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I would like to make a list that contains value of '1's than other random number.

let's say I want 60% of '1's appear out of 10 elements. so 6 elements of 1s and 4 elements with random. here is how I approached.

import numpy as np
import random

# to avoid generating 1s..
list_60 = np.random.randint(2,11, 10) 

array([ 6,  2,  8,  6,  6,  3,  5, 10,  6,  8])


count = 0
percentage = int(len(list_60)*(0.6) + 0.5)
for i in range(0,len(list_60)):
    if count < percentage:
        list_60[i]=0
    count += 1

list_60
array([ 1,  1,  1,  1,  1,  1,  5, 10,  6,  8])


random.shuffle(list_60)
array([ 1,  1,  1,  6,  1,  5,  1,  1,  8, 10])

Precedure STEPS:

  1. create randint from 2 to 10.
  2. loop each element and based on the percentage. and change the element to 1s.
  3. shuffle the list for more variation.

I don't think this is smart way of generating more 1s. Is there fancy/smart way to create weighted randomness?

Any help would be appreciated.

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  • Does it need to be an exact amount of 1s in the result; or do you want to choose each value independently, with a bias? Also, why are you using Numpy for this? Commented Feb 15, 2022 at 1:29
  • Yes it has to be exact amount of 1s. 60% of 10 elements in the list. So 6 elements of 1s should be there in the list. I used numpy library to generate randoms. Commented Feb 15, 2022 at 1:34
  • thanks just edited! Commented Feb 15, 2022 at 1:41
  • Initialize the list to have exactly six 1's, then add four more random numbers. Commented Feb 15, 2022 at 2:01

1 Answer 1

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You can get a random subset of indices, and then set those indices with 1:

import numpy as np

arr = np.random.randint(2, 11, 10)
index = np.random.choice(len(arr), int(len(arr) * 0.6), replace=False)
arr[index] = 1

print(arr)

You can also do this without numpy:

import random

arr = [random.randint(2, 11) for _ in range(10)]
index = random.sample(range(len(arr)), int(len(arr) * 0.6))
for i in index:
    arr[i] = 1

print(arr)

The above two implementations use 10 + 6 random bits. You technically only need 4 + 4 (4 for the random numbers, and 4 for their random positions (thanks to @KellyBundy for noticing this). You can achieve this in numpy with:

import numpy as np

arr = np.ones(10)
index = np.random.choice(len(arr), int(len(arr) * 0.4), replace=False)
arr[index] = np.random.randint(2, 11, len(index))

print(arr)

Or even simpler using plain python:

import random

arr = [1] * 10
for i in random.sample(range(len(arr)), int(len(arr) * 0.4)):
    arr[i] = random.randint(2, 11)

print(arr)
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4 Comments

How about minimizing the use of random data? Like, instead of 10+6, only 4+4 random numbers.
The shuffle one looks like 4+10 to me, though I guess the 10 might use decreasing numbers of random bits. What I meant is this, and I'm curious about the best numpy way to do that (I'm a numpy noob).
you're right, still 10+4, lemme give it another crack
edited my response to fix the error

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