小编Dum*_*Guy的帖子

我有一个numpy数组

[[  5.80084178e-05   1.20779787e-02  -2.65970238e-02]
 [ -1.36810406e-02   6.85722519e-02  -2.60280724e-01]
 [  4.21996519e-01  -1.43644036e-01   2.12904690e-01]
 [  3.03098198e-02   1.50170659e-02  -1.09683402e-01]
 [ -1.50776089e-03   7.22369575e-03  -3.71181228e-02]
 [ -3.04448275e-01  -3.66987035e-01   1.44618682e-01]
 [ -1.46744916e-01   3.47112167e-01   3.09550267e-01]
 [  1.16567762e-03   1.72858807e-02  -9.39297514e-02]
 [  1.25896836e-04   1.61310167e-02  -6.00253128e-02]
 [  1.65062798e-02   1.96933143e-02  -4.26540031e-02]
 [ -3.78020965e-03   7.51770012e-03  -3.67852984e-02]]

我想从这些中选择任意两个随机行,所以输出将是 -

[[ -1.36810406e-02   6.85722519e-02  -2.60280724e-01]
[  1.16567762e-03   1.72858807e-02  -9.39297514e-02]]

我有3D阵列

 A = [[x1 y1 z1]
      [x2 y2 z2]
      [x3 y3 z3]]

我必须找到每个点之间的欧几里德距离,这样我才能得到只有3个距离的输出(row0,row1),(row1,row2)和(row0,row2).

我有一些代码

dist = scipy.spatial.distance.cdist(A,A, 'euclidean')

但它会以矩阵形式给出距离

dist= [[0  a   b]
       [a  0   c]
       [b  c   0]]

我想要结果[a b c].