joining outputs corresponding to their original list

Question

Hi I have a nested for loop which processes each element in the list and gives it right alignment and whitespace at the same time.

I have successfully managed to output the individual elements, but I am currently having trouble joining the outputs into their original lists.

Below is my code.

input = [[1, 1, 1], [1, 2, 3], [1, 3, 6]]
space = len(str(input[-1][-1]))

for row in input:
    for e in row:
        new_element = '{:>{}d}'.format(e, space + 1)
        print(new_element)

>>> [[1, 1, 1], [1, 2, 3], [1, 3, 6]]
 # current output
 1
 1
 1
 1
 2
 3
 1
 3
 6 

 # desired output
 1 1 1
 1 2 3
 1 3 6   

I have little clue how to reassemble the outputs into their original groupings. What methods can I use?

Answer 1

you are pretty close! You shouldn't use "input" as the name of your list because it is a python function. I changed it to input2. Try this:

input2 = [[1, 1, 1], [1, 2, 3], [1, 3, 6]]

for row in input2:
    print(" ".join([str(x) for x in row]))

The "join" method joins items in a list by the characters provided in the quotes. You can call join on the list comprehension for list of lists to display each list in the list of lists how you'd like to view it:

out:

>>> for row in input2:
        print(" ".join([str(x) for x in row]))
1 1 1
1 2 3
1 3 6

if I changed the join to a comma:

>>> for row in input2:
...     print(",".join([str(x) for x in row]))
1,1,1
1,2,3
1,3,6 

If you'd like to retain the whitespace in front of each element and store them into a list of their own (I think I understand):

z = []
for row in input2:
    z.append(" ".join([str(x) for x in row]))
print(z)

###['1 1 1', '1 2 3', '1 3 6']

for y in z:
    print(y)

1 1 1
1 2 3
1 3 6

Now you understand "join" a little better!