Suppose I have this string:

'0123456abcde78sdfnwjeabcde8923kasjsducuabcded999'

How can I move the substring 'abcde' two characters to the right to produce:

'012345678abcdesdfnwje89abcde23kasjsducud9abcde99'

So this:

s = '0123456abcde78'
t = 'abcde'
n = 2  # (right shift)

Function(s, t, n) 

Should give:

'012345678abcde'

Edge cases that can be safely ignored:

• fewer characters than shift amount following substring ('abcde1')
• consecutive occurrences of the substring ('abcdeabcde123')

CodePudding user response：

Python strings are immutable, so you will have to create a new string from:

• the beginning up to the index of t in s
• n characters starting from the end of t (so at `s.index(t) len(t))
• t
• the end of the string starting at n characters past the end of t

In Python is could be:

def function(s, t, n):
    ix = s.index(t)
    length = len(t)
    return s[:ix]  s[ix   length: ix   length   n]   t   s[ix   length   n:]

Error and corner case handling are omitted per your requirement...

CodePudding user response：

Make a new string according to the position of the index of the string you are looking for. Something among the following lines:

s = '0123456abcde78'
t = 'abcde'
n = 2  # (right shift)
tmp = s.index(t)
new_s = s[:tmp]   s[tmp len(t):tmp len(t) n]   t   s[tmp len(t) n:]

CodePudding user response：

I would use the regex-equivalent of replace - re.sub:

import re

s = "0123456abcde78sdfnwjeabcde8923kasjsducuabcded999"
t = 'abcde'
n = 2
print(re.sub(rf"({t})(.{{{n}}})", r"\2\1", s))

Gives:

012345678abcdesdfnwje89abcde23kasjsducud9abcde99

Explanation:

• pattern:

  • ( - First matching group:
    • {t} - replaces t using f-strings - literally match the full substring.
  • ) - End first group
  • ( - Second matching group:
    • . - any character
    • {{ - escaped curly braces in f-string - to denote number of repetitions.
    • {n} - f-string replacement of n - the number of repetitions.
    • }} - escaped closing braces.
  • ) - End second group

• replacement:

  • Simply replace the order of the above groups.

Demo and explanations (without the f-strings) in regex101.com

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To do a left shift instead, just change the order of the groups (i.e. first match any n characters and then match t):

re.sub(rf"(.{{{n}}})({t})", r"\2\1", s)