如何垂直打印数字列表？

Question

I am trying to solve an exercise. I should write a function that takes a list of integers, converts it to a string of numbers displayed vertically.

mylist = [5,69,2090]

the function should return/print the following string:

    2
    0
  6 9
5 9 0

I tried to solve it with the following code, but it doesn't help me

def printstring(mylist):
    h = len(mylist)//3
    for i in range(h):
        print(mylist[i],mylist[h+i],l[h+i+1])

Answer 1

lst = [5,69,2090]

n = max(len(str(i)) for i in lst)
for i in zip(*['{:>{n}}'.format(i, n=n) for i in lst]):
    print(' '.join(i))

Prints:

    2
    0
  6 9
5 9 0

---

For lst = [1,123,12,12345,1234] :

      1  
      2 1
  1   3 2
  2 1 4 3
1 3 2 5 4

Answer 2

First convert your list items to a string then find the maximum length. Pad your strings to fit that length and print them:

lst = [5, 69, 2090]
lst = list(map(str, lst))
max_lenght = max(map(len, lst))
padded_lst = [item.rjust(max_lenght, " ") for item in lst]
for row in zip(*padded_lst):
    print(" ".join(row))

    2
    0
  6 9
5 9 0

Answer 3

I'd apply a few steps here. First, determine the maximum length of digits for each number by converting them to strings. Secondly, left pad each number string to that length using rjust . Finally, rotate the matrix 90 degrees using the classic zip(*m) trick and join it all back into a printable square.

>>> lst = [5,69,2090]
>>> size = max(map(len, map(str, lst)))
>>> m = [str(x).rjust(size, " ") for x in lst]
>>> print("\n".join([" ".join(x) for x in zip(*m)]))
    2
    0
  6 9
5 9 0

Answer 4

A not so much readable two liner

sl = ['{:>{}d}'.format(s, len(str(max(mylist)))) for s in mylist]
print('\n'.join(' '.join(s) for s in zip(*sl)))