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numPy综合使用

numPy综合使用

作者: 空山新雨后丶 | 来源:发表于2017-10-24 16:19 被阅读0次
>>> from numpy import *
>>> rMat1 = mat(random.rand(4,4))
>>> rMat1
matrix([[ 0.64942483,  0.13354212,  0.73462188,  0.84064478],
        [ 0.98905725,  0.47277942,  0.46831185,  0.04350675],
        [ 0.1914044 ,  0.85404567,  0.52689656,  0.38911737],
        [ 0.50723587,  0.33690424,  0.96063439,  0.42176084]])
>>> rMat2 = rMat1.I
>>> rMat2
matrix([[ 0.48364538,  1.12727011, -0.43000817, -0.68354892],
        [-0.31685954,  0.26538305,  1.31309351, -0.6072805 ],
        [-0.85113497, -0.46638126, -0.46684178,  2.17528291],
        [ 1.61005598, -0.50544909,  0.53156445, -1.27640031]])
>>> rMat1*rMat2
matrix([[  1.00000000e+00,   1.11022302e-16,  -5.55111512e-17,
           0.00000000e+00],
        [  2.77555756e-17,   1.00000000e+00,  -1.04083409e-17,
           6.93889390e-18],
        [  1.11022302e-16,   2.77555756e-17,   1.00000000e+00,
          -1.11022302e-16],
        [  0.00000000e+00,   8.32667268e-17,  -5.55111512e-17,
           1.00000000e+00]])
>>> rMat1*rMat2 - eye(4)
matrix([[  0.00000000e+00,   1.11022302e-16,  -5.55111512e-17,
           0.00000000e+00],
        [  2.77555756e-17,   0.00000000e+00,  -1.04083409e-17,
           6.93889390e-18],
        [  1.11022302e-16,   2.77555756e-17,   0.00000000e+00,
          -1.11022302e-16],
        [  0.00000000e+00,   8.32667268e-17,  -5.55111512e-17,
           0.00000000e+00]])

.I 操作符实现了矩阵求逆的运算,eye(4)产生4×4的单位矩阵,相减可以得到误差。

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