How do you remove a column from a structured numpy array?

Question

I have another basic question, that I haven\'t been able to find the answer for, but it seems like something that should be easy to do.

Ok, imagine you have a structured

Answer 1

It's not quite a single function call, but the following shows one way to drop the i-th field:

In [67]: a
Out[67]: 
array([(1.0, 2.0, 3.0), (4.0, 5.0, 6.0)], 
      dtype=[('A', '<f8'), ('B', '<f8'), ('C', '<f8')])

In [68]: i = 1   # Drop the 'B' field

In [69]: names = list(a.dtype.names)

In [70]: names
Out[70]: ['A', 'B', 'C']

In [71]: new_names = names[:i] + names[i+1:]

In [72]: new_names
Out[72]: ['A', 'C']

In [73]: b = a[new_names]

In [74]: b
Out[74]: 
array([(1.0, 3.0), (4.0, 6.0)], 
      dtype=[('A', '<f8'), ('C', '<f8')])

Wrapped up as a function:

def remove_field_num(a, i):
    names = list(a.dtype.names)
    new_names = names[:i] + names[i+1:]
    b = a[new_names]
    return b

It might be more natural to remove a given field name:

def remove_field_name(a, name):
    names = list(a.dtype.names)
    if name in names:
        names.remove(name)
    b = a[names]
    return b

Also, check out the drop_rec_fields function that is part of the mlab module of matplotlib.

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Update: See my answer at How to remove a column from a structured numpy array *without copying it*? for a method to create a view of subsets of the fields of a structured array without making a copy of the array.

Answer 2

Having googled my way here and learned what I needed to know from Warren's answer, I couldn't resist posting a more succinct version, with the added option to remove multiple fields efficiently in one go:

def rmfield( a, *fieldnames_to_remove ):
    return a[ [ name for name in a.dtype.names if name not in fieldnames_to_remove ] ]

Examples:

a = rmfield(a, 'foo')
a = rmfield(a, 'foo', 'bar')  # remove multiple fields at once

Or if we're really going to golf it, the following is equivalent:

rmfield=lambda a,*f:a[[n for n in a.dtype.names if n not in f]]