It looks like you just need a basic integer array indexing:

filter_indices = [1,3,5]
np.array([11,13,155,22,0xff,32,56,88])[filter_indices] 

Use slicing:

>>> arr = np.array([['1A34', 'RBP', 0.0, 1.0, 0.0, 0.0, 0.0, 0.0],
       ['1A9N', 'RBP', 0.0456267, 0.0539268, 0.331932, 0.0464031,
        4.41336e-06, 0.522107],
       ['1AQ3', 'RBP', 0.0444479, 0.201112, 0.268581, 0.0049757,
        1.28505e-12, 0.480883],
       ['1AQ4', 'RBP', 0.0177232, 0.363746, 0.308995, 0.00169861, 0.0,
        0.307837]], dtype=object)
>>> arr[:,:5:2]
array([['1A34', 0.0, 0.0],
       ['1A9N', 0.0456267, 0.331932],
       ['1AQ3', 0.0444479, 0.268581],
       ['1AQ4', 0.0177232, 0.308995]], dtype=object)

If the column indices are irregular then you can do something like this:

>>> indices = [0, 3, 4]
>>> arr[:, indices]
array([['1A34', 1.0, 0.0],
       ['1A9N', 0.0539268, 0.331932],
       ['1AQ3', 0.201112, 0.268581],
       ['1AQ4', 0.363746, 0.308995]], dtype=object)

Note that there's a subtle but substantial difference between slicing (which is basic indexing) and using a sequence for indexing (also known as advanced indexing or fancy indexing). When using a slice such as arr[:, :5:2], no data is copied, and we get a view of the original array. This implies that mutating the result of arr[:, :5:2] will affect arr itself. With fancy indexing arr[:, [0, 3, 4]] is guaranteed to be a copy: this takes up more memory, and mutating this result will not affect arr.

Numpy uses multiple indexing, so instead of A[1][2][3], you can--and should--use A[1,2,3].

You might then think you could do A[:, second, third], but the numpy indices are broadcast, and broadcasting second and third (two one-dimensional sequences) ends up being the numpy equivalent of zip, so the result has shape (5, 2).

What you really want is to index with, in effect, the outer product of second and third. You can do this with broadcasting by making one of them, say second into a two-dimensional array with shape (2,1). Then the shape that results from broadcasting second and third together is (2,2).

For example:

In [8]: import numpy as np

In [9]: a = np.arange(125).reshape(5,5,5)

In [10]: second = [1,2]

In [11]: third = [3,4]

In [12]: s = a[:, np.array(second).reshape(-1,1), third]

In [13]: s.shape
Out[13]: (5, 2, 2)

Note that, in this specific example, the values in second and third are sequential. If that is typical, you can simply use slices:

In [14]: s2 = a[:, 1:3, 3:5]

In [15]: s2.shape
Out[15]: (5, 2, 2)

In [16]: np.all(s == s2)
Out[16]: True

There are a couple very important difference in those two methods.

• The first method would also work with indices that are not equivalent to slices. For example, it would work if second = [0, 2, 3]. (Sometimes you'll see this style of indexing referred to as "fancy indexing".)
• In the first method (using broadcasting and "fancy indexing"), the data is a copy of the original array. In the second method (using only slices), the array s2 is a view into the same block of memory used by a. An in-place change in one will change them both.

You can use boolean array indexing. First, create a mask of indices you want to set to 0 and then apply the mask to array and assign the replacement value (e.g., 0 in your case).

mask = b>np.arange(a.shape[1])[:,None]
a[~mask]=0

output:

array([[1, 2, 3],
       [4, 0, 6],
       [0, 0, 0]])

Here is a method that use array broadcast:

x = np.random.randint(1, 63, 10)
y = np.random.randint(1, 63, 10)

dy, dx = [grid.astype(int) for grid in np.mgrid[-1:1:3j, -1:1:3j]]
Y = dy[None, :, :] + y[:, None, None]
X = dx[None, :, :] + x[:, None, None]

then you can use a[Y, X] to select blocks from a. Here is an example code:

img = np.zeros((64, 64))
img[Y, X] = 1

Here is graph ploted by pyplot.imshow():

We can use broadcasting to create a mask of places to be edited with two sets of comparisons against start and end arrays and then simply assign with boolean-indexing for a vectorized solution -

# Range array for the length of columns
r = np.arange(b.shape[1])

# Broadcasting magic to give us the mask of places
mask = (start[:,None] <= r) & (end[:,None] >= r)

# Boolean-index to select and assign 
b[:len(mask)][mask] = a[:len(mask)][mask]

Sample run -

In [222]: a = np.arange(50).reshape(5,10)
     ...: b = np.zeros(a.shape,dtype=int)
     ...: start = np.array([1,4,7])
     ...: end = np.array([5,6,9]) # different from sample for variety
     ...: 

# Mask of places to be edited
In [223]: mask = (start[:,None] <= r) & (end[:,None] >= r)

In [225]: print mask
[[False  True  True  True  True  True False False False False]
 [False False False False  True  True  True False False False]
 [False False False False False False False  True  True  True]]

In [226]: b[:len(mask)][mask] = a[:len(mask)][mask]

In [227]: a
Out[227]: 
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44, 45, 46, 47, 48, 49]])

In [228]: b
Out[228]: 
array([[ 0,  1,  2,  3,  4,  5,  0,  0,  0,  0],
       [ 0,  0,  0,  0, 14, 15, 16,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  0, 27, 28, 29],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0]])

Use take():

In [87]: m = np.random.random((6, 2))

In [88]: m
Out[88]: 
array([[ 0.6641412 ,  0.31556053],
       [ 0.11480163,  0.00143887],
       [ 0.4677745 ,  0.43055324],
       [ 0.49749099,  0.15678506],
       [ 0.48024596,  0.65701218],
       [ 0.48952677,  0.97089177]])

In [89]: m.take([0, 2, 5], axis=0)
Out[89]: 
array([[ 0.6641412 ,  0.31556053],
       [ 0.4677745 ,  0.43055324],
       [ 0.48952677,  0.97089177]])