Shifting String characters

Question

Recently I got a task to shift characters of a string as shown below sample

s = 'abcd'
right shift by 2
output: 'cdab'

I used below code and it works fine.

l1=[]
l2=[]
s='abcs'
l1=list(s)
loop = 0
rightshift = 2
while loop < rightshift:
     l2=[]
     l2=l1[1:]
     l2.append(l1[0])
     l1=[]
     l1=l2
     loop += 1
 print('After shift ' , l1)

However, if we have a large number like 544645655, it takes a lot of time. Do we have any other alternative method?

Some more examples(where s='abcd')
bcda 1st shift
cdab 2nd shift
dabc 3rd shift
abcd 4th shift
bcda 5th shift

Does this answer your question? [Rotating strings in Python](https://stackoverflow.com/questions/48607319/rotating-strings-in-python) — mkrieger1, Apr 18 '20 at 09:33
have added some samples in the question, there was some issue with the question earlier — Pradeep Shanbhag, Apr 18 '20 at 11:05
one simple optimization is to realize that shifting by the length of the string is a no-op, so you only need to shift by the remainder of `N / len(s)`. Which you can simply do by using the modulo operator — njzk2, Apr 18 '20 at 11:08

score 2 · Answer 1 · edited Apr 18 '20 at 10:31

2

The easiest way to move characters is:

N = 3
s = s[N:] + s[:N]

or if you prefer to drive on the opposite side if the road:

s = s[-N:] + s[:-N]

Play with those and use the one that makes you happy.

edited Apr 18 '20 at 10:31

Jon Clements

138,671
33
247
280

answered Apr 18 '20 at 09:36

lenik

23,228
4
34
43

For 'abcd' and N = 3, it gives 'dabc', not what OP wants as given in his example – Zain Arshad Apr 18 '20 at 09:44
1

@ZainArshad once you get the idea, you may figure out the minor details by yourself – lenik Apr 18 '20 at 09:46
It get the idea, but answer should solve the problem. To give the idea we use comment section under the question – Zain Arshad Apr 18 '20 at 09:47
1

@ZainArshad you CANNOT get result posted by OP: 'abcd' with right shift by 3 will NEVER be 'cdab' – lenik Apr 18 '20 at 09:48
I know but you are left-shifting, you can use negative of N to do a right shift – Zain Arshad Apr 18 '20 at 09:50

Emil M George · Answer 2 · 2020-04-18T17:08:43.237

You can use modular arithmetic for reducing number of shifts for large numbers:

s = "abcd"
leftshift = 544645655             # using left shift as per sample output in question
                                  # for rightshift, use negative number

N = leftshift % len(s)            # reduce shifts using modulo, explained below
print((s+s)[N:N+len(s)])

You can also use: print(s[N:] + s[:N]) for the print, as other answers show.

Output:

dabc

Explanation:

For numbers greater than 4 (string length), the pattern just repeats. You can use modular arithmetic to convert it to a number within [0,3].

a % b below denotes the modulus operation (It is equivalent to remainder of division a / b for positive numbers)

abcd   # leftshift = 0   (0 % 4 = 0)     [leftshift % len(s) = N]
bcda   # leftshift = 1   (1 % 4 = 1)
cdab   # leftshift = 2   (2 % 4 = 2)
dabc   # leftshift = 3   (3 % 4 = 3)

abcd   # leftshift = 4   (4 % 4 = 0)
bcda   # leftshift = 5   (5 % 4 = 1)
cdab   # leftshift = 6   (6 % 4 = 2)
dabc   # leftshift = 7   (7 % 4 = 3)

abcd   # leftshift = 8   (8 % 4 = 0)
  ^ repeating pattern              ^ you only have to shift by this number

So to shift by 544645655, we only have to shift by 544645655 % len(s) = 3

>>> s = "abcd"
>>> (s+s)
'abcdabcd'

We take the slice of length len(s) at position 0 like this:

'abcdabcd'
 ^^^^
>>> (s+s)[:len(s)]
'abcd'

Now all we need is move this slice like this:

abcdabcd
   ^^^^                 
>>> (s+s)[3:3+len(s)]   # 3 is the leftshift count
'dabc'

Put in the modular expression N = leftshift % len(s) instead of 3:

>>> leftshift = 5
>>> N = leftshift % len(s)   # N = 1
>>> (s+s)[N:N+len(s)]
'bcda'

Sorry, I have corrected the question,Sample = bcda 1st shift, cdab 2nd shift, dabc 3rd shift, abcd 4th shift, bcda 5th shift — Pradeep Shanbhag, Apr 18 '20 at 11:00
@PradeepShanbhag The new sample shows left shift. It's okay, you can use negative values for `rightshift` variable to get left shift. — Emil M George, Apr 18 '20 at 16:26

Zain Arshad · Answer 3 · 2020-04-18T10:23:50.953

-1

If your string is abcd, then performing a right shift of 3 would convert the string into bcda not cdab(It is right shift of 2 not 3).

As stated in @lenik's answer, you can use string splicing to get this done, use this simple method that will do right shift as well as left shift:

def shift(s, op, n):
    if op == 'left':
        return s[n:] + s[:n]
    elif op == 'right':
        return s[-n:] + s[:-n]
    else:
        print('Invalid operation')


print(shift('abcd', 'right', 3))

Output: bcda

edited Apr 18 '20 at 10:23

answered Apr 18 '20 at 09:56

Zain Arshad

1,885
1
11
26

bad idea to use `string` and `type` as variables. – lenik Apr 18 '20 at 10:18

Shifting String characters

3 Answers3