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python - 按类别创建项目列表的受限排列

转载 作者:太空宇宙 更新时间:2023-11-03 11:10:57 25 4
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我正在尝试创建一系列项目列表的限制排列。每个项目都有一个类别,我需要找到项目组合,以使每个组合不包含来自同一类别的多个项目。为了说明,这里有一些示例数据:

   Name      | Category
==========|==========
1. Orange | fruit
2. Apple | fruit
3. GI-Joe | toy
4. VCR | electronics
5. Racquet | sporting goods

组合将被限制为三个长度,我不需要每个长度的每个组合。因此,上述列表的一组组合可能是:

(Orange, GI-Joe, VCR)
(Orange, GI-Joe, Racquet)
(Orange, VCR, Racquet)
(Apple, GI-Joe, VCR)
(Apple, GI-Joe, Racquet)
... and so on.

我经常在各种列表上这样做。列表的长度永远不会超过 40 个项目,但可以理解的是,这可能会产生数千种组合(尽管每个列表可能会有大约 10 个独特的类别,这在一定程度上有所限制)

我想出了一些伪 python 来说明如何递归地实现它。自从我学习组合学以来已经太久了,但据我所知,这本质上是集合组合的子集,类似于 C(列表长度,所需大小)。可能有一些库模块可以使它更清洁(或至少更高效)

我想知道是否有比我现有的方法更好的方法(也许以某种方式使用 itertools.combinations):

# For the sake of this problem, let's assume the items are hashable so they
# can be added to a set.

def combinate(items, size=3):
assert size >=2, "You jerk, don't try it."
def _combinate(index, candidate):
if len(candidate) == size:
results.add(candidate)
return
candidate_cats = set(x.category for x in candidate)
for i in range(index, len(items)):
item = items[i]
if item.category not in candidate_cats:
_combinate(i, candidate + (item, ))

results = set()
for i, item in enumerate(items[:(1-size)]):
_combinate(i, (item, ))

return results

最佳答案

朴素的方法:

#!/usr/bin/env python

import itertools

items = {
'fruits' : ('Orange', 'Apple'),
'toys' : ('GI-Joe', ),
'electronics' : ('VCR', ),
'sporting_goods' : ('Racquet', )
}

def combinate(items, size=3):
if size > len(items):
raise Exception("Lower the `size` or add more products, dude!")

for cats in itertools.combinations(items.keys(), size):
cat_items = [[products for products in items[cat]] for cat in cats]
for x in itertools.product(*cat_items):
yield zip(cats, x)

if __name__ == '__main__':
for x in combinate(items):
print x

将产生:

# ==> 
#
# [('electronics', 'VCR'), ('toys', 'GI-Joe'), ('sporting_goods', 'Racquet')]
# [('electronics', 'VCR'), ('toys', 'GI-Joe'), ('fruits', 'Orange')]
# [('electronics', 'VCR'), ('toys', 'GI-Joe'), ('fruits', 'Apple')]
# [('electronics', 'VCR'), ('sporting_goods', 'Racquet'), ('fruits', 'Orange')]
# [('electronics', 'VCR'), ('sporting_goods', 'Racquet'), ('fruits', 'Apple')]
# [('toys', 'GI-Joe'), ('sporting_goods', 'Racquet'), ('fruits', 'Orange')]
# [('toys', 'GI-Joe'), ('sporting_goods', 'Racquet'), ('fruits', 'Apple')]

关于python - 按类别创建项目列表的受限排列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3686521/

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