How to create increment 2D array in python

Question

• How to create increment 2D list

• if input = 3 then output is

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

• if input = 4 then output is

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

Code is below to print every number to Zero

• How to increment of [x][y]

w = 3
matrix = [[0 for x in range(w)] for y in range(w)] 
matrix

or

rows = 3
cols = 3

matrix = []
for i in range(rows):
  row = []
  for j in range(cols):
    row.append(0)
  matrix.append(row)

print(matrix)

Answer 1

You could try with a list comprehension:

matrix = [[x + 1 for x in range(y + 1)] + [0] * (w - y - 1) for y in range(w)]
print(matrix)

Output for 3:

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

Output for 4:

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

If you want to edit the value of this list in the future, use:

matrix = [[x + 1 for x in range(y + 1)] + [0 for x in range(w - y - 1)] for y in range(w)]

Because multiplying a list creates unexpected modifications for modifying.

More about this here.

Edit:

If you don't want a list comprehension, try:

matrix = []
for y in range(w):
    l1 = []
    for x in range(y + 1):
        l1.append(x + 1)
    l1.extend([0] * (w - y - 1))
    matrix.append(l1)

Answer 2

Using double-comprehension for 2D list:

>>> [list(range(1, i+1))+[0]*(n-i) for i in range(1,n+1)]

OUTPUT:

[[1, 0], [1, 2]] #n=2
[[1, 0, 0], [1, 2, 0], [1, 2, 3]]  #n=3
[[1, 0, 0, 0], [1, 2, 0, 0], [1, 2, 3, 0], [1, 2, 3, 4]] #n=4

Non-comprehension solution:

result = []
for i in range(1, n+1):
    row = list(range(1, i+1))+[0]*(n-i)
    result.append(row)