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is there any way to calculate L2 norm of multiple 2d matrices at once, in python?

for example, I have a matrix of dimensions (a,b,c,d). I want to calculate L2 norm of all d matrices of dimensions (a,b,c). Is there any way to use numpy.linalg.norm with out any looping structure?
I mean, the resultant array should be 1 x d

How about this?
import numpy as np
mat = np.arange(2*3*4*5).reshape(2,3,4,5) # create 4d array
mat2 = np.moveaxis(mat,-1,0) # bring last axis to the front
*outarr, = map(np.linalg.norm,mat2) # use map

Unravel 1D list back to 3D array

Basically, is there a way to transform a 1D list that has been "flattened" through the numpy.ravel() function back to it's original 3D form ? I know the dimensions, and one might ask why I just don't use the original 3D array in the first place, instead of converting it - but there reasons for that.
I just need to know if I can actually create the same 3D array from a 1D array that was created by using numpy.ravel() on the 3D array.
Basically the 3D array was created like this:
import numpy as np
nx = 50
ny = 40
nz = 150
x = np.linspace(1, 51, nx)
y = np.linspace(1, 41, ny)
z = np.linspace(1, 151, nz)
x_bc = x[:, np.newaxis, np.newaxis]
y_bc = y[np.newaxis, :, np.newaxis]
z_bc = z[np.newaxis, np.newaxis, :]
arr = x_bc + y_bc + z_bc
And nope, I can't just do this to get it back, since calculations has been done to it in the mean time, and then converted to a 1D array in the mean time as well. So the data in this array is not the same as the one I actually want to convert back.

Just reshape it back to the original shape?
raveled = np.ravel(arr)
new_arr = raveled.reshape(*arr.shape)

Does numpy.reshape do what you want?

Numpy inner product of 2 column vectors

How can I take an inner product of 2 column vectors in python's numpy
Below code does not work
import numpy as np
x = np.array([[1], [2]])
np.inner(x, x)
It returned
array([[1, 2],
[2, 4]])`
instead of 5

The inner product of a vector with dimensions 2x1 (2 rows, 1 column) with another vector of dimension 2x1 (2 rows, 1 column) is a matrix with dimensions 2x2 (2 rows, 2 columns). When you take the inner product of any tensor the inner most dimensions must match (which is 1 in this case) and the result is a tensor with the dimensions matching the outter, i.e.; a 2x1 * 1x2 = 2x2.
What you want to do is transpose both such that when you multiply the dimensions are 1x2 * 2x1 = 1x1.
More generally, multiplying anything with dimensions NxM by something with dimensionsMxK, yields something with dimensions NxK. Note the inner dimensions must both be M. For more, review your matrix multiplication rules
The np.inner function will automatically transpose the second argument, thus when you pass in two 2x1, you get a 2x2, but if you pass in two 1x2 you will get a 1x1.
Try this:
import numpy as np
x = np.array([[1], [2]])
np.inner(np.transpose(x), np.transpose(x))
or simply define your x as row vectors initially.
import numpy as np
x = np.array([1,2])
np.inner(x, x)

i think you mean to have:
x= np.array([1,2])
in order to get 5 as output, your vector needs to be 1xN not Nx1 if you want to apply np.inner on it

Try the following it will work
np.dot(np.transpose(a),a))

make sure col_vector has shape (N,1) where N is the number of elements
then simply sum one to one multiplication result
np.sum(col_vector*col_vector)

Numpy operation for euclidean distance between multidimensional arrays

I have two numpy arrays. 'A' of size w,h,2 and 'B' with n,2.
In other words, A is a 2-dimensional array of 2D vectors while B is a 1D array of 2D vectors.
What i want as a result is an array of size w,h,n. The last dimension is an n-dimensional vector where each of the components is the euclidean distance between the corresponding vector from A (denoted by the first two dimensions w and h) and the nth vector of B.
I know that i can just loop through w, h and n in python manually and calculate the distance for each element, but i like to know if there is a smart way to do that with numpy operations to increase performance.
I found some similar questions but unfortunately all of those use input arrays of the same dimensionality.

Approach #1
You could reshape A to 2D, use Scipy's cdist that expects 2D arrays as inputs, get those euclidean distances and finally reshape back to 3D.
Thus, an implementation would be -
from scipy.spatial.distance import cdist
out = cdist(A.reshape(-1,2),B).reshape(w,h,-1)
Approach #2
Since, the axis of reduction is of length 2 only, we can just slice the input arrays to save memory on intermediate arrays, like so -
np.sqrt((A[...,0,None] - B[:,0])**2 + (A[...,1,None] - B[:,1])**2)
Explanation on A[...,0,None] and A[...,1,None] :
With that None we are just introducing a new axis at the end of sliced A. Well, let's take a small example -
In [54]: A = np.random.randint(0,9,(4,5,2))
In [55]: A[...,0].shape
Out[55]: (4, 5)
In [56]: A[...,0,None].shape
Out[56]: (4, 5, 1)
In [57]: B = np.random.randint(0,9,(3,2))
In [58]: B[:,0].shape
Out[58]: (3,)
So, we have :
A[...,0,None] : 4 x 5 x 1
B[:,0] : 3
That is essentially :
A[...,0,None] : 4 x 5 x 1
B[:,0] : 1 x 1 x 3
When the subtraction is performed, the singleton dims are broadcasted corresponding to the dimensions of the other participating arrays -
A[...,0,None] - B : 4 x 5 x 3
We repeat this for the second index along the last axis. We add these two arrays after squaring and finally a square-root to get the final eucl. distances.

Scale rows of 3D-tensor

I have an n-by-3-by-3 numpy array A and an n-by-3 numpy array B. I'd now like to multiply every row of every one of the n 3-by-3 matrices with the corresponding scalar in B, i.e.,
import numpy as np
A = np.random.rand(10, 3, 3)
B = np.random.rand(10, 3)
for a, b in zip(A, B):
a = (a.T * b).T
print(a)
Can this be done without the loop as well?

You can use NumPy broadcasting to let the elementwise multiplication happen in a vectorized manner after extending B to 3D after adding a singleton dimension at the end with np.newaxis or its alias/shorthand None. Thus, the implementation would be A*B[:,:,None] or simply A*B[...,None].