How to use zfill to make length equivalent to 2 digit length of n

Question

I have asked a previously one question where I was impressed with unpacking approach shared by one person I am bit playing with bit to print patterns

I want to pad the single digit with zero to make it equivalent to 2 digit number length

I tried zfill but not throwing me error : AttributeError: 'int' object has no attribute 'zfill'

Code :

n=15
for i in range(1,n+1):
  print(*range(1,n+1))
  n = n -1

Expected output :

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15
01 02 03 04 05 06 07 08 09 10 11 12 13 14
01 02 03 04 05 06 07 08 09 10 11 12 13
01 02 03 04 05 06 07 08 09 10 11 12
01 02 03 04 05 06 07 08 09 10 11
01 02 03 04 05 06 07 08 09 10
01 02 03 04 05 06 07 08 09
01 02 03 04 05 06 07 08
01 02 03 04 05 06 07
01 02 03 04 05 06
01 02 03 04 05
01 02 03 04
01 02 03
01 02
01

For `zfill`, you would need to do `print(*str(j).zfill(2) for j in range(1,n+1))` — Ben Grossmann, Jan 17 '22 at 16:49
[`zfill`](https://docs.python.org/3/library/stdtypes.html#str.zfill) only exists for strings: `str(value).zfill(2)`. — Tom Aarsen, Jan 17 '22 at 16:49
@c Alternatively, `print(*map(lambda x: str(x+1).zfill(2),range(n)))` — Ben Grossmann, Jan 17 '22 at 16:52

S.B · Accepted Answer · 2022-01-17T16:51:47.567

You can use format string syntax 02.

change the line : print(*range(1,n+1)) to print(*(f"{i:02}" for i in range(1, n + 1)))

Full code:

n = 15
for i in range(1, n + 1):
    print(*(f"{i:02}" for i in range(1, n + 1)))
    n = n - 1

output:

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15
01 02 03 04 05 06 07 08 09 10 11 12 13 14
01 02 03 04 05 06 07 08 09 10 11 12 13
01 02 03 04 05 06 07 08 09 10 11 12
01 02 03 04 05 06 07 08 09 10 11
01 02 03 04 05 06 07 08 09 10
01 02 03 04 05 06 07 08 09
01 02 03 04 05 06 07 08
01 02 03 04 05 06 07
01 02 03 04 05 06
01 02 03 04 05
01 02 03 04
01 02 03
01 02
01

Another option is to use zfill like :

print(*(str(i).zfill(2) for i in range(1, n + 1)))

you need to convert the i into str because zfill is a method of string objects.

@coreke2042 `print(*(str(i).zfill(2) for i in range(1, n + 1)))` I will add it to the answer. — S.B, Jan 17 '22 at 16:50

score 0 · Answer 2 · answered Jan 17 '22 at 16:50

0

There are many ways to do this. Here's one of them:

n = 15

for i in range(1, n+1):
    print(' '.join(f'{j:02d}' for j in range(1, n-i+2)))

answered Jan 17 '22 at 16:50

DarkKnight

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How to use zfill to make length equivalent to 2 digit length of n

2 Answers2