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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)

Fastest way to mask rows of a 2D numpy array given a boolean vector of same length?

I have a numpy boolean vector of shape 1 x N, and an 2d array with shape 160 x N. What is a fast way of subsetting the columns of the 2d array such that for each index of the boolean vector that has True in it, the column is kept, and for each index of the boolean vector that has False in it, the column is discarded?
If you call the vector mask and the array features, i've found the following to be far too slow: np.array([f[mask] for f in features])
Is there a better way? I feel like there has to be, right?

You can try this,
new_array = 2d_array[:,bool_array==True]
So depending on the axes you can select which one you want to remove. In case you get a 1-d array, then you can just reshape it and get the required array. This method will be faster also.

How do I take an array of row locations and an array of column locations and make a new array of the combined points?

I'd like to take a 1x3 matrix of 'row' values, where a = ([1],[3],[5]) and a 3x1 matrix of column values, b = ([1,4,7]) , and create a 3x3 matrix of (row, column) values so that the final matrix would look like
([(1,1) (1,4) (1,7)],
[(3,1) (3,4) (3,7)],
[(5,1) (5,4) (5,7)])
Is there a way to do this without using a for loop?
import numpy as np
a = np.array([1,4,7]).reshape((3,1))
b = np.array([1,3,5]).reshape((1,3))

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)

Covariance with a columns

If I have a numpy array X with X.shape=(m,n) and a second column vector y with y.shape=(m,1), how can I calculate the covariance of each column of X with y wihtout using a for loop? I expect the result to be of shape (m,1) or (1,m).

Assuming that the output is meant to be of shape (1,n) i.e. a scalar each for covariance operation for each column of A with B and thus for n columns ending up with n such scalars, you can use two approaches here that use covariance formula.
Approach #1: With Broadcasting
np.sum((A - A.mean(0))*(B - B.mean(0)),0)/B.size
Approach #2: With Matrix-multiplication
np.dot((B - B.mean(0)).T,(A - A.mean(0)))/B.size