[英]Compare several (but not all) elements in a list of namedtuples
我有一個很長的namedtuple列表(目前它可以達到10.000行,但將來可能會更多)。
我需要將每個namedtuple的幾個元素與列表中的所有其他namedtuple進行比較。 我正在尋找一種有效且通用的方法 。
為了簡單起見,我將用蛋糕做一個比喻,這應該使理解問題更加容易。
有一個namedtuple列表,其中每個namedtuple是一塊蛋糕:
Cake = namedtuple('Cake',
['cake_id',
'ingredient1', 'ingredient2', 'ingredient3',
'baking_time', 'cake_price']
)
cake_price
和baking_time
都很重要。 如果蛋糕的成分相同,我想從列表中刪除不相關的那些。 因此,任何蛋糕(具有相同的成分)都是相同的或更昂貴的,並且烘烤相同或更長的時間都沒有關系(下面有一個詳細的示例)。
最好的方法是什么?
到目前為止,我所做的是按cake_price
和baking_time
對named_tuples列表進行排序:
sorted_cakes = sorted(list_of_cakes, key=lambda c: (c.cake_price, c.baking_time))
然后創建一個新列表,我要添加所有蛋糕,只要以前添加的蛋糕沒有相同的成分,則價格便宜且烘焙速度更快。
list_of_good_cakes = []
for cake in sorted_cakes:
if interesting_cake(cake, list_of_good_cakes):
list_of_good_cakes.append(cake)
def interesting_cake(current_cake, list_of_good_cakes):
is_interesting = True
if list_of_good_cakes: #first cake to be directly appended
for included_cake in list_of_good_cakes:
if (current_cake.ingredient1 == included_cake.ingredient1 and
current_cake.ingredient2 == included_cake.ingredient2 and
current_cake.ingredient3 == included_cake.ingredient3 and
current_cake.baking_time >= included_cake.baking_time):
if current_cake.cake_price >= included_cake.cake_price:
is_interesting = False
return is_interesting
(我知道嵌套循環遠非最優,但我想不出其他任何方法...)
有
list_of_cakes = [cake_1, cake_2, cake_3, cake_4, cake_5]
哪里
cake_1 = Cake('cake_id'=1,
'ingredient1'='dark chocolate',
'ingredient2'='cookies',
'ingredient3'='strawberries',
'baking_time'=60, 'cake_price'=20)
cake_2 = Cake('cake_id'=2,
'ingredient1'='dark chocolate',
'ingredient2'='cookies',
'ingredient3'='strawberries',
'baking_time'=80, 'cake_price'=20)
cake_3 = Cake('cake_id'=3,
'ingredient1'='white chocolate',
'ingredient2'='bananas',
'ingredient3'='strawberries',
'baking_time'=150, 'cake_price'=100)
cake_4 = Cake('cake_id'=4,
'ingredient1'='dark chocolate',
'ingredient2'='cookies',
'ingredient3'='strawberries',
'baking_time'=40, 'cake_price'=30)
cake_5 = Cake('cake_id'=5,
'ingredient1'='dark chocolate',
'ingredient2'='cookies',
'ingredient3'='strawberries',
'baking_time'=10, 'cake_price'=80)
預期結果將是:
list_of_relevant_cakes = [cake_1, cake_3, cake_4, cake_5]
您的方法的運行時間將大致與
len(list_of_cakes) * len(list_of_relevant_cakes)
...如果您有很多蛋糕並且其中很多與蛋糕有關,則可能會變得很大。
我們可以利用以下事實對此加以改進:具有相同成分的每個蛋糕簇都可能小得多。 首先,我們需要一個函數來檢查新蛋糕是否與具有相同成分的現有,已經優化的集群相對應:
from copy import copy
def update_cluster(cakes, new):
for c in copy(cakes):
if c.baking_time <= new.baking_time and c.cake_price <= new.cake_price:
break
elif c.baking_time >= new.baking_time and c.cake_price >= new.cake_price:
cakes.discard(c)
else:
cakes.add(new)
這樣做是針對new
蛋糕,在cakes
的副本中對照每個蛋糕c
進行檢查,然后:
如果其烘烤時間和價格都大於或等於現有蛋糕,則立即退出(您可以return
而不是break
,但我希望明確說明控制流程)。
如果其烘烤時間和價格均小於或等於現有蛋糕,則從群集中刪除該現有蛋糕
如果它超過所有現有的蛋糕(並到達for
語句的else
子句),則將其添加到集群中。
一旦有了,我們就可以用它來過濾蛋糕了:
def select_from(cakes):
clusters = {}
for cake in cakes:
key = cake.ingredient1, cake.ingredient2, cake.ingredient3
if key in clusters:
update_cluster(clusters[key], cake)
else:
clusters[key] = {cake}
return [c for v in clusters.values() for c in v]
它在起作用:
>>> select_from(list_of_cakes)
[Cake(cake_id=1, ingredient1='dark chocolate', ingredient2='cookies', ingredient3='strawberries', baking_time=60, cake_price=20),
Cake(cake_id=4, ingredient1='dark chocolate', ingredient2='cookies', ingredient3='strawberries', baking_time=40, cake_price=30),
Cake(cake_id=5, ingredient1='dark chocolate', ingredient2='cookies', ingredient3='strawberries', baking_time=10, cake_price=80),
Cake(cake_id=3, ingredient1='white chocolate', ingredient2='bananas', ingredient3='strawberries', baking_time=150, cake_price=100)]
該解決方案的運行時間大致與
len(list_of_cakes) * len(typical_cluster_size)
我對隨機蛋糕列表進行了一些測試,每個蛋糕都使用您的五種不同食材,隨機價格和烘烤時間進行選擇,
這種方法始終產生與您相同的結果(盡管未排序)
它的運行速度要快得多-在我的機器上運行100,000個隨機蛋糕的時間為0.2秒,而您的機器為3秒。
未經測試的代碼,但應該有助於指出一種更好的方法:
equivalence_fields = operator.attrgetter('ingredient1', 'ingredient2', 'ingrediant3')
relevant_fields = operator.attrgetter('baking_time', 'cake_price')
def irrelevent(cake1, cake2):
"""cake1 is irrelevant if it is both
more expensive and takes longer to bake.
"""
return cake1.cake_price > cake2.cake_price and cake1.baking_time > cake2.bake_time
# Group equivalent cakes together
equivalent_cakes = collections.defaultdict(list)
for cake in cakes:
feature = equivalence_fields(cake)
equivalent_cakes[feature].append(cake)
# Weed-out irrelevant cakes within an equivalence class
for feature, group equivalent_cakes.items():
best = min(group, key=relevant_fields)
group[:] = [cake for cake in group if not irrelevant(cake, best)]
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