不确定for循环迭代？(Unsure about for-loop iteration?)

 通过使用产品与总和触发器。 身份 ，你可以减少三角形的数量。 函数调用。 在下面， func1和func2计算相同的值，但func2减少了触发的调用。 功能。 
 
import numpy as np

def func1(a, b, c, d):
    A = np.cos(a) * np.cos(b) - np.sin(a) * np.sin(b) * np.sin(c - d)
    return A

def func2(a, b, c, d):
    s = np.sin(c - d)
    A = 0.5*((1 - s)*np.cos(a - b) + (1 + s)*np.cos(a + b))
    return A
 
 这是N = 5000的时序比较： 
 
In [48]: %timeit func1(a, b, c, d)
1000 loops, best of 3: 374 µs per loop

In [49]: %timeit func2(a, b, c, d)
1000 loops, best of 3: 241 µs per loop

By using the product-to-sum trig. identities, you can reduce the number of trig. function calls. In the following, func1 and func2 compute the same value, but func2 makes fewer calls to trig. functions.
 
import numpy as np

def func1(a, b, c, d):
    A = np.cos(a) * np.cos(b) - np.sin(a) * np.sin(b) * np.sin(c - d)
    return A

def func2(a, b, c, d):
    s = np.sin(c - d)
    A = 0.5*((1 - s)*np.cos(a - b) + (1 + s)*np.cos(a + b))
    return A
 
Here's a timing comparison with N = 5000:
 
In [48]: %timeit func1(a, b, c, d)
1000 loops, best of 3: 374 µs per loop

In [49]: %timeit func2(a, b, c, d)
1000 loops, best of 3: 241 µs per loop