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.

CodePudding user response：

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(10), 4):
    arr[i] = random.randint(2, 11)

print(arr)