Numpy: Perform Multiplication-like Addition - python

I wanted to define my own addition operator that takes an Nx1 vector (call it A) and a 1xN vector (B) such that the element in the i^th row and j^th column is the sum of the i^th element in A and the j^th element in B. An example is illustrated here.
I was able to write the following code for the function (and it is correct as far as I know).
def test_fn(a, b):
a_len = a.shape[0]
b_len = b.shape[1]
prod = np.array([[0]*a_len]*b_len)
for i in range(a_len):
for j in range(b_len):
prod[i, j] = a[i, 0] + b[0, j]
return prod
However, the vectors I am working with contain thousands of elements, and the function above is quite slow. I was wondering if there was a better way to approach this problem, or if there was a numpy function that could be of use. Any help would be appreciated.

According to numpy's broadcasting rules, you can use a+b to implement your own defined operator.
The first rule of broadcasting is that if all input arrays do not have the same number of dimensions, a “1” will be repeatedly prepended to the shapes of the smaller arrays until all the arrays have the same number of dimensions.
The second rule of broadcasting ensures that arrays with a size of 1 along a particular dimension act as if they had the size of the array with the largest shape along that dimension. The value of the array element is assumed to be the same along that dimension for the “broadcast” array.

Related

How can you do an outer summation over only one dimension of a numpy 2D array?

I have a (square) 2 dimensional numpy array where I would like to compare (subtract) all of the values within each row to each other but not to other rows so the output should be a 3D array.
matrix = np.array([[10,1,32],[32,4,15],[6,3,1]])
Output should be a 3x3x3 array which looks like:
output = [[[0,-9,22],[0,-28,-17],[0,-3,-5]], [[9,0,31],[28,0,11],[3,0,-2]], [[-22,-31,0],[17,-11,0],[5,2,0]]]
I.e. for output[0], for each of the 3 rows of matrix, subtract that row's zeroth element from every other, for output[1] subtract each row's first element etc.
This seems to me like a reduced version of numpy's ufunc.outer functionality which should be possible with
tryouter = np.subtract(matrix, matrix)
and then taking some clever slice and/or transposition.
Indeed, if you do this, one finds that: output[i,j] = tryouter[i,j,i]
This looks like it should be solvable by using np.transpose to switch the 1 and 2 axes and then taking the arrays on the new 0,1 diagonal but I can't work out how to do this with numpy diagonal or any slicing method.
Is there a way to do this or is there a simpler approach to this whole problem built into numpy?
Thanks :)
You're close, you can do it with broadcasting:
out = matrix[None, :, :] - matrix.T[:, :, None]
Here .T is the same as np.transpose, and using None as an index introduces a new dummy dimension of size 1.

How to multiply each row of an array with all rows of an array element-wise in Python

I need to multiply each row of an array A with all rows of an array B element-wise. For instance, let's say we have the following arrays:
A = np.array([[1,5],[3,6]])
B = np.array([[4,2],[8,2]])
I want to get the following array C:
C = np.array([[4,10],[8,10],[12,12],[24,12]])
I could do this by using for loop but I think there could be a better way to do it.
EDIT: I thought of repeating and tiling but my arrays are not that small. It could create some memory problem.
Leverage broadcasting extending dims for A to 3D with None/np.newaxis, perform the elementwise multiplication and reshape back to 2D -
(A[:,None]*B).reshape(-1,B.shape[1])
which essentially would be -
(A[:,None,:]*B[None,:,:]).reshape(-1,B.shape[1])
Schematically put, it's :
A : M x 1 x N
B : 1 x K x N
out : M x K x N
Final reshape to merge last two axes and give us (M x K*N) shaped 2D array.
We can also use einsum to perform the extension to 3D and elementwise multiplication in one function call -
np.einsum('ij,kj->ikj',A,B).reshape(-1,B.shape[1])

Vectorized Evaluation of a Function, Broadcasting and Element Wise Operations

Given this...
I have to explain what this code does, knowing that it performs the vectorized evaluation of F, using broadcasting and element wise operations concepts...
def F(x_pos, alpha):
D = x_pos.reshape(1,-1) - x_pos.reshape(-1,1)
return (1./alpha) * (alpha.reshape(1,-1) * R(D)).sum(axis=1)
My explanation is:
In the first line of the function F receives x_pos and alpha as parameters (both numpy arrays), in the second line the matrix D is calculated by means of broadcasting (basic operations such as addition in arrays numpy are performed elementwise, ie, element by element, but it is also possible with arranys of different size if numpy can transform them into others of the same size, this conversion is called broadcasting), subtracting an array of order 1xN with another of order Nx1, resulting in the matrix D of order NxN containing x_j - x_1, x_j - x_2, etc. as elements, finally, in the last line the reciprocal of alpha is calculated (which clearly is an arrangement), where each element is multiplied by the sum of the R evaluation of each cell of the matrix D multiplied by alpha_j horizontally (due to axis = 1 in the argument)
Questions:
Considering I'm new to Python, is my explanation OK?
The code has an error or not? Because I don't see that the "j must be different from 1, 2, ..., n" in each sum is taken into consideration in the code... and If it's in fact wrong... How can I fix the code so it do exactly the same thing as stated as in the image?
Few comments/improvements/fixes could be suggested here.
1] The first step could be alternatively done with just introducing a new axis and subtracting with itself, like so -
D = x_pos[:,None] - x_pos
In my opinion, this is a cleaner option. The performance benefit might be just marginal.
2] In the second line, I think it needs a fix as we need to avoid computations for the diagonal elements of R(D). So, If I got that correctly, the corrected code would be -
vals = R(D)
np.fill_diagonal(vals,0)
out = (1./alpha) * (alpha.reshape(1,-1) * vals).sum(axis=1)
Now, let's make the code a bit more idiomatic/cleaner.
At that line, we could write : (alpha * vals) instead of alpha.reshape(1,-1) * vals. This is because the shapes are already aligned for broadcasting as shown in a schematic diagram below -
alpha : n
vals : n x n
Thus, alpha would be automatically extended to 2D with its elements broadcasted along the first axis for the length of vals and then elementwise multiplications being generated with it. Again, this is meant as a cleaner code.
There's a further performance improvement possible here with (alpha.reshape(1,-1) * vals).sum(axis=1) being replaceable with a matrix-multiplicatiion using np.dot as alpha.dot(vals). The benefit on performance should be noticeable with this step.
So, the second step reduces to -
out = (1./alpha) * alpha.dot(vals)

Matrix vector multiplication along array axes

in a current project I have a large multidimensional array of shape (I,J,K,N) and a square matrix of dim N.
I need to perform a matrix vector multiplication of the last axis of the array with the square matrix.
So the obvious solution would be:
for i in range(I):
for j in range(J):
for k in range(K):
arr[i,j,k] = mat.dot(arr[i,j,k])
but of course this is rather slow. So I also tried numpy's tensordot but had little success.
I would expect that something like:
arr = tensordot(mat,arr,axes=((0,1),(3)))
should do the trick but I get a shape mismatch error.
Has someone a better solution or knows how to correctly use tensordot?
Thank you!
This should do what your loops, but with vectorized looping:
from numpy.core.umath_tests import matrix_multiply
arr[..., np.newaxis] = matrix_multiply(mat, arr[..., np.newaxis])
matrix_multiply and its sister inner1d are hidden, undocumented, gems of numpy, although a full set of linear algebra gufuncs should see the light with numpy 1.8. matrix_multiply does matrix multiplication on the last two dimensions of its inputs, and broadcasting on the rest. The only tricky part is setting an additional dimension, so that it sees column vectors when multiplying, and adding it also on assignment back into array, so that there is no shape mismatch.
I think your for loop is wrong, and for this case dot seems to be enough:
# a is your IJKN
# b is your NN
c = dot(a, b)
Here c will be a IJKN array. If you want to sum over the last dimension to get the IJK array:
arr = dot(a,b).sum(axis=3)
BUT I'm NOT SURE IF THIS IS WHAT YOU WANT...

Null matrix with constant diagonal, with same shape as another matrix

I'm wondering if there is a simple way to multiply a numpy matrix by a scalar. Essentially I want all values to be multiplied by the constant 40. This would be an nxn matrix with 40's on the diagonal, but I'm wondering if there is a simpler function to use to scale this matrix. Or how would I go about making a matrix with the same shape as my other matrix and fill in its diagonal?
Sorry if this seems a bit basic, but for some reason I couldn't find this in the doc.
If you want a matrix with 40 on the diagonal and zeros everywhere else, you can use NumPy's function fill_diagonal() on a matrix of zeros. You can thus directly do:
N = 100; value = 40
b = np.zeros((N, N))
np.fill_diagonal(b, value)
This involves only setting elements to a certain value, and is therefore likely to be faster than code involving multiplying all the elements of a matrix by a constant. This approach also has the advantage of showing explicitly that you fill the diagonal with a specific value.
If you want the diagonal matrix b to be of the same size as another matrix a, you can use the following shortcut (no need for an explicit size N):
b = np.zeros_like(a)
np.fill_diagonal(b, value)
Easy:
N = 100
a = np.eye(N) # Diagonal Identity 100x100 array
b = 40*a # Multiply by a scalar
If you actually want a numpy matrix vs an array, you can do a = np.asmatrix(np.eye(N)) instead. But in general * is element-wise multiplication in numpy.

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