使用给定项创建表

Question

这只是使用传入的项创建一个表。我希望使这一功能更短和/或更有效率。我所拥有的似乎是多余的。

def apriori_creator(items, k):
    """Function that creates apriori table using supported candidates"""
    table = []
    the_length = len(items)
    new_k = k-2 # first k-2 items as described in algorithm in book for 'apriori gen'
    for x in range(the_length):
        next = x + 1
        for y in range(next, the_length):
            combined = items[x].union(items[y]) # merge
            l_sub_one = list(items[x])
            l_sub_one.sort()
            l_sub_two = list(items[y])
            l_sub_two.sort()
            first = l_sub_one[:new_k] # getting the first k-2 items of the set
            second = l_sub_two[:new_k]
            if first == second: 
                table.append(combined)
    return table

Answer 1

为了提高效率，您可以从内部循环中提取一些内容：

• 只依赖于x的计算应该在外部循环中。
• combined的计算应该在使用它的if下面。
• 使用enumerate和直接迭代代替range更简单，速度也更快。
• l_sub_one = sorted(items[x])的操作与这两行相同: l_sub_one = list(itemsX) l_sub_one.sort()

经修订的守则：

def apriori_creator(items, k):
    """Function that creates apriori table using supported candidates"""
    table = []
    new_k = k-2 # first k-2 items as described in algorithm in book for 'apriori gen'
    for x, item_x in enumerate(items):
        l_sub_one = sorted(item_x)
        first = l_sub_one[:new_k] # getting the first k-2 items of the set
        for item_y in items[x+1:]:
            l_sub_two = sorted(item_y)
            second = l_sub_two[:new_k]
            if first == second: 
                combined = item_x.union(item_y) # merge
                table.append(combined)
    return table

还可能有进一步的改进。描述您正在做的事情的一种方法是生成共享sorted(item)[:new_k]相同值的所有不同对项。您的代码通过考虑所有可能的配对来做到这一点。一种更有效的方法首先是将项目按该值分组。在字典的帮助下，这只需要对项目进行一次传递。然后，您只需要生成对与每一组。

这是再次修改的代码。我使用collections.defaultdict进行分组，itertools.combinations用于配对。

from collections import defaultdict
from itertools import combinations

def apriori_creator(items, k):
    """Function that creates apriori table using supported candidates"""
    new_k = k-2 # first k-2 items as described in algorithm in book for 'apriori gen'
    groups = defaultdict(list)  # group by the first k-2 items of the set
    for item in items:
        l_sub_one = sorted(item)
        key = tuple(l_sub_one[:new_k])
        groups[key].append(item)

    table = []
    for group in groups.values():
        if len(group) > 1:
            for item_x, item_y in combinations(group, 2): # all pairs
                combined = item_x.union(item_y) # merge
                table.append(combined)
    return table     

Answer 2

看来你希望items是sets的名单，我想你可以利用更合适的数据结构，比如SortedSet。

from sortedcontainers import SortedSet

def apriori_creator(items, k):
    table = []
    items = [SortedSet(item) for item in items]
    for x, item in enumerate(items):
        for later_item in items[x+1:]:
            if item[:k-2] == later_item[:k-2]:
                # Since the first k-2 elements are the same, we should only
                # need to merge the remaining two elements, at most.
                table.append(item.union(later_item[k-2:]))
    return table