经过多次计算步骤后，基于退出 numpy 数组创建新的 numpy 数组

Question

我有一个像这样的二维数组 numpy ：

s = np.array([[ 5.,  4., np.nan,  1., np.nan],
       [np.nan,  4.,  4.,  2.,  2.],
       [ 3., np.nan, np.nan,  5.,  5.],
       [np.nan,  3.,  4.,  4., np.nan]])

#now i want to create a new np array s1 base on s like this

s1= np.empty((len(s),len(s)))
for i  in range(len(s)):
    a = np.abs(s - s[i])
    a = np.nanmean(a, axis=1)
    w = 1 / (a + 0.001)
    s1[i] = w
s1
array([[1000.        ,    1.99600798,    0.33322226,    0.49975012],
       [   1.99600798, 1000.        ,    0.33322226,    0.999001  ],
       [   0.33322226,    0.33322226, 1000.        ,    0.999001  ],
       [   0.49975012,    0.999001  ,    0.999001  , 1000.        ]])

#without use for loop i write like this 

def f(x,y):
    a = np.abs(s[y]-s[x])
    a = np.nanmean(a)
    if np.isnan(a):
        return 0
    w = 1/(a+0.001) #not let 1/0
    return w

s1 = np.fromfunction(np.vectorize(f),(len(s),len(s)),dtype='int')
s1
array([[1000.        ,    1.99600798,    0.33322226,    0.49975012],
       [   1.99600798, 1000.        ,    0.33322226,    0.999001  ],
       [   0.33322226,    0.33322226, 1000.        ,    0.999001  ],
       [   0.49975012,    0.999001  ,    0.999001  , 1000.        ]])

首先我想问的是我的 np.fromfunction 对吗？ 其次，有没有其他方法可以用numpy重写这段代码而不使用for循环？

Answer 1

怎么样

>>> 1/(
        1e-3 + np.nanmean(
            np.abs(s - s[:, None,:]),
            axis=2
        )
    )
array([[1.00000000e+03, 1.99600798e+00, 3.33222259e-01, 4.99750125e-01],
       [1.99600798e+00, 1.00000000e+03, 3.33222259e-01, 9.99000999e-01],
       [3.33222259e-01, 3.33222259e-01, 1.00000000e+03, 9.99000999e-01],
       [4.99750125e-01, 9.99000999e-01, 9.99000999e-01, 1.00000000e+03]])

或者以完全 numpy 惯用的方式

>>> np.reciprocal(1e-3 + np.nanmean(np.abs(s - s[:, np.newaxis, :]), axis=2))
array([[1.00000000e+03, 1.99600798e+00, 3.33222259e-01, 4.99750125e-01],
       [1.99600798e+00, 1.00000000e+03, 3.33222259e-01, 9.99000999e-01],
       [3.33222259e-01, 3.33222259e-01, 1.00000000e+03, 9.99000999e-01],
       [4.99750125e-01, 9.99000999e-01, 9.99000999e-01, 1.00000000e+03]])

_{其中，截至 2021 年 12 月，np.newaxis 为 None。}

Answer 2

您可能正在寻找纯粹的 numpy 解决方案：

repeat_s_by_array = np.repeat(s[np.newaxis,:,:], repeats=len(s), axis=0) 
repeat_s_by_rows = np.repeat(s, repeats=len(s), axis=0).reshape(len(s),len(s),-1)
abs_vals = np.abs(repeat_s_by_array - repeat_s_by_rows)
mean = np.nanmean(abs_vals, axis=-1)
s1 = 1 / (mean + 0.001)

或者你可以做一个单行：

s1 = 1 / (np.array([np.nanmean(np.abs(s - s[i]), axis=1) for i in range(len(s))]) + 0.001)