Question

我有2个np数组，其中包含区间[0,1]中的值。我想创建第三个数组，包含最多的＃34;自信＆＃34;值，意味着采用元素，数组中的数字更接近1或0.请考虑以下示例：

[0.7,0.12,1,0.5]
[0.1,0.99,0.001,0.49]

所以我构造的数组将是：

[0.1,0.99,1,0.49]

Answer 1

你可以试试这个：

c=np.array([a[i] if min(1-a[i],a[i])<min(1-b[i],b[i]) else b[i] for i in range(len(a))])

结果是：

array([ 0.1 ,  0.99,  1.  ,  0.49])

Answer 2

import numpy as np
A = np.array([0.7,0.12,1,0.5])
B = np.array([0.1,0.99,0.001,0.49])

maxi = np.maximum(A,B)
mini = np.minimum(A,B)
# Find where the maximum is closer to 1 than the minimum is to 0
idx = 1-maxi < mini

maxi*idx + mini*~idx

返回

array([ 0.1 ,  0.99,  1.  ,  0.49])

Answer 3

说明你的信心的另一种方式＆＃34;衡量是要问两个数字中的哪一个远离0.5。也就是说，两个数字x中的哪一个产生最大的abs(0.5 - x)。以下解决方案使用原始数组作为列构建2D数组c。然后我们构造并应用基于abs(0.5 - c)的布尔掩码：

import numpy as np

a = np.array([0.7,0.12,1,0.5])
b = np.array([0.1,0.99,0.001,0.49])

# Combine
c = np.concatenate((a, b)).reshape((2, len(a))).T
# Create mask
b_or_a = np.asarray(np.argmax(np.abs((0.5 - c)), axis=1), dtype=bool)
mask = np.zeros(c.shape, dtype=bool)
mask[:, 0] = ~b_or_a
mask[:, 1] =  b_or_a
# Applt mask
d = c[mask]
print(d)  # [ 0.1   0.99  1.    0.49]

自信地聚合2个NumPy阵列

3 个答案: