For instance if you want to remove the third column from an array of shape (2, 3) :

import numpy as np
a = np.ones((2, 3))
b = np.delete(a, 2, axis=1)

Note that delete does not work in-place, so a is unmodified. If you want to keep working on a do :

a = np.delete(a, 2, axis=1)

This will assign the new array to the same variable.

The second argument of numpy.delete is not the value to be deleted. It is the index (or indices) to be deleted. Take a look at the examples in the docstring, such as this one:

In [25]: arr = np.array([[1,2,3,4], [5,6,7,8], [9,10,11,12]])

In [26]: arr
Out[26]: 
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])

In [27]: np.delete(arr, 1, axis=0)
Out[27]: 
array([[ 1,  2,  3,  4],
       [ 9, 10, 11, 12]])

The value 1 in that function call, combined with axis=0, means "delete the row with index 1".

It's because delete function, returns a copy of changed array.
You need to do this:

import numpy as np

m = np.array([[1], [1], [3]])
m = np.delete(m, 2)
print(m)

Don't call np.delete in a loop. It would be quicker to use boolean indexing:

In [6]: A[X.astype(bool).any(axis=0)]
Out[6]: array([[3, 4, 5]])

X.astype(bool) turns 0 into False and any non-zero value into True:

In [9]: X.astype(bool).any(axis=0)
Out[9]: array([False,  True, False], dtype=bool)

the call to .any(axis=0) returns True if any value in a column of X.astype(bool) is True, and False otherwise.

Removing items from a list (or array) while looping over the same list is a classic pitfall. The problem is cause by the fact that removing items from the list changes the meaning of ordinal indexing so that if you use ordinal indexing to remove other items, you may end up removing the wrong items, or get IndexError if you try to index beyond the valid range for the modified list.

In your case, when you loop through rows of A and also delete rows of A:

for j in xrange(np.shape(X)[1]):
    if X[i,j] == 0.0:
        A = np.delete(A, j, axis=0)

each time you modify A the index into A changes. So

        A = np.delete(A, 0, axis=0)

deletes the first row of the original A, but now the new A only has two rows. So

        A = np.delete(A, 2, axis=0)

raises an IndexError since 2 refers to the third row, and the new A does not have a third row. This problem is only exacerbated if the i-loop causes A = np.delete(A, j, axis=0) to be called multiple times too.

I'm not totally sure what you're trying to do programmatically, but I think the problem might lie in the way you're selecting your values. The indices of a specific value in radiocut are going to be different than in values because you've deleted a bunch of stuff in front of it. It might be easier to do all the filtering in one step with numpy's logical array functions, e.g.:

radioloud = np.delete(values, np.logical_and(quiet, indices))

To me, it appears that you are simply trying to delete an index that is out of the array; hence the lack of change ... From your code len(scale) gives only 17 .

For the record as the doc indicates, numpy.delete(arr,obj) will try to delete the element returned by arr[obj] for a 1-D array so :

• numpy.delete(arr,0)
• numpy.delete(arr,[0])
• numpy.delete(arr,0.0)
• numpy.delete(arr,[0.0])

will all delete arr[0] which is the zero-th element of that 1-D array.

numpy.delete() is not an in-place operation, and returns a new copy of your array that has been operated on. As a result, ponto itself is never changed, you just return a changed copy of it and do nothing with it. Assign a new variable to your numpy.delete() statements, and print that. Also note try-except-pass is inherently evil and you could in fact have 100 things wrong in your code without knowing as it will hide all the errors.

You can find the documentation concerning numpy.delete which explains this behavior here.