Python：为相对支持先验算法生成候选项目集

Question

Please note: Title of this question might be ambiguous so I request other users to please edit it. I was not able to come up with a suitable title which fits this problem.

The problem discussed above is a part of an algorithm called RSAA (Relative Support Apriori Algorithm), here's the research paper link: http://dl.acm.org/citation.cfm?id=937663

Problem : I am implementing algorithms like apriori using python, and while doing so I am facing an issue where I have generate patterns (candidate itemsets) like these at each step of the algorithm.

• At each step the length of the sublists in the main list should be incremented by 1.
• Output of one step is going to be the input for the next step.
• Sublists in the main list can occur in any order, and numbers inside sublists can occur in any order.

Here's the example:

Input:

input = [[5, 3], [5, 4], [5, 6], [7, 6]]

Output should be:

output = [[5,3,4], [5,3,6], [4,5,6], [5,6,7]]

Each sublist of output list (^) must have only 3 items (example: [5,3,4]) .

The approach to solve this problem should be generic , because in the next step:

Input:

input = [[5,3,4], [5,3,6], [4,5,6], [5,6,7]]

Output:

output = [[5,3,4,6], [4,5,6,7]]

Each sublist of output list (^) must have only 4 items.

( [5,3,4,6] is formed by joining [5,3,4] and [5,3,6]. We can't join [5,3,4] and [5,6,7] because doing so would create [5,3,4,6,7] which will be of length = 5 )

Answer 1

I think your requirement is included in apriori. I wrote a blog about the algorithm, but unfortunately in chinese. Here is the link http://www.zealseeker.com/archives/apriori-algorithm-python/
Here is the snippets (also hosted in chinese)

has_infrequent_subset and apriori_gen may be the two functions you want.

If the code is useful for you, comment my answer and I'll be glade to continue help you.

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update

It is easy to get the intersection and difference of two sequence in python.

a = set([5, 6])
b = set([6, 7])
c = a & b # get the itersection
if len(c) == len(a) - 1: 
  return a | b # their union