cMi*_*nor 4 python numpy row-major-order column-major-order
有像矩阵一样
ma = [[0.343, 0.351, 0.306], [0.145, 0.368, 0.487]]
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我想得到一个像这样的矢量:
[0.343, 0.145, 0.351, 0.368, 0.306, 0.487]
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试图得到它,我正在使用numpy,reshape但它不起作用.
a = np.array(ma)
>>> print a.shape
(2, 3)
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但我得到:
c = a.reshape(3, 2, order='F')
>>> print c
array([[ 0.343, 0.368],
[ 0.145, 0.306],
[ 0.351, 0.487]])
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对于任何矩阵大小,最好的方法是什么?我的意思是,例如,如果矩阵不是平方的:
[[0.404, 0.571, 0.025],
[0.076, 0.694, 0.230],
[0.606, 0.333, 0.061],
[0.595, 0.267, 0.138]]
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我想得到:
[0.404, 0.076, 0.606, 0.595, 0.571, 0.694, 0.333, 0.267, 0.025, 0.230, 0.061, 0.138]
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小智 5
你可以通过转置矩阵然后使用numpy 的 ravel 函数来得到你想要的:
mat = np.random.rand(3,2)
print np.ravel(mat.T)
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您可以使用ravel()扁平化阵列.
>>> a.T.ravel()
array([ 0.343, 0.145, 0.351, 0.368, 0.306, 0.487])
# Or specify Fortran order.
>>> a.ravel('F')
array([ 0.343, 0.145, 0.351, 0.368, 0.306, 0.487])
a = np.random.rand(4,2)
>>> a
array([[ 0.59507926, 0.25011282],
[ 0.68171766, 0.41653172],
[ 0.83888691, 0.22479481],
[ 0.04540208, 0.23490886]])
>>> a.T.ravel() # or a.ravel('F')
array([ 0.59507926, 0.68171766, 0.83888691, 0.04540208, 0.25011282,
0.41653172, 0.22479481, 0.23490886])
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