Question

尝试实现apriori算法，并使其能够提取所有事务中一起出现的子集。

这就是我所拥有的：

subsets = [set(['Breakfast & Brunch', 'Restaurants']), set(['American (Traditional)', 'Breakfast & Brunch']), set(['American (Traditional)', 'Restaurants']), set(['American (Traditional)', 'Breakfast & Brunch']), set(['Breakfast & Brunch', 'Restaurants']), set(['American (Traditional)', 'Restaurants'])]

例如set(['Breakfast & Brunch', 'Restaurants'])出现两次我需要跟踪出现的次数以及相应的模式。

我试图使用：

from collections import Counter

support_set = Counter()
# some code that generated the list above

support_set.update(subsets)

但它会产生此错误：

  supported = itemsets_support(transactions, candidates)
  File "apriori.py", line 77, in itemsets_support
    support_set.update(subsets)
  File"/usr/local/Cellar/python/2.7.12/Frameworks/Python.framework/Versions/2.7/lib/python2.7/collections.py", line 567, in update
    self[elem] = self_get(elem, 0) + 1
TypeError: unhashable type: 'set'

有什么想法吗？

Answer 1

您可以将这些集转换为可以缓存的frozenset个实例：

>>> from collections import Counter
>>> subsets = [set(['Breakfast & Brunch', 'Restaurants']), set(['American (Traditional)', 'Breakfast & Brunch']), set(['American (Traditional)', 'Restaurants']), set(['American (Traditional)', 'Breakfast & Brunch']), set(['Breakfast & Brunch', 'Restaurants']), set(['American (Traditional)', 'Restaurants'])]
>>> c = Counter(frozenset(s) for s in subsets)
>>> c
Counter({frozenset(['American (Traditional)', 'Restaurants']): 2, frozenset(['Breakfast & Brunch', 'Restaurants']): 2, frozenset(['American (Traditional)', 'Breakfast & Brunch']): 2})

如何计算Python中列表一部分的集合的出现次数？

1 个答案: