I have an array of shape (B, N, 3) and an array of indexes with shape (B, M) with M<N
I want to get, for each item in B, M elements at the indexes indicated by the array of indexes, getting an array of shape (B, M, 3)
What I have tried with no success
array[indexes]
array[indexes.to_list()]
I am sure there is a fancy way of doing this but I haven't found it. Any advice?
How do I combine N, 2D numpy arrays (of dimension R x C) to create a 3D numpy array of shape (N, R, C)? Right now, the N-2D numpy arrays are contained inside a list, and I want that to become a 3D numpy array. Let's say X is my list of 2D numpy arrays, if I just do np.array(X), I get something of shape (N,). If I do np.vstack(X), I get something of shape (N x R, C). How do I solve this problem?
You can use np.stack:
test = np.stack([np.ones([2, 3]) for _ in range(4)])
print(test.shape) # (4, 2, 3)
you could just use :
np.array([np.array(x) for x in ArrayList])
I've got some numpy 2d arrays:
x, of shape(N,T)
W, of shape(V,D)
they are described as the following:
"Minibatches of size N where each sequence has length T. We assume a vocabulary of V words, assigning each to a vector of dimension D."(This is a question from cs231 A3.)
I want an output array of shape(N, T, D), where i can match the N elements to the desired vectors.
First I came out with the solution using a loop to run through all the elements in the first row of x:
for n in range(N):
out[n, :, :] = W[x[n, :]]
Then I go on to experiment with the second solution:
out = W[x]
Both solutions gave me the right answer, but why does the second solution work? Why can I index a 3d array in a 2d array?
I have stored a number of 2d arrays in a 3d array and I need to multiply each one with a vector. so I have stored all those vectors in a 2d array. It's like this:
A = np.random.random((L, M, N))
B = np.random.random((L, M))
and I need to multiply each A[l] by B[l] which results in a Nx1 array and the output of the whole operation would be a LxN 2d array. Is there a function that can do this or do I need a loop?
An option is np.einsum
import numpy as np
output = np.einsum("ijk, ij -> ik", A, B)
This results in a (L, N) sized array containing matrix products of all the A[i].T.dot(B[i])
If I have an array, A, with shape (n, m, o) and an array, B, with shape (n, m), is there a way to divide each array at A[n, m] by the scalar at B[n, m] without a list comprehension?
>>> A.shape
(4,173,1469)
>>> B.shape
(4,173)
>>> # Better way to do:
>>> np.array([[A[i, j] / B[i, j] for j in range(len(B[i]))] for i in range(len(B))])
The problem with a list comprehension is that it is slow, it doesn't return an array (so you have to np.array(_) it, which makes it even slower), it is hard to read, and the whole point of numpy was to move loops from Python to C++ or Fortran.
If A was of shape (n) and B was a scalar (of shape ( )), then this would be trivial: A / B, but this property does not scale with dimensions
>>> A / B
ValueError: operands could not be broadcast together with shapes (4,173,1469) (4,173)
I am looking for a fast way to do this (preferably not by tiling B to an array of shape (n, m, o), and preferably using native numpy tools).
You are absolutely right, there is a better way, I think you are getting the spirit of numpy.
The solution in your case is that you have to add a new dimension to B that consists of one entry in that dimension:
so if your A is of shape (n,m,o) your B has to be of shape (n,m,1) and then you can use the native broadcasting to get your operation "A/B" done.
You can just add that dimension to be by adding a "newaxis" to B there.
import numpy as np
A = np.ones(10,5,3)
B = np.ones(10,5)
Result = A/B[:,:,np.newaxis]
B[:,:,np.newaxis] --> this will turn B into an array of shape of (10,5,1)
From here, the rules of broadcasting are:
When operating on two arrays, NumPy compares their shapes
element-wise. It starts with the trailing dimensions, and works its
way forward. Two dimensions are compatible when
they are equal, or
one of them is 1
Your dimensions are n,m,o and n,m so not compatible.
The / division operator will work using broadcasting if you use:
o,n,m divided by n,m
n,m,o divided by n,m,1