import numpy as np
a=np.array([[1,2,3], [4,5,6], [7,8,9]])
k = [0, 1, 2]
print np.delete(a, k, 1)
This returns
[]
But, the result I really want is
[[2,3],
[4,6],
[7,8]]
I want to delete the first element (indexed as 0) from a[0], the second (indexed as 1) from a[1], and the third (indexed as 2) from a[2].
Any thoughts?
Here's an approach using boolean indexing -
m,n = a.shape
out = a[np.arange(n) != np.array(k)[:,None]].reshape(m,-1)
If you would like to persist with np.delete, you could calculate the linear indices and then delete those after flattening the input array, like so -
m,n = a.shape
del_idx = np.arange(n)*m + k
out = np.delete(a.ravel(),del_idx,axis=0).reshape(m,-1)
Sample run -
In [94]: a
Out[94]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
In [95]: k = [0, 2, 1]
In [96]: m,n = a.shape
In [97]: a[np.arange(n) != np.array(k)[:,None]].reshape(m,-1)
Out[97]:
array([[2, 3],
[4, 5],
[7, 9]])
In [98]: del_idx = np.arange(n)*m + k
In [99]: np.delete(a.ravel(),del_idx,axis=0).reshape(m,-1)
Out[99]:
array([[2, 3],
[4, 5],
[7, 9]])