如何计算Python中列表一部分的集合的出现次数？

Question

Trying to implement the apriori algorithm and made it to the point where I can extract the subsets occurring together in all transactions.

This is what I have:

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'])]

For example set(['Breakfast & Brunch', 'Restaurants']) occurs twice and I need to keep track of the numbers of occurrences together with the corresponding patterns.

I tried to use:

from collections import Counter

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

support_set.update(subsets)

but it generates this error:

  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'

Any idea?

Answer 1

You can turn the sets to frozenset instances which are hashable:

>>> 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})