Tuplify a number

Question

def tuplify(number):
    x = str(number)
    if len(x) <= 0:
        return 0
    lst = []
    for i in range(0,10):
        if i == x[len(x)-1]:
            lst.append(i)
        return tuple(lst)

This is my code and I can't get the output I want.

tuplify(102486) which should give me (1, 0, 2, 4, 8, 6)

You probably should return the result outside of your loop. Also, what's wrong with `tuple(str(number))`? — Frédéric Hamidi, Apr 15 '14 at 07:41
@FrédéricHamidi That should be an answer, obviously. Brilliant. — unwind, Apr 15 '14 at 07:43
@FrédéricHamidi the slight problem being it will return `('1','2','3')` not `(1,2,3)` — Burhan Khalid, Apr 15 '14 at 07:52
@BurhanKhalid, absolutely, but the questioner's current code tries to generate a tuple of strings, so I ran with that. — Frédéric Hamidi, Apr 15 '14 at 07:53

thefourtheye · Answer 1 · 2014-04-15T07:55:37.490

You can convert that to a tuple of numbers like this

print tuple(map(int, str(102486)))
# (1, 0, 2, 4, 8, 6)

But note that, leading zeros in the number will NOT be retained.

Apart from the fact that you are comparing a string and a number,

if i == x[len(x)-1]:

you are always comparing i with the last element.

Also, you wanted to return a tuple, but if the string is empty, then you are returning zero

if len(x) <= 0:
    return 0

instead, an empty tuple would have been appropriate

if len(x) == 0:
    return ()

Prashant Gaur · Answer 2 · 2014-04-15T09:00:10.813

2

Edited :

>>> p = 123434455
>>> tuple(int(i) for i in str(p))
(1, 2, 3, 4, 3, 4, 4, 5, 5)
>>>

edited Apr 15 '14 at 09:00

answered Apr 15 '14 at 07:46

Prashant Gaur

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Does not help with his problems, and is even the wrong answer. – RemcoGerlich Apr 15 '14 at 07:48

score 1 · Answer 3 · answered Apr 15 '14 at 07:45

You have several problems in your code, one is this:

 if i == x[len(x)-1]:

As i is an integer, but x is a string, this is never going to be true. 1 is not equal to "1".

You,r return tuple(lst) is also indented so that it is inside the for loop, but it shouldn't be.

But the for loop is completely strange too -- you loop over numbers 0 to 9, and always compare to the last character of x, why? What is it trying to do?

There are one-liner ways to do this but I don't think those help you.

score 0 · Answer 4 · edited May 23 '17 at 11:57

0

I would look at :

you can cast your number to a string, split it and convert each char back to a int

edited May 23 '17 at 11:57

Community

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answered Apr 15 '14 at 07:46

nobe4

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score 0 · Answer 5 · answered Apr 15 '14 at 07:49

You have a few issues with your code.

First you don't need to check the length, because there is no such thing as a "blank" number; so what you need to check is if a number was sent or not. Also you are returning 0 which is not a tuple.

Second, you don't need a list because you can step through a string directly.

Finally, your result will be a tuple of strings, when you need a tuple of integers.

Combining this, you get:

def tuplify(number):
    if not number:
       return ()  # this is an empty tuple
    x = str(number)
    return tuple(map(int,x))

Actually `()` is an empty tuple, `(,)` is a syntax error. Confusingly, `(1,)` is the way to write a one-element tuple. — RemcoGerlich, Apr 15 '14 at 07:50

score 0 · Answer 6 · answered Apr 15 '14 at 07:57

0

Try this ,

>>> def tuplify(number):
    t=tuple()
    number=str(number)
    for i in number:
        t+=tuple(i)
    return t

>>> tuplify(123456)
('1', '2', '3', '4', '5', '6')
>>>

answered Apr 15 '14 at 07:57

Nishant Nawarkhede

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Tuplify a number

6 Answers6