On Apr 17, 1:14 am, [EMAIL PROTECTED] wrote:
> Once in while I too have something to ask. This is a little problem
> that comes from a Scheme Book (I have left this thread because this
> post contains too much Python code for a Scheme
> newsgroup):http://groups.google.com/group/comp.lang.scheme/browse_thread/thread/...
>
> The function multiremberandco is hard (for me still) if done in that
> little Scheme subset, but it's very easy with Python. It collects two
> lists from a given one, at the end it applies the generic given fun
> function to the two collected lists and returns its result:
>
> def multiremberandco1(el, seq, fun):
> l1, l2 = [], []
> for x in seq:
> if x == el:
> l2.append(el)
> else:
> l1.append(el)
> return fun(l1, l2)
>
Hi bearophile,
Couldn't you also use count rather than the explicit loop to
get something like:
def multiremberandco1(el, seq, fun):
l1, l2 = [], []
c = seq.count(e1)
l1 = [el] * c
l2 = [el] * (len(seq) - c)
return fun(l1, l2)
I don't have the book so can't comment on its suitability,
but there you go.
- Paddy.
--
http://mail.python.org/mailman/listinfo/python-list
