How can I make this matrix in python without using numpy?-CodePudding

I have to make a matrix thats N by N and the example im given looks like this:

So what I get from the example is that its gonna take the number N is (4 in this example since its 4 by 4) and print the number on the top row first column then fill it with zeros and then go down one line and print N -1 in the first two columns and then zeros. My code looks like this atm:

def fill_matrix(n): #Fills the matrix with 0s
    # Llena la matriz
    for r in range(n):
        row = []
        for c in range(n):
            row.append(0)
        matrix.append(fila)
    return matrix

def print_matrix(matriz): #Prints the matrix
    rows = len(matriz)
    columns = len(matriz[0])
    for f in range(row):
        for c in range(columns):
            print ("=" %matrix[f][c], end="")
        print()
       
# Programa principal
side = int(input("Input the size of the matrix: ")) #Input of N by N
while side < 1:
    print("Size must be bigger than 0")
    side = int(input("Input the size of the matrix: "))
matrix = []
fill_matrix(side)
print_matrix(matrix)

How can I make this matrix look like the one in the exercise?

CodePudding user response：

Use list comprehension:

N = 4
>>> [[N-i]*(i 1) [0]*(N-i-1) for i in range(N)]
[[4, 0, 0, 0], [3, 3, 0, 0], [2, 2, 2, 0], [1, 1, 1, 1]]

In a function:

def fill_matrix(N):
    return [[N-i]*(i 1) [0]*(N-i-1) for i in range(N)]

def print_matrix(m):
    print("\n".join(["\t".join(map(str, row)) for row in m]))

>>> fill_matrix(6)
[[6, 0, 0, 0, 0, 0],
 [5, 5, 0, 0, 0, 0],
 [4, 4, 4, 0, 0, 0],
 [3, 3, 3, 3, 0, 0],
 [2, 2, 2, 2, 2, 0],
 [1, 1, 1, 1, 1, 1]]

>>> print_matrix(fill_matrix(6))
6   0   0   0   0   0
5   5   0   0   0   0
4   4   4   0   0   0
3   3   3   3   0   0
2   2   2   2   2   0
1   1   1   1   1   1

The ith row consists of:

The number N-i repeated i 1 times
0 repeated N-(i 1) times

CodePudding user response：

Similar to the excellent approach by @not_speshal, slight more readability can be achieved with reversed(range())

n = 4

[[i 1]*(n-i) [0]*(i) for i in reversed(range(n))]

[[4, 0, 0, 0], 
 [3, 3, 0, 0], 
 [2, 2, 2, 0], 
 [1, 1, 1, 1]]

CodePudding user response：

not_speshal has already given you the list-comprehension solution, but you might find it easier to understand the logic by writing a full loop. Let's analyze the matrix you want:

Row #   Value   Count
0       4       1
1       3       2
2       2       3
3       1       4
# General case:
i       (N - i) (i   1)

So your ith row needs to contain the number N - i, i 1 times, and then the rest should be zeros.

def fill_matrix(N):
    # First, let's create a matrix of zeros
    matrix = [ [0] * N for _ in range(N) ]
    
    # Now, iterate over rows
    for i in range(N):
        # Set the first (i   1) values to (N - i)
        for j in range(i   1):
            matrix[i][j] = N - i

    return matrix

To run this:

>>> fill_matrix(4)
 [[4, 0, 0, 0], 
  [3, 3, 0, 0], 
  [2, 2, 2, 0], 
  [1, 1, 1, 1]]

If you want to reduce the iterations by getting rid of the zeros-matrix creation, but you will have to append zeros to fill out each row after the for j in range(i 1) loop.

def fill_matrix(N):
    # An empty list that will contain our rows
    matrix = []

    # Now, iterate over rows
    for i in range(N):
        row = []
        # Set the first (i   1) values to (N - i)
        for _ in range(i   1):
            row.append(N - i)
        # Set the remaining values to 0
        for _ in range(N - len(row)):
            row.append(0)
        matrix.append(row)
    return matrix