假设我有以下的meshgrids ```omega```, ```phiy```, 和 ```phix```。
```python
import numpy as np
phi = np.linspace(-0.75, 0.75, num=4)
omega = np.linspace(-100, 100, num=5)
omega, phiy, phix = np.meshgrid(omega, phi, phi, sparse=True, indexing='ij')
rand = np.random.rand(5, 4)

def f(j, k):
    return np.argmin(np.abs(phi[j]**2 + np.sin(phi[k]) - phi))
C = np.empty((5, 4, 4))
for i in range(5):
    for j in range(4):
        for k in range(4):
            C[i, j, k] = rand[i, f(j, k)]

<details>
<summary>英文:</summary>
Suppose I have the following meshgrids ```omega```, ```phiy```, and ```phix```.

Then I want to get the following ```C```

Are there any more compact and better ways to get it, using some tools for meshgrids?
</details>
# 答案1
**得分**: 1
你可以通过消除 `for` 循环并仅使用向量化操作来更快地计算你的 `f` 函数：
```python
# f_mat[i,j] contains f(i,j)
f_mat = np.abs((phi[None].T**2 + np.sin(phi))[...,None] - phi).argmin(-1)
C = rand[:,f_mat]