How to increment and decrement with a string Python?

Question

Hope you all are doing well in these times.

here's my code:


def ab(n):


    first = 0
    last = -1 

    endprod = n[first] + n[last]
    endprod2 = n[first+1] + n[last-1]
    endprod3 = n[first+2] + n[last-2]
    endprod4 = n[first+3] + n[last-3]
    endprod5 = n[first+4] + n[last-4]

        
    endprod100 = endprod[::-1] + endprod2[::-1] + endprod3[::-1]+ endprod4[::-1]+ endprod5[::-1]

    return endprod100

I was able do to it, however mine isn't a loop. Is there a way to convert my code into a for loop. So, increment by 1 and decrement by 1.

Thanks,

Sorry about that, I edited that part. I'm just curious to know how I would increment it by 1 and decrement by 1 like you see in the code, without having to manually write it. Thanks :) — Mr. Green
– Mr. Green, Commented Oct 27, 2022 at 16:45
Your question is already answered, but I want to add that "endprod = n[first] + n[last]" followed by "endprod[::-1]" seems wasteful when you could do directly: "endprod = n[last] + n[first]" . — Swifty
– Swifty, Commented Oct 27, 2022 at 16:58

Riccardo Bucco · Accepted Answer · 2022-10-27 16:50:15Z

1

Try this:

def ab(n):
    result = ''
    for j in range(len(n) // 2):
        result += n[-j-1] + n[j]
    if len(n) % 2 == 1:
        result += n[len(n) // 2]
    return result

You also need the part

if len(n) % 2 == 1:
    result += n[len(n) // 2]

because your input string might have an odd number of characters

Examples:

>>> ab('0123456789')
'9081726354'
>>> ab('01234956789')
'90817263549'

If you want to reuse your original logic:

def ab(n):
    result = ''
    first = 0
    last = -1
    for j in range(len(n) // 2):
        result += n[last-j] + n[first+j]
    if len(n) % 2 == 1:
        result += n[len(n) // 2]
    return result

edited Oct 27, 2022 at 16:50

answered Oct 27, 2022 at 16:44

Riccardo Bucco

15.5k4 gold badges29 silver badges57 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Mr. Green Over a year ago

Thanks but would it be possible to convert my logic into a for loop ?

Mr. Green Over a year ago

I'm just curious to know how I would increment it by 1 and decrement by 1 like you see in the code, without having to manually write it. Thanks :)

Riccardo Bucco Over a year ago

@Mr.Green Please check the new solution (which is by the way completely identical to the other one, you just need to drop the constants first and last, which are basically useless

Mr. Green Over a year ago

Thanks Ricardo, could you just put comments, why have if len(n)%2 ==1: result + = n[len(n)//2] ?? thanks

Swifty · Accepted Answer · 2022-10-27 17:39:37Z

1

You could also recurse it:

def ab(s):
    if len(s)>2:
        return s[-1]+s[0]+ab(s[1:-1])
    else:
        return s

But the last part of Riccardo's answer fits your question more closely.

edited Oct 27, 2022 at 17:39

answered Oct 27, 2022 at 17:04

Swifty

3,4642 gold badges6 silver badges25 bronze badges

Comments

amirhm · Accepted Answer · 2022-10-27 17:00:11Z

0

you need to split your string for your loop, means first you broke your string to half then build your string, you could use zip to iterate over multiple iteratable. something like this:

def ab(s):
    out = ""
    for v0,v1 in zip(s[:len(s)//2 -1 :-1], s[:len(s)//2 ]):
        out += v0 + v1
    return out

the better version you should write without loop. like this:

out = "".join(map(lambda x: "".join(x), zip(s[:len(s)//2 -1 :-1], s[:len(s)//2 ])))

edited Oct 27, 2022 at 17:00

answered Oct 27, 2022 at 16:54

amirhm

1,23911 silver badges15 bronze badges

Comments

Savir · Accepted Answer · 2022-10-27 17:28:13Z

There are already a lot of good answers that are probably clearer, easier to read and much better suited for learning purposes than what I'm about to write. BUT... something I wanted to bring up (maybe just telling myself, because I tend to forget it) is that sometimes destroying the original argument can facilitate this sort of things.

In this case, you could keep "poping" left and right, shrinking your string until you exhaust it

def swap_destr(my_str):
    result = ""
    while len(my_str) > 1:
        result += my_str[-1]  # Last char
        result += my_str[0]   # First char
        my_str = my_str[1:-1] # Shrink it!
    return result + my_str  # + my_str just in case there 1 char left

print(swap_destr("0123456789"))
print(swap_destr("123456789"))

# Outputs:
# 9081726354
# 918273645

This is also a good problem to see (and play with) recursion:

def swap_recur(my_str):
    if len(my_str) <= 1:
        return my_str
    return my_str[-1] + my_str[0] + swap_recur(my_str[1:-1])

print(swap_recur("0123456789"))
print(swap_recur("123456789"))

# Outputs:
# 9081726354
# 918273645

To think it's only after reading yours that I realized there was no need to differentiate between len(str) == 1 and len(str) == 0... I must be very tired :)
Haha... well... sometimes explicit is better than implicit, right? 😉 ( @Swifty )

Collectives™ on Stack Overflow

How to increment and decrement with a string Python?

4 Answers 4

4 Comments

Comments

Comments

3 Comments

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

Comments

Comments

3 Comments

Related