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Talk:Combinations with repetitions: Difference between revisions
Talk:Combinations with repetitions (view source)
Revision as of 02:37, 19 November 2010
, 13 years ago→Task definition: How about this definition?
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:::: [1, 2]
::If it were "sampling with replacement" I imagine that that result would have also included [1,1] and [3,3]. Since it did not, I do not think we can have (a,a,a) as a result for set n= ('a,a,b,c,d'), k=3. But perhaps the results from the deleted example should be reposted as a part of the task description? --[[User:Rdm|Rdm]] 21:58, 18 November 2010 (UTC)
==How about this definition?==
From the [http://docs.python.org/py3k/library/itertools.html#itertools.combinations_with_replacement Python documentation]:
'''itertools.combinations_with_replacement(iterable, r)'''
:Return r length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.
:Combinations are emitted in lexicographic sort order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.
:Elements are treated as unique based on their position, not on their value. So if the input elements are unique, the generated combinations will also be unique.
:Equivalent to:
:<lang python>def combinations_with_replacement(iterable, r):
# combinations_with_replacement('ABC', 2) --> AA AB AC BB BC CC
pool = tuple(iterable)
n = len(pool)
if not n and r:
return
indices = [0] * r
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != n - 1:
break
else:
return
indices[i:] = [indices[i] + 1] * (r - i)
yield tuple(pool[i] for i in indices)</lang>
:The code for combinations_with_replacement() can be also expressed as a subsequence of [http://docs.python.org/py3k/library/itertools.html#itertools.product product()] after filtering entries where the elements are not in sorted order (according to their position in the input pool):
:<lang python>def combinations_with_replacement(iterable, r):
pool = tuple(iterable)
n = len(pool)
for indices in product(range(n), repeat=r):
if sorted(indices) == list(indices):
yield tuple(pool[i] for i in indices)</lang>
:The number of items returned is <code>(n+r-1)! / r! / (n-1)! when n > 0</code>.
We would need to '''change the page name''' to ''Combinations with replacement'', but I think the above Python does what the [[wp:Combination#Example of counting multicombinations|wp entry]] is trying to describe as:
:<lang python>>>> from itertools import combinations_with_replacement
>>> len(list(combinations_with_replacement('1234567890', 3)))
220</lang>
--[[User:Paddy3118|Paddy3118]] 02:37, 19 November 2010 (UTC)
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