Numpy: Create new array from 1d integer array where each element in integer array represents the number of elements in the new array

Question

This seems like it should be straightforward, but I'm stumped (also fairly new to numpy.)

I have a 1d array of integers a.

I want to generate a new 1d array b such that:

the number of elements in b equals the sum of the elements in a
The values in b are equal to some arbitrary constant divided by the corresponding element in a.

That's a mouthful, so here's an example of what I want to make it more concrete:

a = array([2,3,3,4])
CONSTANT = 120
.
.
.
b = array([60,60,
           40,40,40,
           40,40,40,
           30,30,30,30])

Any help is appreciated!

so the number a[i] is the time you repeat and also the divisor? a = [1,2,3] -> b = [c/1, c/2,c/2 , c/3,c/3,c/3,] ? — Ulises Bussi
– Ulises Bussi, Commented Oct 28, 2021 at 0:07

Ulises Bussi · Accepted Answer · 2021-10-28 00:12:45Z

3

I think a pretty clear way to do this is

import numpy as np
a = np.array([2,3,3,4]) 
constant = 120

#numpy.repeat(x,t) repeats the val x t times you can use x and t as vectors of same len
b = np.repeat(constant/a , a)

answered Oct 28, 2021 at 0:12

Ulises Bussi

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3 Comments

erip Over a year ago

This is a much better answer than mine. Nice one!

Ulises Bussi Over a year ago

sorry I edited it so make it more clear and straightforward

Alex Parthemer Over a year ago

Thank you both! Learned 2 new useful functions.

erip · Accepted Answer · 2021-10-28 00:09:39Z

0

You can do this with np.concatenate and good old fashioned zip:

>>> elements = CONSTANT / a
>>> np.concatenate([np.array([e]*n) for e, n in zip(elements, a)])
array([60., 60., 40., 40., 40., 40., 40., 40., 30., 30., 30., 30.])

answered Oct 28, 2021 at 0:09

erip

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2 Comments

Alex Parthemer Over a year ago

Perfect, Thank you!

erip Over a year ago

@AlexParthemer I think you should prefer the other answer. That uses more idiomatic numpy which will probably be better optimized.

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