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/a059f78eb4457d08/
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)
data = [1, 'a', 3, 'a', 4, 5, 6, 'a']
print multiremberandco1('a', data, lambda l1,l2: (len(l1), len(l2)))
More compact:
def multiremberandco2(el, seq, fun):
l1, l2 = [], []
for x in seq:
[l1, l2][x == el].append(el)
return fun(l1, l2)
A bit cleaner (but I don't like it much):
def multiremberandco3(el, seq, fun):
l1, l2 = [], []
for x in seq:
(l2 if x == el else l1).append(el)
return fun(l1, l2)
For fun I have tried to make it lazy, if may be useful if seq is a
very long iterable. So I've used tee:
from itertools import ifilter, tee
def multiremberandco4(el, iseq, fun):
iseq1, iseq2 = tee(iseq)
iterable1 = ifilter(lambda x: x == el, iseq1)
iterable2 = ifilter(lambda x: x != el, iseq2)
return fun(iterable1, iterable2)
def leniter(seq):
count = 0
for el in seq:
count += 1
return count
idata = iter(data)
print multiremberandco4('a', idata, lambda l1,l2: (leniter(l1),
leniter(l2)))
But from the docs: >in general, if one iterator is going to use most
or all of the data before the other iterator, it is faster to use
list() instead of tee().<
So I have tried to create two iterables for the fun function scanning
seq once only, but I haven't succed so far (to do it I have tried to
use two coroutines with the enhanced generators, sending el to one or
to the other according to the value of x == el, this time I don't show
my failed versions), do you have suggestions?
(Note: in some situations it may be useful to create a "splitting"
function that given an iterable and a fitering predicate, returns two
lazy generators, of the items that satisfy the predicate and of the
items that don't satisfy it, so this exercise isn't totally useless).
Bye,
bearophile
--
http://mail.python.org/mailman/listinfo/python-list
