独立循环的列表理解

Question

我试图将列表理解用于嵌套的循环太少。这是我没有列表理解的解决方案：

import numpy as np
n_steps = 20
x_steps = [int(i) for i in np.linspace(10, 60, n_steps)]
y_steps = [int(i) for i in np.linspace(25, 150, n_steps)]
steps = [(x_steps[i], y_steps[i]) for i in range(len(x_steps))]

如您所见，我想要steps = [(10, 25), (13, 31), ...]

我的问题是，有一种用列表理解或类似方法在一行中做到这一点的优雅而pythonic的方法吗？在我心中，我有这样的东西：

steps = [(int(x), int(j)) for x in np.linspace(10, 60, n_steps) and j in np.linspace(25, 150, n_steps)]

Answer 1

使用zip

import numpy as np
n_steps = 20
steps = [(int(i),int(j)) for i,j in zip(np.linspace(10, 60, n_steps),np.linspace(25, 150, n_steps))]

Answer 2

邮政编码是您的答案：）

list(zip(x_steps, y_steps))