How to create a uniform structured array in numpy?

Question

I could create the following strucutured array in numpy:

dt = np.dtype([('n', 'i4'),('x', 'f8'), ('y', 'f8'), ('z', 'f8')])
arr = np.array((1, 5.0, 6.0, 7.0))

This creates the array:

array((1, 5., 6., 7.), dtype=[('n', '<i4'), ('x', '<f8'), ('y', '<f8'), ('z', '<f8')])

In dt the last three columns are all floats f8 is there a shorter way to declare dt when there are multiple consecutive columns of the same type?

dt = np.dtype([('n', 'i4'), *zip('xyz', ['f8']*3)])?

Corralien
– Corralien

2022-01-15 13:45:29 +00:00
Commented Jan 15, 2022 at 13:45 — Corralien
– Corralien, Commented Jan 15, 2022 at 13:45

Corralien · Accepted Answer · 2022-01-15 13:58:27Z

1

Not faster, but you can create an helper function to generate dtype:

def gen_dtype(names, override=None, default='f8'):
    d = dict(zip(names, [default]*len(names)))
    d.update(override)
    return np.dtype(list(d.items()))

dt  = gen_dtype(list('nxyz'), {'n': 'i4'})

Output:

>>> dt
dtype([('n', '<i4'), ('x', '<f8'), ('y', '<f8'), ('z', '<f8')])

answered Jan 15, 2022 at 13:58

Corralien

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kcw78 · Accepted Answer · 2022-01-15 20:06:51Z

0

An alternative method is to use the dictionary format to define the dtype:

dt2 = np.dtype( {'names': ['n','x','y','z'],
                 'formats': ['i4'] + ['f8']*3} )

answered Jan 15, 2022 at 20:06

kcw78

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