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How to build a diagonal matrix using a 2d network in numpy?

 3 years ago
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How to build a diagonal matrix using a 2d network in numpy?

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Using np.diag I'm able to construct a 2-D array where a 1-D array is input is returned on the diagonal. But how to do the same if I have n-D array as input?

This works

foo = np.random.randint(2, size=(36))
print foo
print np.diag(foo)
[1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 1 0]
[[1 0 0 ..., 0 0 0]
 [0 1 0 ..., 0 0 0]
 [0 0 1 ..., 0 0 0]
 ...,
 [0 0 0 ..., 1 0 0]
 [0 0 0 ..., 0 1 0]
 [0 0 0 ..., 0 0 0]]

This won't

foo = np.random.randint(2, size=(2,36))
print foo
[[1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0]
 [0 1 1 1 0 0 1 1 1 1 0 1 0 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 1 0 1 1 1 1 0 1]]
do_something(foo)

Should return

array([[[ 1.,  0.,  0., ...,  0.,  0.,  0.],
        [ 0.,  1.,  0., ...,  0.,  0.,  0.],
        [ 0.,  0.,  0., ...,  0.,  0.,  0.],
        ...,
        [ 0.,  0.,  0., ...,  1.,  0.,  0.],
        [ 0.,  0.,  0., ...,  0.,  0.,  0.],
        [ 0.,  0.,  0., ...,  0.,  0.,  0.]],

       [[ 0.,  0.,  0., ...,  0.,  0.,  0.],
        [ 0.,  1.,  0., ...,  0.,  0.,  0.],
        [ 0.,  0.,  1., ...,  0.,  0.,  0.],
        ...,
        [ 0.,  0.,  0., ...,  1.,  0.,  0.],
        [ 0.,  0.,  0., ...,  0.,  0.,  0.],
        [ 0.,  0.,  0., ...,  0.,  0.,  1.]]])

EDIT

Some test based on answers of Alan and ajcr in this post and Saulo Castro and jaime where ajcr refers to. As usual, it all depends on your input. My input has usually the following shape:

M = np.random.randint(2, size=(1000, 36))

with the functions as follows:

def Alan(M):
    M = np.asarray(M)
    depth, size = M.shape
    x = np.zeros((depth,size,size))
    for i in range(depth):
        x[i].flat[slice(0,None,1+size)] = M[i]
    return x

def ajcr(M):
    return np.eye(M.shape[1]) * M[:,np.newaxis,:]

def Saulo(M):
    b = np.zeros((M.shape[0], M.shape[1], M.shape[1]))
    diag = np.arange(M.shape[1])
    b[:, diag, diag] = M
    return b

def jaime(M):
    b = np.zeros((M.shape[0], M.shape[1]*M.shape[1]))
    b[:, ::M.shape[1]+1] = M
    return b.reshape(M.shape[0], M.shape[1], M.shape[1])

result for me in the following

%timeit Alan(M)
100 loops, best of 3: 2.22 ms per loop
%timeit ajcr(M)
100 loops, best of 3: 5.1 ms per loop
%timeit Saulo(M)
100 loops, best of 3: 4.33 ms per loop
%timeit jaime(M)
100 loops, best of 3: 2.07 ms per loop


def makediag3d(a):
    a = np.asarray(a)
    depth, size = a.shape
    x = np.zeros((depth,size,size))
    for i in range(depth):
        x[i].flat[slice(0,None,1+size)] = a[i]
    return x

report

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