I have a 2D array A:
28 39 52
77 80 66
7 18 24
9 97 68
And a vector array of column indexes B:
1
0
2
0
How, in a pythonian way, using base Python or Numpy, can I select the elements from A which DO NOT correspond to the column indexes in B?
I should get this 2D array which contains the elements of A, Not corresponding to the column indexes stored in B:
28 52
80 66
7 18
97 68
You can make use of broadcasting and a row-wise mask to select elements not contained in your array for each row:
Setup
B = np.array([1, 0, 2, 0])
cols = np.arange(A.shape[1])
Now use broadcasting to create a mask, and index your array.
mask = B[:, None] != cols
A[mask].reshape(-1, 2)
array([[28, 52],
[80, 66],
[ 7, 18],
[97, 68]])
A spin off of my answer to your other question,
Replace 2D array elements with zeros, using a column index vector
We can make a boolean mask with the same indexing used before:
In [124]: mask = np.ones(A.shape, dtype=bool)
In [126]: mask[np.arange(4), B] = False
In [127]: mask
Out[127]:
array([[ True, False, True],
[False, True, True],
[ True, True, False],
[False, True, True]])
Indexing an array with a boolean mask produces a 1d array, since in the most general case such a mask could select a different number of elements in each row.
In [128]: A[mask]
Out[128]: array([28, 52, 80, 66, 7, 18, 97, 68])
In this case the result can be reshaped back to 2d:
In [129]: A[mask].reshape(4,2)
Out[129]:
array([[28, 52],
[80, 66],
[ 7, 18],
[97, 68]])
Since you allowed for 'base Python' here's list comprehension answer:
In [136]: [[y for i,y in enumerate(x) if i!=b] for b,x in zip(B,A)]
Out[136]: [[28, 52], [80, 66], [7, 18], [97, 68]]
If all the 0's in the other A come from the insertion, then we can also get the mask (Out[127]) with
In [142]: A!=0
Out[142]:
array([[ True, False, True],
[False, True, True],
[ True, True, False],
[False, True, True]])
arange:arr[np.arange(arr.shape[0]), cols-1][A[i][B[i]] for i in range(len(A))]is what you want in plain python.