如何将 python 矩阵中的 12 个随机 0 更改为 1

Question

My program is suppose to change 12 zeros on random positions to 1 in python 6x6 matrix. This is my code.

from random import randint
M = []
i = 0

for i in range(6):
    M.append([])
    for j in range(6):
        M[i].append(0)

while i < 12:
    index = randint(0,5)
    index2 = randint(0,5)
    if (M[index][index2] == 0):
        M[index][index2] = 1
    i+=1

print(M)

So my matrix is going to look like this at the beggining

[[0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]

So I randomly chose an array and the element in chosen array. The problem is that different number of zeros is changed every time. Not 12 like I want them to change. Does anyone know how to solve this problem?

Answer 1

The problem is that the two indices index1 and index2 are randomly chosen with replacement, so a pair of indices can appear more than once. This means that one value will be set to 1 two or more times.

One slightly complicated way to do this is to create a list of all possible indices and choose a random set from this.

from random import randint, sample
import itertools
M = []
i = 0

for i in range(6):
    M.append([])
    for j in range(6):
        M[i].append(0)

indices = list(itertools.product(range(6), range(6))
random_indicies = sample(indices, 12)
for ind in random_indicies:
    if (M[ind[0]][ind[1]] == 0):
        M[ind[0]][ind[1]] = 1
    i+=1

print(M)

This creates a list indices which contains all pairs of possible indicies. sample picks 12 from these without replacement, which is iterated over and used in the matrix.

A much slower, but functional way is to just keep looping until 12 indices are set to 1

from random import randint
M = []
i = 0

for i in range(6):
    M.append([])
    for j in range(6):
        M[i].append(0)

while i < 12:
    index = randint(0,5)
    index2 = randint(0,5)
    if (M[index][index2] == 0):
        M[index][index2] = 1
        i+=1

print(M)

Which just skips i+=1 if an element in M isn't changed, so it keeps iterating until 12 elements in M are changed.

Answer 2

You could instead make a flat array of length 36, randomly choose 12 indeces, change values at indeces to 1, and then reshape to 6x6 in the end:

import numpy as np

array = np.zeros(36)
indeces = np.random.choice(np.arange(36), 12, replace = False)
array[indeces] = 1
M = np.reshape(array, (6,6))
print(M)

Answer 3

Two issues in the code:

1. just before the while loop the value of "i" var is "5", from previous for loop
2. you should increment the "i" var in the while loop, only if one value has been changed from 0 to 1

Here is the code

from random import randint
M = []
i = 0

for i in range(6):
    M.append([])
    for j in range(6):
        M[i].append(0)

print(M)

i=0
while i < 12:
    index = randint(0,5)
    index2 = randint(0,5)
    if (M[index][index2] == 0):
        M[index][index2] = 1
        i+=1

print(M)

and the output

[[0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
[[0, 1, 1, 1, 0, 1], [0, 1, 0, 1, 1, 1], [0, 0, 0, 1, 0, 0], [1, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0], [1, 0, 0, 0, 0, 0]]

Answer 4

So you have two issues with your code. First off, you use the variable i in your for loop. By the time that's finished running, Python has stored the variable and assigned the value 5 to it. So when you run your while-loop, you're instead starting at 5 rather than zero like how you would like. A simple fix would be to use a different variable or assign i=0 again.

The second problem is your if statement and i+=1 . This makes your while loop subject to repeating probabilities, with a 1/36 chance of having only 1 zero change! Including the i+=1 statement inside the if statement should fix your issue.