Re: substitution
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.
Your flowing pythonicity (if len(x):) gave me lots of inspiration.
I bet this will win the prize ;)
-
def mySubst(reps,string):
if not(len(reps)):
return string
a,b,c = string.partition(reps[0][0])
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
--
http://mail.python.org/mailman/listinfo/python-list
Re: substitution
Cleaned. You can remove the 'else's if you want,
because 'return' breaks the instruction flow.
Should also work with other sequence types.
def mySubst(reps,seq):
if reps:
a,b,c = string.partition(reps[0][0])
if b:
return mySubst(reps,a) + reps[0][1] + mySubst (reps,c)
else:
return mySubst(reps[1:], seq)
else:
return seq
print mySubst( ( ('foo','bar'), ('bar','qux'), ('qux','foo') ), 'foobarquxfoo')
print mySubst( ( ('foo','bar'), ('bar','qux'), ('qux','foo') ),
'foobarquxxxfoo')
---
Wyrmskull
--
http://mail.python.org/mailman/listinfo/python-list
Re: substitution
Nvm, my bad, I misunderstood the split instruction. No difference :) --- Wyrmskull P.S Sorry about the P.M., I misclicked on a GUI -- http://mail.python.org/mailman/listinfo/python-list
