[英]block matrix assignment in Python
From this mwe: 从这个mwe:
a=np.zeros((5,5))
b=np.zeros((2,2))
a=np.matrix(a)
b=np.matrix(b)
b[0,0]=4
b[1,1]=9
b[0,1]=7
indice=[2,3]
# 1
c=a[indice,:][:,indice]
c=b
print c
# 2
a[indice,:][:,indice]=b
print a[indice,:][:,indice]
I get: 我明白了:
>>> c
matrix([[ 4., 7.],
[ 0., 9.]])
and: 和:
>>> a[indice,:][:,indice]
matrix([[ 0., 0.],
[ 0., 0.]])
I do not understand why the values of a remain zeroes. 我不明白为什么的值保持零。 If a similar operation is done in two steps, everything works fine: 如果通过两个步骤完成类似的操作,一切正常:
>>> for k in range(len(indice)):
... a[indice[k],indice]=b[k,:]
I obtain: 我获得:
>>> a
matrix([[ 0., 0., 0., 0., 0.],
[ 0., 4., 0., 7., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 9., 0.],
[ 0., 0., 0., 0., 0.]])
It's because a[indice,:][:,indice]
isn't a view into the array but is a separate copy - 这是因为a[indice,:][:,indice]
不是数组的视图,而是一个单独的副本 -
In [142]: np.may_share_memory(a, a[indice,:][:,indice])
Out[142]: False
To solve it, we can use np.ix_
- 要解决它,我们可以使用np.ix_
-
a[np.ix_(indice, indice)] = b
Verify result - 验证结果 -
In [145]: a
Out[145]:
matrix([[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.]])
In [146]: b
Out[146]:
matrix([[ 4., 7.],
[ 0., 9.]])
In [147]: a[np.ix_(indice, indice)] = b
In [148]: a
Out[148]:
matrix([[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 4., 7., 0.],
[ 0., 0., 0., 9., 0.],
[ 0., 0., 0., 0., 0.]])
In [149]: a[indice,:][:,indice]
Out[149]:
matrix([[ 4., 7.],
[ 0., 9.]])
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