[mrhlth]

Hi there guys;

im supposed to return this table:

1 2 3
0 3 4
0 0 5

in the form of a list of lists, so the function listoflists will return as so: [[1,2,3],[3,4],[5]]

my attempt:
ive planned to first create 3 arrays of size 3 full of zeroes, as such: [[0,0,0],[0,0,0],[0,0,0]]
and then further append them with the numbers i need, but this didnt really work out well for me, as i needed to delete the extra zeroes but couldnt, and the loops used to append everything into the lists confused the hell out of me. any help will be appreciated, thanks!

[SuperOscar]

Originally Posted by mrhlth

Hi there guys;

im supposed to return this table:

1 2 3
0 3 4
0 0 5

in the form of a list of lists, so the function listoflists will return as so: [[1,2,3],[3,4],[5]]

Since you didn’t specify what’s being sent TO the function, I guess this’ll do:

def listoflists():
    return [[1, 2, 3], [3, 4], [5]]

[mrhlth]

Originally Posted by SuperOscar

Since you didn’t specify what’s being sent TO the function, I guess this’ll do:

def listoflists():
    return [[1, 2, 3], [3, 4], [5]]

ah my bad! the parameter for listoflists is int n, when n = 4, the table is:
1 2 3
0 3 4
0 0 5

and if n = 5, table:
1 2 3 4
0 3 4 5
0 0 5 6
0 0 0 7

so the input for this is n, and it returns a list of lists based on a table created, with dimensions n-1 x n-1, sorry for the inconvenience!

[b49P23TIvg]

Following program is close but runs into index error. I found the bug and fixed it. Can you? I found a simple dependence of A[i][j] on i and j. The line marked with ***1 is unneeded, and the dependence of a[i][j] on with the previous row can be eliminated on the line marked ***2 . Can you find a simplification?

def f(n):
    '''
        whacko array creator
        >>> f(4) == [[1,2,3],[0,3,4],[0,0,5]]
        True
    '''
    A = [[0 for i in range(n - 1)] for j in range(n - 1)]
    A[0][:] = range(1, n) # first row  ******1
    for i in range(1, n - 1): # (yes, this range would need to change as well to answer the second question)
        for j in range(i, n):
            A[i][j] = A[i-1][j] + 1  # ******2
    return A

[mrhlth]

Originally Posted by b49P23TIvg

Following program is close but runs into index error. I found the bug and fixed it. Can you? I found a simple dependence of A[i][j] on i and j. The line marked with ***1 is unneeded, and the dependence of a[i][j] on with the previous row can be eliminated on the line marked ***2 . Can you find a simplification?

def f(n):
    '''
        whacko array creator
        >>> f(4) == [[1,2,3],[0,3,4],[0,0,5]]
        True
    '''
    A = [[0 for i in range(n - 1)] for j in range(n - 1)]
    A[0][:] = range(1, n) # first row  ******1
    for i in range(1, n - 1): # (yes, this range would need to change as well to answer the second question)
        for j in range(i, n):
            A[i][j] = A[i-1][j] + 1  # ******2
    return A

Thank you SO MUCH for your help! i found the bug and fixed it, now its giving me the correct answers! Thank you again!