PROSAGA码农传奇-大数据安全-我们怎样才能找到支持并对规则有先见之明？

找到正确的最小支持和最小置信度值并以0个频繁项目集或0个关联规则结束的问题非常普遍。读

这个

如果你需要复习什么是支持和信心的确切含义。

我们先来看看你的交易数据：


    
      summary(trans)
transactions as itemMatrix in sparse format with
 2531 rows (elements/itemsets/transactions) and
 6632 columns (items) and a density of 0.0005951533 
most frequent items:
AR845311 AR800369 AR828249 AR839869 AR831167  (Other)
      84       35       31       29       24     9787 
element (itemset/transaction) length distribution:
sizes
   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21
 767 509 306 238 160 112 100  52  69  50  31  27  18  12  13  15   9  10   7   5   4
 23  24  25  27  28  32  34  36  48
  3   4   2   3   1   1   1   1   1 
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.000   1.000   2.000   3.947   5.000  48.000 
</code>


  
    要处理的第一个问题是最低限度的支持。摘要说你最常用的项目（
     
      AR845311
    
    ）在数据集中出现84次。您的物品通常支持率很低
  

   
    
      summary(itemFrequency(trans))
  Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
  0.0003951 0.0003951 0.0003951 0.0005952 0.0003951 0.0331900 
</code>
  

  
    你用一分钟。支持0.6，但最常见的单项只有0.033的支持！你需要减少支持。如果要查找数据中至少出现10次的项目集/规则，则可以将最低支持设置为：
  

   
    
       10/length(trans)
 [1] 0.003951008
</code>
  

  
    第二个问题是您的数据非常稀疏（摘要显示密度约为0.0006）。这意味着您的交易相当短（即只包含少量项目）。
  

   
    
      table(size(trans))
  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21
767 509 306 238 160 112 100  52  69  50  31  27  18  12  13  15   9  10   7   5   4
 23  24  25  27  28  32  34  36  48
  3   4   2   3   1   1   1   1   1 
</code>
  

  
    短交易意味着规则的可信度可能很低。对于你的数据，事实证明它非常低，所以我先使用0。
  

   
    
      rules <- apriori(trans, 
parameter = list(support = 0.004, confidence = 0, target = “rules”))
Apriori
Parameter specification:
 confidence minval smax arem  aval originalSupport maxtime support minlen maxlen
          0    0.1    1 none FALSE            TRUE       5   0.004      1     10
 target   ext
  rules FALSE
Algorithmic control:
 filter tree heap memopt load sort verbose
    0.1 TRUE TRUE  FALSE TRUE    2    TRUE
Absolute minimum support count: 10 
set item appearances …[0 item(s)] done [0.00s].
set transactions …[6632 item(s), 2531 transaction(s)] done [0.00s].
sorting and recoding items … [40 item(s)] done [0.00s].
creating transaction tree … done [0.00s].
checking subsets of size 1 2 done [0.00s].
writing … [46 rule(s)] done [0.00s].
creating S4 object  … done [0.00s].
summary(rules)
set of 46 rules
rule length distribution (lhs + rhs):sizes
 1  2
40  6 
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   1.00    1.00    1.00    1.13    1.00    2.00 
summary of quality measures:
    support           confidence            lift            count
 Min.   :0.004346   Min.   :0.004346   Min.   : 1.000   Min.   :11.00
 1st Qu.:0.004741   1st Qu.:0.004840   1st Qu.: 1.000   1st Qu.:12.00
 Median :0.005531   Median :0.005729   Median : 1.000   Median :14.00
 Mean   :0.006803   Mean   :0.057301   Mean   : 3.316   Mean   :17.22
 3rd Qu.:0.007112   3rd Qu.:0.008890   3rd Qu.: 1.000   3rd Qu.:18.00
 Max.   :0.033188   Max.   :0.705882   Max.   :21.269   Max.   :84.00  
mining info:
  data ntransactions support confidence
 trans          2531   0.004          0
</code>
  

  
    结果表明，至少有一条规则的置信度为0.7。您可以更高的信心再次运行APRIORI。以下是最值得信赖的规则：
  

   
    
      inspect(head(rules, by = “confidence”))
    lhs           rhs        support     confidence lift     count
[1] {AR835501} => {AR845311} 0.004741209 0.7058824  21.26891 12
[2] {AR743988} => {AR845311} 0.004346108 0.6470588  19.49650 11
[3] {AR800369} => {AR845311} 0.007111814 0.5142857  15.49592 18
[4] {AR845311} => {AR800369} 0.007111814 0.2142857  15.49592 18
[5] {AR845311} => {AR835501} 0.004741209 0.1428571  21.26891 12
[6] {AR845311} => {AR743988} 0.004346108 0.1309524  19.49650 11 
</code>
  

  
    可以找到有关如何使用关联规则挖掘的完整示例
    
      这里
    
    。
  

  
    希望这可以帮助！