Question

这是我的代码缩短版本：

`dist_array = ssd.cdist(test[y], training)`

打印测试[y]是

[  0.00000000e+00   1.79900000e+01   1.03800000e+01   1.22800000e+02
   1.00100000e+03   1.18400000e-01   2.77600000e-01   3.00100000e-01
   1.47100000e-01   2.41900000e-01   7.87100000e-02   1.09500000e+00
   9.05300000e-01   8.58900000e+00   1.53400000e+02   6.39900000e-03
   4.90400000e-02   5.37300000e-02   1.58700000e-02   3.00300000e-02
   6.19300000e-03   2.53800000e+01   1.73300000e+01   1.84600000e+02
   2.01900000e+03   1.62200000e-01   6.65600000e-01   7.11900000e-01
   2.65400000e-01   4.60100000e-01   1.18900000e-01]

印刷（缩短）的培训是：

[[  0.00000000e+00   1.92100000e+01   1.85700000e+01 ...,   2.09100000e-01
    3.53700000e-01   8.29400000e-02]
 [  0.00000000e+00   1.47100000e+01   2.15900000e+01 ...,   1.83400000e-01
    3.69800000e-01   1.09400000e-01]
 [  1.00000000e+00   1.30500000e+01   1.93100000e+01 ...,   1.11100000e-02
    2.43900000e-01   6.28900000e-02]
 ..., 
 [  0.00000000e+00   1.66000000e+01   2.80800000e+01 ...,   1.41800000e-01
    2.21800000e-01   7.82000000e-02]
 [  0.00000000e+00   2.06000000e+01   2.93300000e+01 ...,   2.65000000e-01
    4.08700000e-01   1.24000000e-01]
 [  1.00000000e+00   7.76000000e+00   2.45400000e+01 ...,   0.00000000e+00
    2.87100000e-01   7.03900000e-02]]

两者都有31列。有没有办法找到阵列XA和阵列XB的每一行之间的距离，并将距离输出到另一个阵列？

非常感谢！

Answer 1

<button *ngFor="let item of items | sort:suffix | filter:filterargs | merge:newItems:false:counter"></button>

Answer 2

此处讨论：http://comments.gmane.org/gmane.comp.python.scientific.user/29106

您需要使用例如二维阵列广播您的一维数组。 newaxis运算符：distance.cdist(ref_1d[numpy.newaxis, :], query_2d)。

Answer 3

from scipy.spatial import distance

p_a = np.array([4, 0])

p_b = np.array([0, 3])

p_a = p_a.reshape(1, -1)

p_b = p_b.reshape(1, -1)

distance.cdist(p_a, p_b, 'euclidean')
Out[58]: array([[ 5.]])

python cdist error ValueError：XA必须是一个二维数组

3 个答案: