Steven D'Aprano wrote:
I've never needed such a split function, and I don't like the name, and
the functionality isn't general enough. I'd prefer something which splits
the input sequence into as many sublists as necessary, according to the
output of the key function. Something like itertools.groupby(), except it
runs through the entire sequence and collates all the elements with
identical keys.
splitby(range(10), lambda n: n%3)
=> [ (0, [0, 3, 6, 9]),
(1, [1, 4, 7]),
(2, [2, 5, 8]) ]
Your split() would be nearly equivalent to this with a key function that
returns a Boolean.
Well, here is my go at doing the original with iterators:
def splitter(source, test=bool):
a, b = itertools.tee((x, test(x)) for x in source)
return (data for data, decision in a if decision), (
data for data, decision in b if not decision)
This has the advantage that it can operate on infinite lists. For
something like splitby for grouping, I seem to need to know the cases
up front:
def _make_gen(particular, src):
return (x for x, c in src if c == particular)
def splitby(source, cases, case):
'''Produce a dict of generators for case(el) for el in source'''
decided = itertools.tee(((x, case(x)) for x in source), len(cases))
return dict((c, _make_gen(c, src))
for c, src in zip(cases, decided))
example:
def classify(n):
'''Least prime factor of a few'''
for prime in [2, 3, 5, 7]:
if n % prime == 0:
return prime
return 0
for k,g in splitby(range(50), (2, 3, 5, 7, 0), classify).items():
print('%s: %s' % (k, list(g)))
0: [1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
2: [0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24,
26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48]
3: [3, 9, 15, 21, 27, 33, 39, 45]
5: [5, 25, 35]
7: [7, 49]
--Scott David Daniels
scott.dani...@acm.org
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