从jSon fi中提取矩阵

2024-05-16 06:42:31 发布

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我使用fSpy软件:开源静态图像相机匹配 要将结果与Rhino 3d一起使用,我将导出到包含摄像机位置矩阵的jSon文件 我需要用Python将提取到一个数字列表中 谢谢你的帮助

enter image description here

由此:

  "cameraTransform": {
    "rows": [
      [
        -0.7469096503244566,
        -0.0048362499036055774,
        -0.6649079522302825,
        -6.756347957309588
      ],
      [
        -0.6649255299159078,
        0.005255128514812499,
        0.7468911723205343,
        7.333944516331069
      ],
      [
        -0.00011797562067967932,
        0.9999744968303753,
        -0.007140852231397498,
        -4.200072574252649
      ],
      [
        0,
        0,
        0,
        1
      ]
    ]
  }

收件人:

0   -0.7469096503244566
1   -0.0048362499036055774
2   -0.6649079522302825
3   -6.756347957309588
4   -0.6649255299159078
5    0.005255128514812499
6   0.7468911723205343
7   7.333944516331069
8   -0.00011797562067967932
9   0.9999744968303753
10   -0.007140852231397498
11   -4.200072574252649
12   0
13   0
14   0
15   1

Tags: 文件图像json列表软件静态矩阵开源
1条回答
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1楼 · 发布于 2024-05-16 06:42:31

将json转换为python格式后,可以尝试以下操作:

cameraTransform= {
        "rows": [
          [
            -0.7469096503244566,
            -0.0048362499036055774,
            -0.6649079522302825,
            -6.756347957309588
          ],
          [
            -0.6649255299159078,
            0.005255128514812499,
            0.7468911723205343,
            7.333944516331069
          ],
          [
            -0.00011797562067967932,
            0.9999744968303753,
            -0.007140852231397498,
            -4.200072574252649
          ],
          [
            0,
            0,
            0,
            1
          ]
        ]
      }

from itertools import chain
for idx, value in enumerate(chain(*cameraTransform['rows'])):
    print(idx, value)

输出:

0 -0.7469096503244566
1 -0.0048362499036055774
2 -0.6649079522302825
3 -6.756347957309588
4 -0.6649255299159078
5 0.005255128514812499
6 0.7468911723205343
7 7.333944516331069
8 -0.00011797562067967932
9 0.9999744968303753
10 -0.007140852231397498
11 -4.200072574252649
12 0
13 0
14 0
15 1

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