How to select a row in a numpy structured array and set a value?

Question

I have a numpy structured array that looks like this:

>>> arr
array([('2020-03-26T21:30',    0, 0.),
    ('2020-03-26T21:31',    1, 0.),
    ('2020-03-26T21:32',    2, 0.), ...,
    ('2020-03-27T22:27', 1497, 0.),
    ('2020-03-27T22:28', 1498, 0.),
    ('2020-03-27T22:29', 1499, 0.)],
    dtype=[('time', '<M8[m]'), ('idx', '<i4'), ('value', '<f4')])

I want to select a certain row and set its value as some value. How can I do this?

I tried this:

>>> arr[ np.where(arr['time'] == np.datetime64('now', 'm')) ]['value'] = 10

But it seems it just extracts the row and makes a new array out of the original array. How can I actually set a value and keep it in the original array?

Try putting the field indexing first

hpaulj
– hpaulj

2020-03-27 15:02:44 +00:00
Commented Mar 27, 2020 at 15:02 — hpaulj
– hpaulj, Commented Mar 27, 2020 at 15:02

senderle · Accepted Answer · 2020-03-27 15:03:56Z

1

In general, if you want to assign values to an existing array, you need to assign the value to a view of the original, rather than a copy. When you use array indexing this way, you'll always get a copy, as you've realized.

In the case of record arrays, you can get a view by accessing the field first:

>>> arr['value'][numpy.where(arr['time'] == b'2020-03-26T21:31')] = 100
>>> arr
array([(b'2020-03-26T21:30',    0,   0.),
       (b'2020-03-26T21:31',    1, 100.),
       (b'2020-03-26T21:32',    2,   0.),
       (b'2020-03-27T22:27', 1497,   0.),
       (b'2020-03-27T22:28', 1498,   0.),
       (b'2020-03-27T22:29', 1499,   0.)],
      dtype=[('time', 'S20'), ('idx', '<i4'), ('value', '<f4')])

answered Mar 27, 2020 at 15:03

senderle

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1 Comment

maynull Over a year ago

Thank you for your help!

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