetter(2),reverse= True))
print(selection)

示例：

input = [([0.2,0.10]),0.69, ([0.3,0.67]),0.70, ([0.5,0.68]),0.70, ([0.3,0.67]),0.65]

我正试图在第三个位置对这样一个数组进行排序。你知道吗

output = [([0.3,0.67]),0.70,  ([0.5,0.68]),0.70, ([0.2,0.10]),0.69, ([0.3,0.67]),0.65 ]

Answer 1

另一种方法是使用lambda排序键进行zip和排序：

首先，使用zip将“第三个位置”与第一个和第二个数字放在一个元组中：

output = list(zip(output[::2], output[1::2]))
#[([0.3, 0.67], 0.7), ([0.5, 0.68], 0.7), ([0.2, 0.1], 0.69), ([0.3, 0.67], 0.65)]

然后排序，使用第三个数字（在元组中，它位于位置2）作为排序键：

output.sort(key = lambda x: x[1])
#[([0.3, 0.67], 0.65), ([0.2, 0.1], 0.69), ([0.3, 0.67], 0.7), ([0.5, 0.68], 0.7)]

Answer 2

第一步可以是创建嵌套列表，将每个2元素添加到新的子列表中：

from itertools import chain
from operator import itemgetter

i = [([0.2,0.10]),0.69, ([0.3,0.67]),0.70, ([0.5,0.68]),0.70, ([0.3,0.67]),0.65]

l = [i[x:x+2] for x in range(0, len(i),2)]
# [[[0.2, 0.1], 0.69], [[0.3, 0.67], 0.7], [[0.5, 0.68], 0.7], [[0.3, 0.67], 0.65]]

然后使用operator.itemgetter按每个子列表中的第二个元素对嵌套列表进行排序，并使用^{}将结果展平：

list(chain(*sorted(l, key = itemgetter(1), reverse=True)))

[[0.3, 0.67], 0.7, [0.5, 0.68], 0.7, [0.2, 0.1], 0.69, [0.3, 0.67], 0.65]