Python理解与字典隐蔽不使用理解

Question

此源代码在python 中使用理解

hidden_layer = [{'weights':[random() for i in range (3)]} for i in range(1)]

所以我转换为不使用理解表达式，下面是我转换的源代码

import random 
for i in range(3):
for i in range (1):
hidden_layer = {'weights':random()}

但它在字典中只有一个值

我错过了什么？

Answer 1

from random import random
hidden_layer = []
for i in range(1):
weights = []
for j in range(3):
weights.append(random())
hidden_layer.append({'weights': weights})
print(hidden_layer)

输出：

[{'weights': [0.4708113985288851, 0.861100368435909, 0.7732951090293462]}]

Answer 2

我打开列表压缩的一个个人技巧是始终从最外层开始到内层。

用代码演示：

from random import random
hidden_layer = [{'weights':[random() for i in range (3)]} for i in range(1)]
print(hidden_layer)

正如我们所看到的，最外层是一个列表，我会首先初始化一个该列表，然后逐个推动内部理解

from random import random
hidden_layer = []
for i in range(1):
hidden_layer.append({'weights':[random() for i in range (3)]})
print(hidden_layer)

接下来，我们可以进一步打开内层，技巧是在嵌套理解中将索引识别为同一索引不会中断循环，但在正常的for-loop中肯定会中断

from random import random
hidden_layer = []
for i in range(1):
new_dict = {'weights': []}
for j in range (3): # remember to use different index variable
new_dict['weights'].append(random())
hidden_layer.append(new_dict)
print(hidden_layer)

我们开始了！