Is there a relatively simple way of removing columns of an (numpy) array and keeping the order of the columns?
As an example, consider this array:
a = np.array([[2, 1, 1, 3],
[2, 1, 1, 3]])
a = np.array([[2, 1, 3],
[2, 1, 3]])
Approach #1 Here's an approach using broadcasting

a[:,~np.triu((a[:,None,:] == a[...,None]).all(0),1).any(0)]
Sample run 
In [115]: a
Out[115]:
array([[2, 1, 3, 5, 1, 3, 7],
[6, 5, 4, 6, 5, 4, 8]])
In [116]: a[:,~np.triu((a[:,None,:] == a[...,None]).all(0),1).any(0)]
Out[116]:
array([[2, 1, 3, 5, 7],
[6, 5, 4, 6, 8]])
Explanation
1) Input array 
In [156]: a
Out[156]:
array([[2, 1, 3, 5, 1, 3, 7],
[6, 5, 4, 6, 5, 4, 8]])
2) Use broadcasting to perform elementwise equality comparison keeping the first axis aligned, which would correspond to the column axis from original 2D array 
In [157]: a[:,None,:] == a[...,None]
Out[157]:
array([[[ True, False, False, False, False, False, False],
[False, True, False, False, True, False, False],
[False, False, True, False, False, True, False],
[False, False, False, True, False, False, False],
[False, True, False, False, True, False, False],
[False, False, True, False, False, True, False],
[False, False, False, False, False, False, True]],
[[ True, False, False, True, False, False, False],
[False, True, False, False, True, False, False],
[False, False, True, False, False, True, False],
[ True, False, False, True, False, False, False],
[False, True, False, False, True, False, False],
[False, False, True, False, False, True, False],
[False, False, False, False, False, False, True]]], dtype=bool)
3) Since we are looking for duplicate cols, let's look for ALL matches along the first axis 
In [158]: (a[:,None,:] == a[...,None]).all(0)
Out[158]:
array([[ True, False, False, False, False, False, False],
[False, True, False, False, True, False, False],
[False, False, True, False, False, True, False],
[False, False, False, True, False, False, False],
[False, True, False, False, True, False, False],
[False, False, True, False, False, True, False],
[False, False, False, False, False, False, True]], dtype=bool)
4) We are looking to keep the first occurrence only, so we can use a upper triangular matrix to set all diagonal and lower triangular elems as False

In [163]: np.triu((a[:,None,:] == a[...,None]).all(0),1)
Out[163]:
array([[False, False, False, False, False, False, False],
[False, False, False, False, True, False, False],
[False, False, False, False, False, True, False],
[False, False, False, False, False, False, False],
[False, False, False, False, False, False, False],
[False, False, False, False, False, False, False],
[False, False, False, False, False, False, False]], dtype=bool)
5) Next up, we look for ANY matches along the first axis indicating the duplicates 
In [164]: (np.triu((a[:,None,:] == a[...,None]).all(0),1)).any(0)
Out[164]: array([False, False, False, False, True, True, False], dtype=bool)
6) We are looking to remove those duplicates, so invert the mask 
In [165]: ~(np.triu((a[:,None,:] == a[...,None]).all(0),1)).any(0)
Out[165]: array([ True, True, True, True, False, False, True], dtype=bool)
7) Finally, we index into the columns of input array with the mask for final output 
In [166]: a[:,~(np.triu((a[:,None,:] == a[...,None]).all(0),1)).any(0)]
Out[166]:
array([[2, 1, 3, 5, 7],
[6, 5, 4, 6, 8]])
Approach #2 With focus on memory efficiency and might even be faster, here's an approach considering each column as an indexing tuple 
lidx = np.ravel_multi_index(a,a.max(1)+1)
out = a[:,np.sort(np.unique(lidx,return_index=1)[1])]
Explanation
1) Input array 
In [203]: a
Out[203]:
array([[2, 1, 3, 5, 1, 3, 7],
[6, 5, 4, 6, 5, 4, 8]])
2) Calculate linear index equivalents for each column 
In [207]: lidx = np.ravel_multi_index(a,a.max(1)+1)
In [208]: lidx
Out[208]: array([24, 14, 31, 51, 14, 31, 71])
3) Get the first occurence of each unique linear index
In [209]: np.unique(lidx,return_index=1)[1]
Out[209]: array([1, 0, 2, 3, 6])
4) Sort those and index into cols of input array for final o/p 
In [210]: np.sort(np.unique(lidx,return_index=1)[1])
Out[210]: array([0, 1, 2, 3, 6])
In [211]: a[:,np.sort(np.unique(lidx,return_index=1)[1])]
Out[211]:
array([[2, 1, 3, 5, 7],
[6, 5, 4, 6, 8]])
For a detailed info on the considerations related to converting to indexing tuples, please refer to this post
.