>>> def correct(s):
for c in s:
if c=='1':
c='I'
if c=='5':
c='S'
if c=='0':
c='O'
else:
c=c
return s
>>> x = "1teruje5z t0"
>>> correct(x)
'1teruje5z t0'W T F? Tried many things, this is simply as can be and still doesnt work.
Quote:doesnt work.
please be more specific, how does it not work?
the argument you pass is s. you iterate over it performing some operations and at the end you return s unchanged.
Moving the return line one tab forward also wont fix the problem cause it will return it after first char iteration, so what is pythonic way to do this?
I didn't suggest changing the indentation. You need to store value of c (changed or not) while iterating over s and then construct new string and return it.
and one more advise - the sooner you get rid of habit using single letter cryptic variable names - the better.
Yeah but how to do it without writing a tons of the code? Should be some lambda or 2-3 lines fast and easy way for this.
def correct(text):
corrections = {'1':'I', '5':'S'}
return ''.join(corrections.get(char, char) for char in text)
print(correct("1teruje5z t0"))Output:
IterujeSz t0
Thats very elegant solution, I will remember this.Cheers
If it is one-char to one-char translation, you can do even more elegant
def correct(text):
corrections = str.maketrans('15', 'IS') # or corrections = str.maketrans({'1':'I', '5':'S'})
return text.translate(corrections)
print(correct("1teruje5z t0"))this will not work for one-char to many-char-string translation