重写numpy中的循环以加快执行速度

Numpy的ufunc都具有external方法，以在两个数组上“交叉”执行操作。因此，为了避免大多数中间计算并尽可能向量化：

def f(M, a):
    return 2 * M / np.add.outer(a, a)

• 回答旧版本的问题（左，因为它仍然有用）：

对于这样的事情，我发现最好总是分步进行，并尝试找到正确的einsum表达式。

# the definition given in the original question,
# before the z / (y + x) update
def f0():
    d = np.empty((3,3))
    for i in range(len(a)):
        x = a[i]
        for j in range(len(a)):
            y = a[j]
            z = M[i][j]
            d[i][j] = 2 * x/(y+z)
    return d

# rewrite things inlined
def f1():
    d = np.empty((3,3))
    for i in range(len(a)):
        for j in range(len(a)):
            d[i, j] = 2 * a[i]/(a[j] + M[i, j])
    return d

# factor out broadcasting
def f2():
    d = np.empty((3,3))
    for i in range(len(a)):
        m = a + M[i, :]
        for j in range(len(a)):
            d[i,j] = 2 * a[i]/m[j]
    return d

# more broadcasting
def f3():
    d = np.empty((3,3))
    m = a + M
    for i in range(len(a)):
        for j in range(len(a)):
            d[i,j] = 2 * a[i]/m[i,j]
    return d

# now turn loops into einsums
def f4():
    d = np.empty((3,3))
    m = 1/(a + M)
    d[:,:] = 2 * np.einsum('i,ij->ij', a, m)
    return d

# collect everything
def f5():
    return np.einsum('i,ij->ij', a, 2 / (a + M))

回答来源：stackoverflow