Pythonic/Numpy way of converting a 1D array into 2D vector array of indexed values

Question

This is what I currently have:

import numpy as np

data = [0.2, 0.6, 0.3, 0.5]
vecs = np.reshape([np.arange(len(data)),data], (2, -1)).transpose()

vecs
array([[ 0.  ,  0.2],
       [ 1.  ,  0.6],
       [ 2.  ,  0.3],
       [ 3.  ,  0.5]])

This gives me the correct data as I want it, but it seems complex. Am I missing a trick?

You could use np.stack instead of reshaping and transposing? — Arvin Kushwaha
– Arvin Kushwaha, Commented Jul 30, 2020 at 9:45

yatu · Accepted Answer · 2020-07-30 09:58:52Z

2

You can simplify with np.stack and transpose:

data = np.array([0.2, 0.6, 0.3, 0.5])

np.stack([np.arange(len(data)), data], axis=1)
array([[0. , 0.2],
       [1. , 0.6],
       [2. , 0.3],
       [3. , 0.5]])

Timings -

a = np.random.random(10000)
%timeit np.stack([np.arange(len(a)), a], axis=1) 
# 26.3 µs ± 1.54 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit np.array([*enumerate(a)])
# 4.51 ms ± 156 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

edited Jul 30, 2020 at 9:58

answered Jul 30, 2020 at 9:45

yatu

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8 Comments

Arvin Kushwaha Over a year ago

I think you have one extra dimension.

yatu Over a year ago

Yup, just realized :) @arvin

Arvin Kushwaha Over a year ago

Might I suggest the use of the axis keyword? I think it'll save you from having to do a transpose while also looking neater. Also, you can directly use range instead of np.arange, though I suppose it's a bit "hacky".

yatu Over a year ago

Thanks for that @arvin Forgot it had that arg. Not hacky, but why not just use np.arange? Numpy has then to cast to float anyways, so we're saving it from some work there

yatu Over a year ago

Timings show quite a difference @arvin . And as I say, range creates a range of integers, which would then have to be cast to float

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Sayandip Dutta · Accepted Answer · 2020-07-30 09:47:11Z

2

You can try enumerate:

>>> np.array([*enumerate(data)])
array([[0. , 0.2],
       [1. , 0.6],
       [2. , 0.3],
       [3. , 0.5]])

answered Jul 30, 2020 at 9:47

Sayandip Dutta

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6 Comments

Konchog Over a year ago

Oh doh. That's very nice. But is it costly?

Sayandip Dutta Over a year ago

Given a choice between arange(len(data)) and enumerate I think enumerate would be much faster.

Arvin Kushwaha Over a year ago

It turns out that it's actually quite a bit slower to utilize enumerate.

Sayandip Dutta Over a year ago

Yep, saw that. It seems like scaling issue for large arrays.

Sayandip Dutta Over a year ago

Right, I was basing my argument on range(len(data)) which returns an iterator rather than arange(len(data)).

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