I have researched about character replacement using dictionaries but I still cannot get my code to work properly. My code goes like this:
def encode(code,msg):
for k in code:
msg = msg.replace(k,code[k])
return msg
Now, when I run the code:
code = {'e':'x','x':'e'}
msg = "Jimi Hendrix"
encode(code,msg)
It gives me "Jimi Hxndrix" instead of "Jimi Hxndrie". How do I get the letter 'x' to be replaced by 'e' also?
You can look at str.translate or do:
''.join(code.get(ch, ch) for ch in msg)
Use maketrans and translate:
from string import maketrans
msg.translate(maketrans(''.join(code.keys()), ''.join(code.values())))
{from: to}.The problem is that you iterate on code instead on msg
Iterating on msg is what is done in Jon Clements' program, which can be written more explicitly as
print ''.join(code[ch] if ch in code else ch for ch in msg)
you're swapping x and e; it is overwriting your previous edits.
You should copy from an old string to a new one (or rather, an array of chars, since as Kalle pointed out strings are "immutable"/not editable), so that you don't overwrite characters you've already replaced:
def encode(code, message):
encoded = [c for c in message]
for i, c in enumerate(message):
try:
encoded[i] = code[c]
except KeyError:
pass
return ''.join(encoded)
the other answers are library functions which do something like this, but they don't explain where you went wrong.
encoded[i] = code[c]
python 3.2
use your own code it is ok.
def encode(code,msg):
for k in code:
msg = msg.replace(k,code[k],1)
return msg
## str.replace(old,new,[,count])
so when you loop in your code = {'e':'x','x':'e'}
first it gets the key "x" then the key "e" because dict is not ordered
so, "Hendrix" turns into "Hendrie" then "Hendrie" turns into Hxndrix and
you are not having your result. but if you add 1 in your code
msg.replace(k,code[k],1) then it only will replace one letter per loop and you
have your result Try.
xfore, then the first occurrence ofeforx. Thus,Jimi HendrixbecomesJimi Hendrie, thenJimi Hxndrie..replace()is not doing what you think it does.msg = msg.replace(k,code[k])followed byprint k,' ',msgon "Jimi Hendrix fixed the axes" , it prints firstly Jimi Hendrie fieed the aees then Jimi Hxndrix fixxd thx axxs . That's because code is a dictionary that is not iterated in the order 'e' then 'x' but in the order 'x' first and 'e' after. And all the occurences of a character in a string are replaced, not only the first one.