substitution

[Peter Otten]

Wyrmskull wrote:

> Peter Otten wrote:

>> def replace_many(s, pairs):
>>     if len(pairs):
>>         a, b = pairs[0]
>>         rest = pairs[1:]
>>         return b.join(replace_many(t, rest) for t in s.split(a))
>>     else:
>>         return s

> Proves wrong, this way x -> y -> z.
> You call replace_many again on the central part of the split
> Specifics indicate that x -> y in the end.

Sorry, I don't understand what you want to say with the above.

> Try with this:
> 
> def mySubst(reps,string):
>     if not(len(reps)):
>         return string
>     current = reps[0][0]
>     a,b,c = string.partition(current)
>     if b:
>         return mySubst(reps,a) + reps[0][1] + mySubst (reps,c)
>     else:
>         return mySubst(reps[1:],string)
> 
> print mySubst( ( ('foo','bar'), ('bar','qux'), ('qux','foo') ),
>  'foobarquxfoo')
> 
> -------
> Wyrmskull <lordkrandel at gmail.com>

I don't see at first glance where the results of replace_many() and 
mySubst() differ. Perhaps you could give an example?

Peter

PS: Please keep the conversation on-list