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How to build a tensor from one tensor of contents and another of indices?

I'm trying to assemble a tensor based on the contents of two other tensors, like so:
I have a 2D tensor called A, with shape I * J, and another 2D tensor called B, with shape M * N, whose elements are indices into the 1st dimension of A.
I want to obtain a 3D tensor C with shape M * N * J such that C[m,n,:] == A[B[m,n],:] for all m in [0, M) and n in [0, N).
I could do this using nested for-loops to iterate over all indices in M and N, assigning the right values to C at each one, but M and N are large so this is quite slow. I suspect there's some nicer, faster way of doing this using clever slicing or a built-in pytorch function, but I don't know what it would be. It looks a bit like somewhere one would use torch.gather(), but that requires all tensors to have the same number of dimensions. Does anyone know how this ought to be done?
EDIT: torch.index_select(input, dim, index) is almost what I want, but it won't work here because it requires that index be a 1D tensor, while my tensor of indices is 2D.

You could achieve this by flattening the first dimensions which let's you index A. A broadcast will be required to recover the final shape
>>> A[B.flatten(),:].reshape(*B.shape, A.size(-1))
Indexing with A[B.flatten(),:] is equivalent to torch.index_select(A, 0, B.flatten()).

Numpy.sum() nD array of (2,2,2...n) to a 1D array of shape (2,)

I want to numpy.sum() my nD arrays (matrix) of shape (2,2,2...n) to 1D arrays of shape (2,).
Basically along the axis=0.
(Sum the first number of each 1d array of shape (2,) and sum the second number of the same arrays into one single resulting 1d array of shape (2,)).
However matrix.sum(axis=0) only seems to work for those of shape (2,2), while I think matrix.sum(axis=(1,2)) works for (2,2,2). But then what about (2,2,2,2) arrays and so on?
The n-dimensions have been confusing me. A generalised solution, along with a brief explanation, would be much appreciated.
EDIT: I've provided an example of the nD arrays I'm trying to sum below. I want to sum along axis=0 to get a (2,) 1D array. numpy.sum(axis=0) seems to only work for the array of shape (2,2)...
#Of shape (2,2):
[[9.99695358e-02 9.99232016e-01]
[9.00030464e-01 7.67983971e-04]].sum(axis=0) seems to work
#Of shape (2,2,2):
[[[2.02737071e-01 7.75883149e-01]
[2.02650032e-08 1.58192237e-02]]
[[7.31718878e-06 1.41793363e-03]
[4.12802168e-03 7.26350831e-06]]].sum(axis=(1,2)) seems to work
#Of shape… (2,2,2,2)
[[[[1.83819962e+00 1.02712560e-02]
[5.05122135e-02 2.80555725e-04]]
[[5.60304309e-07 5.44521143e-04]
[2.41592380e-03 1.49436734e-05]]]
[[[7.04398015e-05 7.66717944e-06]
[1.76843337e-05 1.98868986e-06]]
[[9.74010599e-02 1.12527543e-07]
[2.61427056e-04 2.70778171e-08]]]].sum(axis=?) # What axis? And how to generalise?

What do you think about reshaping x to 2D and summing along the second axis?
x.reshape(x.shape[0], -1).sum(axis=1)

Transpose a 1-dimensional array in Numpy without casting to matrix

My goal is to to turn a row vector into a column vector and vice versa. The documentation for numpy.ndarray.transpose says:
For a 1-D array, this has no effect. (To change between column and row vectors, first cast the 1-D array into a matrix object.)
However, when I try this:
my_array = np.array([1,2,3])
my_array_T = np.transpose(np.matrix(myArray))
I do get the wanted result, albeit in matrix form (matrix([[66],[640],[44]])), but I also get this warning:
PendingDeprecationWarning: the matrix subclass is not the recommended way to represent matrices or deal with linear algebra (see https://docs.scipy.org/doc/numpy/user/numpy-for-matlab-users.html). Please adjust your code to use regular ndarray.
my_array_T = np.transpose(np.matrix(my_array))
How can I properly transpose an ndarray then?

A 1D array is itself once transposed, contrary to Matlab where a 1D array doesn't exist and is at least 2D.
What you want is to reshape it:
my_array.reshape(-1, 1)
Or:
my_array.reshape(1, -1)
Depending on what kind of vector you want (column or row vector).
The -1 is a broadcast-like, using all possible elements, and the 1 creates the second required dimension.

If your array is my_array and you want to convert it to a column vector you can do:
my_array.reshape(-1, 1)
For a row vector you can use
my_array.reshape(1, -1)
Both of these can also be transposed and that would work as expected.

IIUC, use reshape
my_array.reshape(my_array.size, -1)

Python: Functions of arrays that return arrays of the same shape

Note: I'm using numpy
import numpy as np
Given 4 arrays of the same (but arbitrary) shape, I am trying to write a function that forms 2x2 matrices from each corresponding element of the arrays, finds the eigenvalues, and returns two arrays of the same shape as the original four, with its elements being eigenvalues (i.e. the resulting arrays would have the same shape as the input, with array1 holding all the first eigenvalues and array2 holding all the second eigenvalues).
I tried doing the following, but unsurprisingly, it gives me an error that says the array is not square.
temp = np.linalg.eig([[m1, m2],[m3, m4]])[0]
I suppose I can make an empty temp variable in the same shape,
temp = np.zeros_like(m1)
and go over each element of the original arrays and repeat the process. My problem is that I want this generalised for arrays of any arbitrary shape (need not be one dimensional). I would guess that finding the shape of the arrays and designing loops to go over each element would not be a very good way of doing it. How do I do this efficiently?

Construct a 2x2x... array:
temp = np.array([[m1, m2], [m3, m4]])
Move the first two dimensions to the end for a ...x2x2 array:
for _ in range(2):
temp = np.rollaxis(temp, 0, temp.ndim)
Call np.linalg.eigvals (which broadcasts) for a ...x2 array of eigenvalues:
eigvals = np.linalg.eigvals(temp)
And split this into an array of first eigenvalues and an array of second eigenvalues:
eigvals1, eigvals2 = eigvals[..., 0], eigvals[..., 1]

How to assign a 1D numpy array to 2D numpy array?

Consider the following simple example:
X = numpy.zeros([10, 4]) # 2D array
x = numpy.arange(0,10) # 1D array
X[:,0] = x # WORKS
X[:,0:1] = x # returns ERROR:
# ValueError: could not broadcast input array from shape (10) into shape (10,1)
X[:,0:1] = (x.reshape(-1, 1)) # WORKS
Can someone explain why numpy has vectors of shape (N,) rather than (N,1) ?
What is the best way to do the casting from 1D array into 2D array?
Why do I need this?
Because I have a code which inserts result x into a 2D array X and the size of x changes from time to time so I have X[:, idx1:idx2] = x which works if x is 2D too but not if x is 1D.

Do you really need to be able to handle both 1D and 2D inputs with the same function? If you know the input is going to be 1D, use
X[:, i] = x
If you know the input is going to be 2D, use
X[:, start:end] = x
If you don't know the input dimensions, I recommend switching between one line or the other with an if, though there might be some indexing trick I'm not aware of that would handle both identically.
Your x has shape (N,) rather than shape (N, 1) (or (1, N)) because numpy isn't built for just matrix math. ndarrays are n-dimensional; they support efficient, consistent vectorized operations for any non-negative number of dimensions (including 0). While this may occasionally make matrix operations a bit less concise (especially in the case of dot for matrix multiplication), it produces more generally applicable code for when your data is naturally 1-dimensional or 3-, 4-, or n-dimensional.

I think you have the answer already included in your question. Numpy allows the arrays be of any dimensionality (while afaik Matlab prefers two dimensions where possible), so you need to be correct with this (and always distinguish between (n,) and (n,1)). By giving one number as one of the indices (like 0 in 3rd row), you reduce the dimensionality by one. By giving a range as one of the indices (like 0:1 in 4th row), you don't reduce the dimensionality.
Line 3 makes perfect sense for me and I would assign to the 2-D array this way.

Here are two tricks that make the code a little shorter.
X = numpy.zeros([10, 4]) # 2D array
x = numpy.arange(0,10) # 1D array
X.T[:1, :] = x
X[:, 2:3] = x[:, None]