我在交易数据中做项目关联。我在R中使用arules包来构建规则。 我正在使用此链接https://1drv.ms/u/s!Ak1rt2E1f2gFgV9t7hMVAn0P4gd0
分享我的示例数据library(arules)
library(arulesViz)
df = read.csv("trans.csv")
trans = as(split(df[,"Item"], df[,"Billno"]), "transactions")
inspect(trans[1:20])
summary(trans)
rules1 = apriori(trans,parameter = list(support = 0.6, confidence = 0.6,
target = "rules"))
summary(rules1) ##Output is "Set of 0 rules"
我的输出为，
Summary(rules1)
一套0规则
我在发布此链接之前提到了https://stats.stackexchange.com/questions/56034/association-analysis-returns-0-useful-rules这个链接。我也试过随机数来获得支持和信心，没有任何作用。
答案 0 :(得分：4)
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。如果您需要复习支持和信心的确切含义，请阅读this。
让我们先看看你的交易数据：
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
要处理的第一个问题是最低限度的支持。摘要说明您的最常见项目（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
你使用分钟。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。如果您想查找数据中至少出现10次的项目集/规则，那么您可以将最低支持设置为：
10/length(trans)
[1] 0.003951008
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
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，所以我首先使用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
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
可以找到关于如何使用关联规则挖掘的完整示例here。
希望这有帮助！