numpy array with n prefilled columns

Question

I need to specialized numpy arrays. Assume I have a function:

    def gen_array(start, end, n_cols):

It should behave like this, generating three columns where each column goes from start (inclusive) to end (exclusive):

>>> gen_array(20, 25, 3)
array([[20, 20, 20],
       [21, 21, 21],
       [22, 22, 22],
       [23, 23, 23],
       [24, 24, 24]])

My rather naïve implementation looks like this:

def gen_array(start, end, n_columns):
    a = np.arange(start, end).reshape(end-start, 1) # create a column vector from start to end
    return np.dot(a, [np.ones(n_columns)])          # replicate across n_columns

(It's okay, though not required, that the np.dot converts values to floats.)

I'm sure there's a better, more efficient and more numpy-ish way to accomplish the same thing. Suggestions?

Update

Buildin on a suggestion by @msi_gerva to use np.tile, my latest best thought is:

def gen_array(start, end, n_cols):
    return np.tile(np.arange(start, end).reshape(-1, 1), (1, n_cols))

... which seems pretty good to me.

One way to do this: a = np.tile(np.arange(20,25,1),(3,1)).T . Not very elegant (using transpose in the end), but at least a one-liner. — msi_gerva
– msi_gerva, Commented Nov 8, 2022 at 17:36
tile looks good - How about reshaping first: a = np.tile(np.arange(20,25)).reshape(-1,1), (1, 3)) — fearless_fool
– fearless_fool, Commented Nov 8, 2022 at 17:45
Instead of dot just multiply, a * np.ones(n_columns). That's a (5,1) times a (3,) broadcasting to (5,3). — hpaulj
– hpaulj, Commented Nov 8, 2022 at 18:59

ILS · Accepted Answer · 2022-11-08 17:43:55Z

2

In addition to numpy.arange and numpy.reshape, use numpy.repeat to extend your data.

import numpy as np

def gen_array(start, end, n_cols):
    return np.arange(start, end).repeat(n_cols).reshape(-1, n_cols)

print(gen_array(20, 25, 3))
# [[20 20 20]
#  [21 21 21]
#  [22 22 22]
#  [23 23 23]
#  [24 24 24]]

answered Nov 8, 2022 at 17:43

ILS

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1 Comment

fearless_fool Over a year ago

That feels more logical (and perhaps more efficient) than using tile(). Thanks...

Okapi575 · Accepted Answer · 2022-11-09 13:43:18Z

0

The simplest I found:

The [:,None] adds a dimension to the array.

np.arange(start, end)[:,None]*np.ones(n_cols)

edited Nov 9, 2022 at 13:43

answered Nov 8, 2022 at 22:06

Okapi575

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2 Comments

fearless_fool Over a year ago

Yes, but that doesn't allow you to specify the number of columns as a parameter. How would you create 25 columns?

Okapi575 Over a year ago

I edited the answer to reflect your variable names, the [:,None] creates the extra dimension, and the np.ones determines the size via broadcasting

w-m · Accepted Answer · 2022-11-09 13:48:49Z

0

np.arange(start, end)[:, np.newaxis].repeat(n_cols, axis=1)

answered Nov 9, 2022 at 13:48

w-m

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