信心在计算关联规则 [英] Confidence calculation in association rule

问题描述

supportData = {('ELF'): 0.75, ('CAT'): 0.75, ('BAT', 'CAT', 'ELF'): 0.5, ('ARK',    'BAT'): 0.25, ('ARK', 'ELF'): 0.25, ('CAT', 'ELF'): 0.5, ('DOG'): 0.25, ('BAT', 'CAT'): 0.5, ('BAT', 'ELF'): 0.75, ('ARK'): 0.5, ('ARK', 'CAT'): 0.5, ('BAT'): 0.75}

L = [('ARK'), ('CAT'), ('CAT'), ('ELF'),('ARK', 'CAT'), ('BAT', 'ELF'), ('BAT', 'CAT'), ('CAT', 'ELF'),('BAT', 'CAT', 'ELF')]
for freqSet in L:

    H =  list(freqSet)

    if len(H) == 1:
        pass
    else:
            for conseq in H:
            freqsetlist = list(freqSet)
            freqsetlist.remove(conseq)
            if len(freqsetlist) == 1:
               conf = supportData[freqSet]/supportData[tuple(freqsetlist)[0]]
               if conf >= 0.1:
                  print freqsetlist,'-->',conseq,'conf:',conf
            else:
               conf = supportData[freqSet]/supportData[tuple(freqsetlist)[:]]
               if conf >= 0.1:
                  print freqsetlist,'-->',conseq,'conf:',conf

Output

KeyError: ('R','K')

有人能说出为什么我收到这个错误？这似乎错误发生时，len个（freqsetlist）是> 1。这与3元

Can someone point out why I am getting this error? It seems the error occur when len(freqsetlist) is > 1. That is when calculating tuple with 3 element

推荐答案

这是对象的再presentation，如果你想要一个不同的再presentation，你将不得不自己构造它：

That is the representation of the object, if you want a different representation, you will have to construct it yourself:

>>> k = ['van']
>>> "({})".format(", ".join(k))
'(van)'

请注意，这意味着你使用一个对象作为程序的一部分，Python的重新presentation，这是一个坏主意，你应该始终建立你所需要的人工，而不是尝试使用Python的重新presentation，其目的是用于调试。

Note that this implies you are using Python's representation of an object as a part of your program, this is a bad idea, and you should always construct what you need manually rather than try and use Python's representation, which is intended for debugging.

编辑：逗号是显示它是一个元组，括号内为默认情况下，意味着操作，而不是元组的分组Python的方式。你可以使自己的元组的子类，并修改 __再版__（） / __ STR __（）如果你真的想要的，但将是难以置信的毫无意义的（和unpythonic在的情况下， __ __再版（），因为它应该评估的对象）。

The comma is Python's way of showing it's a tuple, as brackets signify grouping of operations rather than tuples by default. You could make your own tuple subclass and change the __repr__()/__str__() if you really wanted, but that would be incredibly pointless (and unpythonic in the case of __repr__() as it should evaluate to the object).

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