Add 1 to every second number

Question

How do I add 1 to every second number?
Example:

2323 -> 2424
1112 -> 1213
3912 -> 31013

That is what I have now:

def plus_to_every_second(integer):
    integer = str(integer)
    integer_new = ''
    for i in integer:
        if integer.find(i) % 2 != 0:
            integer_new += str(int(i) + 1)
        else:
            integer_new += i

    return integer_new

For some reason, it does not work. But what is the actual solution?

Edit the question to explain "does not work". What happens, what should happen? — Michael Butscher, Feb 26 '22 at 21:10
For one thing, `.find()` returns the _first_ occurrence of the target string. So in your second example string "1112", `.find()` will return 0 for all three occurrences of "1". — John Gordon, Feb 26 '22 at 21:10
Does this answer your question? [Accessing the index in 'for' loops?](https://stackoverflow.com/questions/522563/accessing-the-index-in-for-loops) — Tomerikoo, Feb 26 '22 at 22:47

score 5 · Accepted Answer · edited May 30 '22 at 14:38

5

Don't use find, that only finds the first occurrence, rather iterate the digits together with the position using enumerate:

IIUC, you could do:

def add1(n):
    return int(''.join(str(int(d)+i%2) for i,d in enumerate(str(n))))

add1(2323)
# 2424

add1(1112)
# 1213

add1(3912)
# 31013

using a classical loop

def add1(n):
    s = str(n)
    out = ''
    for i,d in enumerate(s):
        out += str(int(d)+i%2)
    return int(out)

edited May 30 '22 at 14:38

Sunderam Dubey

1
11
20
40

answered Feb 26 '22 at 21:15

mozway

194,879
13
39
75

Could you please code it several lines? – Stamend Feb 26 '22 at 21:46
5

@Stamend sure but know that the comprehension is more pythonic and more efficient. You should try to understand it ;) – mozway Feb 26 '22 at 22:03

score 1 · Answer 2 · edited Feb 26 '22 at 23:06

This modified function should work:

def plus_to_every_second(integer):
    integer = str(integer)
    integer_new = ''
    for i in range(len(integer)):
      if i % 2 == 1:
        integer_new += str(int(integer[i]) + 1)
      else:
        integer_new += integer[i]
    return integer_new

First, as you did, we convert integer to a str and create a new variable, integer_new which holds an empty String.

However, in our for loop, we should iterate through integer so that we have access to the index.

If the index is odd (every second number), we then convert the character at that index to a number, add 1 to it, convert it back to a string, and then add it into integer_new.

If it's not at an odd index, then we just add the character into integer_new.

score 0 · Answer 3 · answered Feb 26 '22 at 22:36

You won't have the best performance, but you can do the following for a quick solution:

>>> number = 3912
>>> digit_lst = list(str(number))
>>> digit_lst
['3', '9', '1', '2']
>>> new_digit_lst = [n if i % 2 == 0 else str(int(n)+1) for i, n in enumerate(digit_lst)]
>>> new_digit_lst
['3', '10', '1', '3']
>>> new_number = int(''.join(new_digit_lst))
>>> new_number
31013

Add 1 to every second number

3 Answers3

using a classical loop