在python中对二维数组进行排序

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)例子：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]我试图在第三个位置对这样的数组进行排序。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

第一步可能是创建一个嵌套列表，将每个2元素添加到一个新的子列表中：

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]]

然后按每个子列表中的第二个元素对嵌套列表进行排序operator.itemgetter，并使用itertools.chain来展平结果：

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]

Answer 2

使用 zip 和 lambda 排序键排序的另一种方法：

首先，使用 zip 将您的“第三个位置”放入带有第一个和第二个数字的元组中：

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，使用您的第三个数字（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)]