3

arulesトランザクションのリストが必要です。リストの各行には、製品の配列が含まれます。すべてのトランザクションに同じ量の製品があるわけではありません。ピボットのように聞こえますが、そうではありません。例はここにあります

aggregate(dvd , by=list("ID"), FUN=c) 私は失敗のようなものが欲しい arguments must have same length

これは私のデータです

> dvd
   ID          Item
1   1   Sixth Sense
2   1         LOTR1
3   1 Harry Potter1
4   1    Green Mile
5   1         LOTR2
6   2     Gladiator
7   2       Patriot
8   2    Braveheart
9   3         LOTR1
10  3         LOTR2
11  4     Gladiator
12  4       Patriot
13  4   Sixth Sense
14  5     Gladiator
15  5       Patriot
16  5   Sixth Sense
17  6     Gladiator
18  6       Patriot
19  6   Sixth Sense
20  7 Harry Potter1
21  7 Harry Potter2
22  8     Gladiator
23  8       Patriot
24  9     Gladiator
25  9       Patriot
26  9   Sixth Sense
27 10   Sixth Sense
28 10          LOTR
29 10     Galdiator
30 10    Green Mile

そのようなリストが必要です

TR1     c("Sixth Sense","LOTR1","Harry Potter1","Green Mile","LOTR2")
TR2     c("Gladiator","Patriot","Braveheart")
TR3     c("LOTR1","LOTR2")
....
4

3 に答える 3

3

arulesのread.transactionsformatには、問題を解決する引数があります。使用法は次のとおりです。

read.transactions(file, format = c("basket", "single"), sep = NULL,
                  cols = NULL, rm.duplicates = FALSE, encoding = "unknown")

format議論を見ますか？「バスケット」または「シングル」のいずれかを使用して、入力データの形式を表すことができます。データを「バスケット」形式に変換しようとしていますが、データのタイプはすでに「単一」です。各行はIDを持つ単一のアイテムで構成されています。read.transactionsを使用formatして「シングル」に設定するだけで、ゴールデンになります。

于 2015-03-19T14:17:57.470 に答える
2

私splitはあなたのために仕事をするだろうと思います。

    DF <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 
4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 9L, 
10L, 10L, 10L, 10L), Item = c("   Sixth Sense", "         LOTR1", 
" Harry Potter1", "    Green Mile", "         LOTR2", "     Gladiator", 
"       Patriot", "    Braveheart", "         LOTR1", "         LOTR2", 
"     Gladiator", "       Patriot", "   Sixth Sense", "     Gladiator", 
"       Patriot", "   Sixth Sense", "     Gladiator", "       Patriot", 
"   Sixth Sense", " Harry Potter1", " Harry Potter2", "     Gladiator", 
"       Patriot", "     Gladiator", "       Patriot", "   Sixth Sense", 
"   Sixth Sense", "          LOTR", "     Galdiator", "    Green Mile"
)), .Names = c("ID", "Item"), class = "data.frame", row.names = c(NA, 
-30L))

    DF <- read.csv(textConnection(txt), header = TRUE, stringsAsFactors = FALSE, strip.white = TRUE)
result <- split(DF$Item, DF$ID)
names(result) <- gsub("(.*)", "TR\\1", names(result))
result
## $TR1
## [1] "Sixth Sense"   "LOTR1"         "Harry Potter1" "Green Mile"    "LOTR2"        
## 
## $TR2
## [1] "Gladiator"  "Patriot"    "Braveheart"
## 
## $TR3
## [1] "LOTR1" "LOTR2"
## 
## $TR4
## [1] "Gladiator"   "Patriot"     "Sixth Sense"
## 
## $TR5
## [1] "Gladiator"   "Patriot"     "Sixth Sense"
## 
## $TR6
## [1] "Gladiator"   "Patriot"     "Sixth Sense"
## 
## $TR7
## [1] "Harry Potter1" "Harry Potter2"
## 
## $TR8
## [1] "Gladiator" "Patriot"  
## 
## $TR9
## [1] "Gladiator"   "Patriot"     "Sixth Sense"
## 
## $TR10
## [1] "Sixth Sense" "LOTR"        "Galdiator"   "Green Mile"
于 2013-03-25T08:14:50.937 に答える
2

aggregateコマンドは機能する可能性がありますが、引数を正しく指定していません。次のようなものが必要になりますwith(DF, aggregate(Item, list(ID), FUN = function(x) c(as.character(x))))。

または、次の式を使用することもできますaggregate。

aggregate(Item ~ ID, DF, c)
#    ID                                                 Item
# 1   1 Sixth Sense, LOTR1, Harry Potter1, Green Mile, LOTR2
# 2  10             Sixth Sense, LOTR, Galdiator, Green Mile
# 3   2                       Gladiator, Patriot, Braveheart
# 4   3                                         LOTR1, LOTR2
# 5   4                      Gladiator, Patriot, Sixth Sense
# 6   5                      Gladiator, Patriot, Sixth Sense
# 7   6                      Gladiator, Patriot, Sixth Sense
# 8   7                         Harry Potter1, Harry Potter2
# 9   8                                   Gladiator, Patriot
# 10  9                      Gladiator, Patriot, Sixth Sense
str(.Last.value)
# 'data.frame':  10 obs. of  2 variables:
# $ ID  : chr  "1" "10" "2" "3" ...
# $ Item:List of 10
#  ..$ 1 : chr  "Sixth Sense" "LOTR1" "Harry Potter1" "Green Mile" ...
#  ..$ 6 : chr  "Sixth Sense" "LOTR" "Galdiator" "Green Mile"
#  ..$ 10: chr  "Gladiator" "Patriot" "Braveheart"
#  ..$ 13: chr  "LOTR1" "LOTR2"
#  ..$ 15: chr  "Gladiator" "Patriot" "Sixth Sense"
#  ..$ 18: chr  "Gladiator" "Patriot" "Sixth Sense"
#  ..$ 21: chr  "Gladiator" "Patriot" "Sixth Sense"
#  ..$ 24: chr  "Harry Potter1" "Harry Potter2"
#  ..$ 26: chr  "Gladiator" "Patriot"
#  ..$ 28: chr  "Gladiator" "Patriot" "Sixth Sense"

または、「data.table」パッケージを使用できます。 

library(data.table)
as.data.table(DF)[, list(list(Item)), by = ID]
#     ID                                               V1
#  1:  1 Sixth Sense,LOTR1,Harry Potter1,Green Mile,LOTR2
#  2:  2                     Gladiator,Patriot,Braveheart
#  3:  3                                      LOTR1,LOTR2
#  4:  4                    Gladiator,Patriot,Sixth Sense
#  5:  5                    Gladiator,Patriot,Sixth Sense
#  6:  6                    Gladiator,Patriot,Sixth Sense
#  7:  7                      Harry Potter1,Harry Potter2
#  8:  8                                Gladiator,Patriot
#  9:  9                    Gladiator,Patriot,Sixth Sense
# 10: 10            Sixth Sense,LOTR,Galdiator,Green Mile
于 2014-12-31T08:56:35.460 に答える