Populate data from a list

Question

I have a list with the following items

l = [11.1, 22.2, 33.3, 11.1, 33.3, 33.3, 22.2, 55.5]

Each item is a multiple of 11.1 and the length of the list is 8. I would like to generate another list of 30 items with values 11.1, 22.2, 33.3, 55.5 present in the original list l.

I would like to know how to populate data from the list l to l_new.

Do you want to generate `l_new` at random with selected values? — Andrej Kesely, Oct 03 '20 at 15:02
@AndrejKesely Yes, you are right. Or, is it possible to generate l_new such that `l` and `l_new` have the same distribution — Natasha, Oct 03 '20 at 16:01
@CarloZanocco I hope the above comment tries to answer your question partially — Natasha, Oct 03 '20 at 16:16
@CarloZanocco Thank you. It almost solved my problem. But is it possible to check if `l` and `l_new` follow the same distribution pattern? — Natasha, Oct 04 '20 at 09:12
Take a look [here for the distributions](https://stackoverflow.com/questions/37559470/what-do-all-the-distributions-available-in-scipy-stats-look-like/37559471#37559471) — Carlo Zanocco, Oct 05 '20 at 06:42

Carlo Zanocco · Accepted Answer · 2020-10-03T16:33:44.167

1

You can use the random module to do it:

import random

l = [11.1, 22.2, 33.3, 11.1, 33.3, 33.3, 22.2, 55.5]

l_new = [random.choice(l) for _ in range(0, 30)]
print(l_new)

#OUTPUT:
#[11.1, 11.1, 22.2, 33.3, 22.2, 11.1, 33.3, 11.1, 55.5, 11.1, 33.3, 22.2, 55.5, 22.2, 22.2, 33.3, 11.1, 11.1, 33.3, 22.2, 33.3, 11.1, 11.1, 33.3, 22.2, 33.3, 33.3, 11.1, 33.3, 22.2]

l_new = random.choices(l, k=30)
print(l_new)

#OUTPUT:
#[11.1, 33.3, 33.3, 55.5, 33.3, 33.3, 55.5, 11.1, 22.2, 11.1, 55.5, 11.1, 11.1, 55.5, 22.2, 22.2, 22.2, 33.3, 11.1, 33.3, 55.5, 55.5, 33.3, 11.1, 11.1, 55.5, 22.2, 22.2, 11.1, 22.2]

The first solution l_new = [random.choice(l) for _ in range(0, 30)] use list comprehension and the random.choice() function that select one item from l for each iteration.

The second solution l_new = random.choices(l, k=30) just call the choices() function and let it generate the list, you have to specify the k that is the number of element to select.

There is another way that require the numpy module:

import numpy

l = [11.1, 22.2, 33.3, 11.1, 33.3, 33.3, 22.2, 55.5]

l_new = list(numpy.random.choice(l, size=30))
print(l_new)

#OUTPUT:
#[11.1, 33.3, 11.1, 22.2, 33.3, 22.2, 22.2, 33.3, 55.5, 33.3, 22.2, 33.3, 22.2, 55.5, 33.3, 33.3, 33.3, 55.5, 33.3, 11.1, 11.1, 11.1, 55.5, 11.1, 33.3, 33.3, 22.2, 22.2, 33.3, 22.2]

The list is generated by numpy.random.choice

edited Oct 03 '20 at 16:33

answered Oct 03 '20 at 16:28

Carlo Zanocco

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I would like to know if we can fix the seed while using `numpy.random.choice`. Thanks a lot for the solution posted. – Natasha Oct 09 '20 at 08:32
This should do it: https://stackoverflow.com/a/52991726/5532710 – Carlo Zanocco Oct 09 '20 at 08:34
`from numpy.random import MT19937 from numpy.random import RandomState, SeedSequence rs = RandomState(MT19937(SeedSequence(123456789))) # Later, you want to restart the stream rs = RandomState(MT19937(SeedSequence(987654321)))` is provided [here](https://numpy.org/doc/stable/reference/random/generated/numpy.random.seed.html). I am not sure how to use this in the answer posted. Could you please help? – Natasha Oct 09 '20 at 08:51
Can you create a new question regarding this please and explain what you want to achieve? – Carlo Zanocco Oct 09 '20 at 08:52
https://stackoverflow.com/questions/64276987/how-to-fix-the-seed-while-using-random-choice – Natasha Oct 09 '20 at 09:07

Populate data from a list

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