您始终可以使用蒙版并相乘:
from numpy import array
z = array([[[0.,0.,0.],[0.,0.]],[[0.,0.]]])
i = array([[1,1,1],[1,0],[0,0]])
v = array([[ 2.,4.,10.,7.],[10.,9.,2.],[ 3.,8.,8.],[ 8.,6.,1.]])
z[0] = ~i.astype(bool)*v
z[1] = i.astype(bool)*v
输出:
z =
array([[[ 0.,[ 0.,1.]],[[ 2.,0.]]])
对于更一般的情况:
for ax in np.unique(i):
mask = i == ax
z[ax] = mask*v
print(z)
输出:
[[[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]
[0. 6. 5. 0.]]
[[0. 5. 0. 4.]
[4. 6. 0. 3.]
[4. 0. 4. 8.]
[7. 0. 0. 7.]]
[[5. 0. 0. 0.]
[0. 0. 8. 0.]
[0. 0. 0. 0.]
[0. 0. 0. 0.]]]
注意:这里我假设i
中的值与z
中的通道相同。如果不同,可以尝试以下方法代替np.unique:
for i_val,ax in zip(np.unique(i),range(z.shape[0]))::
mask = i == i_val
z[ax] = mask*v
编辑
单线是:
>>> z = np.array([*map(i.__eq__,np.unique(i))])*np.array([v]*z.shape[0])
>>> z
array([[[0.,5.,4.],[4.,3.],[7.,7.]],[[5.,0.]]])
,
可以使用numpy.where
:
z[0] = np.where(i,z[0],v)
z[1] = np.where(1 - i,z[1],v)
输出:
array([[[ 0.,0.]]])
在更新之后,如果i
拥有要更改的z
的索引(与示例中的0和1相反),则可以轻松地将其缩放为:
for N in range(z.shape[0]):
z[N] = np.where(i == N,v,z[N])
或者如果您真的想避免循环变态:
z = np.where(i == (np.where(np.ones(len(z.reshape(-1))) == 1)[0] // (z.shape[1] * z.shape[2])).reshape(z.shape),z)
输出:
array([[[0.,0.]]])
,
没有numpy的事情会更加复杂:
r = list(zip(i,v))
z[0] = [[l2[i] if l1[i]==0 else 0 for i in range(4)] for l1,l2 in r ]
z[1] = [[l2[i] if l1[i]==1 else 0 for i in range(4)] for l1,l2 in r ]
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