numpy: how to fill multiple fields in a structured array at once

Question

Very simple question: I have a structured array with multiple columns and I'd like to fill only some of them (but more than one) with another preexisting array.

This is what I'm trying:

strc = np.zeros(4, dtype=[('x', int), ('y', int), ('z', int)])
x = np.array([2, 3])
strc[['x', 'y']][0] = x

This gives me this future warning:

main:1: FutureWarning: Numpy has detected that you (may be) writing to an array returned by numpy.diagonal or by selecting multiple fields in a record array. This code will likely break in a future numpy release -- see numpy.diagonal or arrays.indexing reference docs for details. The quick fix is to make an explicit copy (e.g., do arr.diagonal().copy() or arr[['f0','f1']].copy()).

But even though this is a warning, the structured array doesn't get filled. So far I'm iterating over both arrays and it works but I guess that's highly inefficient. Is there a better way?

stackoverflow.com/questions/3058602/… points out that trying to index several fields produces a copy, not a view. If you are working with more rows than fields, there's nothing wrong with iterating over fields. — hpaulj
– hpaulj, Commented Feb 17, 2014 at 5:59

HYRY · Accepted Answer · 2014-02-17 01:32:08Z

12

If all the field have the same dtype, you can create a view:

import numpy as np
strc = np.zeros(4, dtype=[('x', int), ('y', int), ('z', int)])
strc_view = strc.view(int).reshape(len(strc), -1)
x = np.array([2, 3])
strc_view[0, [0, 1]] = x

If you want a common solution that can create columns views of any structured array, you can try:

import numpy as np
strc = np.zeros(3, dtype=[('x', int), ('y', float), ('z', int), ('t', "i8")])

def fields_view(arr, fields):
    dtype2 = np.dtype({name:arr.dtype.fields[name] for name in fields})
    return np.ndarray(arr.shape, dtype2, arr, 0, arr.strides)

v1 = fields_view(strc, ["x", "z"])
v1[0] = 10, 100

v2 = fields_view(strc, ["y", "z"])
v2[1:] = [(3.14, 7)]

v3 = fields_view(strc, ["x", "t"])

v3[1:] = [(1000, 2**16)]

print strc

here is the output:

[(10, 0.0, 100, 0L) (1000, 3.14, 7, 65536L) (1000, 3.14, 7, 65536L)]

edited Feb 17, 2014 at 1:32

answered Feb 17, 2014 at 0:24

HYRY

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1 Comment

Federico Barabas Over a year ago

Thank you HYRY, I think your solution is too much for my little unefficiency uneasyness but it may be useful for someone else

Friedrich -- Слава Україні · Accepted Answer · 2024-06-20 15:54:38Z

0

Your example gives now (numpy v1.18.5) the error ValueError: setting an array element with a sequence.. But it will work with

strc[['x', 'y']][0] = tuple(x)

Another example

A = np.zeros(4, dtype=[('x', int), ('y', float), ('z', 'U3')])

A[['x', 'z']][0] = 44, 'foo'

print(*A, sep="\n")
# (44, 0., 'foo')
# (0, 0., '')
# (0, 0., '')
# (0, 0., '')

Note also that when first indexing the row A[0][['x', 'z']] = 44, 'foo', this will raise IndexError: invalid index; but A[0:1][['x', 'z']] = 44, 'foo' will work.

edited Jun 20, 2024 at 15:54

answered Jun 20, 2024 at 15:13

Friedrich -- Слава Україні

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