通过对角填充元素将列表转换为矩阵

Question

I have a sorted list which looks as below:

mylist = [-2, -2, 1, 1, 4, 4, 3, 3, 3]

The list is sorted in ascending order based upon the number of times it appears. In case of a tie, the list is sorted based upon the values.

I need to convert this list into a square matrix of equal chunks(3*3 in this case) such that the numbers are placed "diagonally" starting from the bottom right corner .

The general case is to divide the list in equal chunks.

Desired Output:

res = [[3, 3, 4],
      [3, 4, 1],
      [1, -2, -2]]

I have written the below code but still not able to get the desired output:

def create_matrix(lst, n):
    for i in range(0, len(lst), n):
        print(i)
        yield lst[i: i+n]

m = create_matrix(mylist, 3)

print(list(m))

One solution could be to place pairs in queue/stack and then pop as needed.

Answer 1

I assume you want to iterate the output matrix this way (NxN = 5x5 example, 0->24 order):

[[ 0,  1,  3,  6, 10],
 [ 2,  4,  7, 11, 15],
 [ 5,  8, 12, 16, 19],
 [ 9, 13, 17, 20, 22],
 [14, 18, 21, 23, 24]]

For each cell, the coordinates (i,j) have their sum equal to the number of the diagonal ( k ) from top-left to bottom right (2*N-1 diagonals in total)

For the N first diagonals, the first item has i=0 , the following ones i=kN where k is the diagonal number.

The last item has i=k with a maximum of N .

j = ki

This gives us the following algorithm to iterate the cells in order:

import math

mylist = [-2, -2, 1, 1, 4, 4, 3, 3, 3]

N = int(math.sqrt(len(mylist))) # 3

out = [[None for _ in range(N)]
       for _ in range(N)]

r = reversed(mylist)

for k in range(2*N-1):
    start = 0 if k<N else k-N+1
    stop = min(k, N-1)
    for i in range(start, stop+1):
        out[i][k-i] = next(r)

print(out)

Output:

[[3, 3, 4],
 [3, 4, 1],
 [1, -2, -2]]

Answer 2

Here's some code to wrap a list backwards up a matrix, which seems to solve this case. Also, your question doesn't make any sense. Diagonally? And how is that list sorted?

def create_matrix(lst, n):
    ret = [[0 for i in range(n)] for i in range(n)]
    for i in range(n*n-1,-1,-1):
        ret[i//n][i%n] = lst[n*n-1-i]
    return ret

Edit: That doesn't work. I don't think anyone knows what you mean by diagonally.