Python: Resize an existing array and fill with zeros

问题:

I think that my issue should be really simple, yet I can not find any help on the Internet whatsoever. I am very new to Python, so it is possible that I am missing something very obvious.

I have an array, S, like this [x x x] (one-dimensional). I now create a diagonal matrix, sigma, with np.diag(S) - so far, so good. Now, I want to resize this new diagonal array so that I can multiply it by another array that I have.

import numpy as np ... shape = np.shape((6, 6)) #This will be some pre-determined size sigma = np.diag(S) #diagonalise the matrix - this works my_sigma = sigma.resize(shape) #Resize the matrix and fill with zeros - returns "None" - why?

However, when I print the contents of my_sigma, I get "None". Can someone please point me in the right direction, because I can not imagine that this should be so complicated.

Thanks in advance for any help!

Casper

Graphical:

I have this:

[x x x]

I want this:

[x 0 0] [0 x 0] [0 0 x] [0 0 0] [0 0 0] [0 0 0] - or some similar size, but the diagonal elements are important.

回答1:

sigma.resize() returns None because it operates in-place. np.resize(sigma, shape), on the other hand, returns the result but instead of padding with zeros, it pads with repeats of the array.

Also, the shape() function returns the shape of the input. If you just want to predefine a shape, just use a tuple.

import numpy as np ... shape = (6, 6) #This will be some pre-determined size sigma = np.diag(S) #diagonalise the matrix - this works sigma.resize(shape) #Resize the matrix and fill with zeros

However, this will first flatten out your original array, and then reconstruct it into the given shape, destroying the original ordering. If you just want to "pad" with zeros, instead of using resize() you can just directly index into a generated zero-matrix.

# This assumes that you have a 2-dimensional array zeros = np.zeros(shape, dtype=np.int32) zeros[:sigma.shape[0], :sigma.shape[1]] = sigma

回答2:

There is a new numpy function in version 1.7.0 numpy.pad that can do this in one-line. Like the other answers, you can construct the diagonal matrix with np.diag before the padding. The tuple ((0,N),(0,0)) used in this answer indicates the "side" of the matrix which to pad.

import numpy as np  A = np.array([1, 2, 3])  N = A.size B = np.pad(np.diag(A), ((0,N),(0,0)), mode='constant')

B is now equal to:

[[1 0 0]  [0 2 0]  [0 0 3]  [0 0 0]  [0 0 0]  [0 0 0]]

回答3:

I see the edit... you do have to create the zeros first and then move some numbers into it. np.diag_indices_from might be useful for you

bigger_sigma = np.zeros(shape, dtype=sigma.dtype) diag_ij = np.diag_indices_from(sigma) bigger_sigma[diag_ij] = sigma[diag_ij] 

回答4:

Another pure python solution is

a = [1, 2, 3] b = [] for i in range(6):     b.append((([0] * i) + a[i:i+1] + ([0] * (len(a) - 1 - i)))[:len(a)])

b is now

[[1