Function giving wrong output in Python

Question

Through this code, i wish to replace all the dots (.) appearing in the string s with the character present at exactly the symmetrically opposite position. For eg: if s=a.bcdcbba, then the . should be replaced by b

i.e:

The element at ith position should be replaced by the element at len(s)-i-1th position. This function gives wrong output for the cases like g.... , .g... etc. Any help ?

def replacedots(s):
 for i in range(0,len(s)):
 if s[i]==".":
 s=s.replace(s[i],s[len(s)-i-1],1)
 return s 

• Padraic Cunningham: what about "foo.bar"?
• joel goldstick: Do you mean the character after the dot?
• Aran-Fey: It's because s.replace doesn't replace the dot at i but the first dot.
• Padraic Cunningham: You need to add what you expect as output for the corner cases
• codehungry95: Thanks every one ! Especially @Dombi Szabolcs .

Answer 1

@chepner's way:

def replacedots(s):
 return ''.join(x if x !='.' else y for x, y in zip(s, reversed(s)))

an alternative:

def replacedots(s):
 return ''.join(c if c != '.' else s[-i - 1] for i, c in enumerate(s))

When the char at position i and the char at position len(s) - i - 1 is a . the dot will remain a dot.

Example:

s = "foo.bar"

• C.LECLERC: Nice ! but isn't s[-i - 1] enought to get the opposite char ?
• Szabolcs Dombi: sure, thank you (next time feel free to edit an answer you can improve) :)
• chepner: Some--me :)--might find it cleaner to use x if x !='.' else y for x, y in zip(s, reversed(s)) and avoid direct indexing. Either way, it is faster to pass a list comprehension instead of a generator to ''.join. (Oddly enough, the zip version is faster than theenumerate version when generators are used, but not when both are used as list comprehensions.)