How to create a list of lists where each sub-list 'increments' as follows: [1, 0, 0], [1, 1, 0], [1, 1, 1] [closed]

Question

This works but is unwieldy and not very 'Pythonic'. I'd also like to be able to run through different values for 'numValues', say 4 to 40...

innerList = []
outerList = []
numValues = 12
loopIter = 0

for i in range(numValues):
    innerList.append(0)

for i in range(numValues):
    copyInnerList = innerList.copy()
    outerList.append(copyInnerList)

for i in range(len(innerList)):
    for j in range(loopIter + 1):
        outerList[i][j] = 1
    loopIter += 1

print(outerList)

Answer 1

numValues = 12
result = [ [1] * i + [0] * (numValues - i) for i in range(1, numValues+1) ]

Answer 2

You can do this as a nested list comprehension, with two iterators over range(numValues) and only setting a 1 when the second iterator is <= the first:

numValues = 4

outerList = [[1 if j <= i else 0 for j in range(numValues)] for i in range(numValues)]
print(outerList)

Output:

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

Answer 3

If numpy is an option, this can be done very easily with np.tril:

import numpy as np

n=5
out = np.ones((n,n))
np.tril(out)

array([[1., 0., 0., 0., 0.],
       [1., 1., 0., 0., 0.],
       [1., 1., 1., 0., 0.],
       [1., 1., 1., 1., 0.],
       [1., 1., 1., 1., 1.]]) 

Answer 4

I think it is a little bit more intuitive using a matrix approach with numpy.

numValues = 5
my_array = np.eye(numValues)

it results in

array([[1., 0., 0., 0., 0.],
       [0., 1., 0., 0., 0.],
       [0., 0., 1., 0., 0.],
       [0., 0., 0., 1., 0.],
       [0., 0., 0., 0., 1.]])

From this matrix you can see that the only think you need to do is to sum the rows

sol =  [np.sum(mat[:i], axis=0) for i in range(numValues+1)][1:]

and you get

 [array([1., 0., 0., 0., 0.]),
 array([1., 1., 0., 0., 0.]),
 array([1., 1., 1., 0., 0.]),
 array([1., 1., 1., 1., 0.]),
 array([1., 1., 1., 1., 1.])]