Question

我有一条轨迹，其中包含一些3D数据的几帧，如下所示（为了重现我的问题，我正在发布整个帧）：

data1=   [[ 89.29,  57.35 , 107.81999 ],
[ 91.37   ,   60.39 , 109.119995],
[ 89.130005 , 61.289997 , 112.12999 ],
[ 89.45    ,  57.729996 ,113.509995],
[ 93.149994 , 58.199997 ,114.20999 ],
[ 92.03999  , 61.21     ,116.44    ],
[ 89.48999  , 58.71     ,117.979996],
[ 92.42     , 56.399998 ,118.84    ],
[ 94.25     , 59.14     ,120.73999 ],
[ 91.44     , 59.62     ,123.28999 ],
[ 91.25     , 55.739998 ,124.      ],
[ 94.95     , 55.829998 ,124.67    ],
[ 94.36     , 58.699997 ,127.079994],
[ 91.5      , 57.05     ,129.05    ],
[ 93.16     , 53.57     ,129.28    ],
[ 96.36     , 55.35     ,130.5     ],
[ 94.15     , 57.53     ,133.      ],
[ 92.24     , 54.42     ,134.18    ],
[ 95.479996 , 52.35     ,134.88    ],
[ 96.81     , 55.429996 ,136.77    ],
[ 93.509995 , 55.73     ,138.76    ],
[ 94.06999  , 51.98     ,139.5     ],
[ 97.63     , 52.929996 ,140.72    ],
[ 96.08     , 55.72     ,142.92    ],
[ 93.63999  , 53.269997 ,144.65    ],
[ 96.149994 , 50.45     ,144.79999 ],
[ 99.10999  , 52.1      ,146.4     ],
[ 97.369995 , 54.16     ,149.      ],
[ 94.2      , 55.65     ,150.56    ]]

我想让行适合我的数据。检查上一个问题（1）的答案后，我借用/编写了以下代码：

def Range(data):
   x=[]
   y=[]
   for i in range(0, len(data.T)):
       x.append(np.mean(data.T[i])- np.std(data.T[i]))
       y.append(np.mean(data.T[i]) + np.std(data.T[i]))
   normx,normy=np.linalg.norm(x),np.linalg.norm(y)
   minimum=normx-(normx+normy)/2
   maximum=normy-(normx+normy)/2
   return(minimum,maximum)

def axis(data):
  minimum,maximum=Range(data)
  datamean = data.mean(axis=0)
  uu, dd, vv = np.linalg.svd(data - datamean)
  linepts = vv[0] * np.mgrid[minimum:maximum:2j][:, np.newaxis]
  linepts += datamean
  return(linepts)

vv[0]的值原来是[ 0.15970461,-0.17264067,0.9719515 ]，非常适合我进一步使用。但是，对于跟踪数据：

 data2=  [ 90.993996 , 62.075    ,108.487   ],
   [ 88.036995 , 59.8525   ,109.3855  ],
   [ 90.5945   , 57.614998 ,111.061005],
   [ 92.17     , 60.6205   ,112.681496],
   [ 88.934006 , 61.622    ,114.4255  ],
   [ 88.077995 , 57.929    ,115.34    ],
   [ 91.642    , 57.3      ,116.81049 ],
   [ 91.431496 , 60.4655   ,118.813   ],
   [ 88.269    , 59.22     ,120.685   ],
   [ 89.883995 , 55.7975   ,121.2585  ],
   [ 93.115    , 57.497    ,122.68849 ],
   [ 91.090004 , 59.724    ,125.11    ],
   [ 89.355    , 56.712498 ,126.7305  ],
   [ 92.6985   , 54.758    ,126.52    ],
   [ 94.4685   , 57.383    ,128.4515  ],
   [ 91.9065   , 57.297997 ,131.2145  ],
   [ 91.99349  , 53.578995 ,131.1695  ],
   [ 95.752495 , 53.376   , 132.0325  ],
   [ 95.24799  , 55.990997, 134.80699 ],
   [ 92.29199  , 54.0455  , 136.277   ],
   [ 94.5055   , 50.9205  , 136.68399 ],
   [ 97.5085   , 52.947   , 137.85399 ],
   [ 95.353    , 54.6695  , 140.651   ],
   [ 94.194496 , 51.2645  , 141.4345  ],
   [ 97.6015   , 49.722   , 141.7245  ],
   [ 99.26149  , 52.813496 ,143.35449 ],
   [ 96.79849  , 53.233498, 146.2645  ],
   [ 96.237    , 49.554  ,  146.97299 ]]

vv[0]的值原来是[-0.18894662 , 0.24432637,-0.9511066 ]，与我的预期结果相反。我不确定为什么会产生这样的结果。谁能帮我解决这个问题？

1 Fitting a line in 3D

Answer 1

第一个主要成分是方向。将其视为无限长的线。翻转此线并不重要，方向保持不变。

换句话说：该算法不考虑对象已移动到的点的顺序。因此对于该算法，通过点向后飞的对象与向前飞一样有效。

3D数据的单值分解（SVD）问题（Python）

1 个答案: