I want to replace every character that isn't "i" in the string "aeiou" with a "!"
I wrote:
def changeWord(word):
for letter in word:
if letter != "i":
word.replace(letter,"!")
return word
This just returns the original. How can I return "!!i!!"?
Strings in Python are immutable, so you cannot change them in place. Check out the documentation of str.replace:
Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced.
So to make it work, do this:
def changeWord(word):
for letter in word:
if letter != "i":
word = word.replace(letter,"!")
return word
O(len(word)^2). Something like ''.join('!' if c != 'i' else c for c in word) would be more efficient for long strings.
Regular expressions are pretty powerful for this kind of thing. This replaces any character that isn't an "i" with "!"
import re
str = "aieou"
print re.sub('[^i]', '!', str)
returns:
!!i!!
something like this using split() and join():
In [4]: strs="aeiou"
In [5]: "i".join("!"*len(x) for x in strs.split("i"))
Out[5]: '!!i!!'
Try this, as a one-liner:
def changeWord(word):
return ''.join(c if c == 'i' else '!' for c in word)
The answer can be concisely expressed using a generator, there's no need to use regular expressions or loops in this case.
Nobody seems to give any love to str.translate:
In [25]: chars = "!"*105 + 'i' + "!"*150
In [26]: 'aeiou'.translate(chars)
Out[26]: '!!i!!'
Hope this helps
str.join and because there is a LOT of translation :)
Generally in python string are immutable that means it can't be changed after its creation. So we have to convert string into mutable and that can be possible using list as it is mutable.
string="Love"
editable = list(string)
editable[2]='s'
string=''.join(editable)
output:Lose
