R 将项目列表改为宽格式

我有一个项目列表的数据框，其中数据框中的每一行都包含LHS和RHS关联规则，以及相应的支持度、置信度和提升度。 以下是数据：

structure(list(rules = structure(c(13L, 4L, 28L, 1L, 24L, 15L
), .Label = c("{butter,jam} => {whole milk}", "{butter,rice} => {whole milk}", 
"{canned fish,hygiene articles} => {whole milk}", "{curd,cereals} => {whole milk}", 
"{domestic eggs,rice} => {whole milk}", "{grapes,onions} => {other vegetables}", 
"{hamburger meat,bottled beer} => {whole milk}", "{hamburger meat,curd} => {whole milk}", 
"{hard cheese,oil} => {other vegetables}", "{herbs,fruit/vegetable juice} => {other vegetables}", 
"{herbs,rolls/buns} => {whole milk}", "{herbs,shopping bags} => {other vegetables}", 
"{liquor,red/blush wine} => {bottled beer}", "{meat,margarine} => {other vegetables}", 
"{napkins,house keeping products} => {whole milk}", "{oil,mustard} => {whole milk}", 
"{onions,butter milk} => {other vegetables}", "{onions,waffles} => {other vegetables}", 
"{pastry,sweet spreads} => {whole milk}", "{pickled vegetables,chocolate} => {whole milk}", 
"{pork,butter milk} => {other vegetables}", "{rice,bottled water} => {whole milk}", 
"{rice,sugar} => {whole milk}", "{soups,bottled beer} => {whole milk}", 
"{tropical fruit,herbs} => {whole milk}", "{turkey,curd} => {other vegetables}", 
"{whipped/sour cream,house keeping products} => {whole milk}", 
"{yogurt,cereals} => {whole milk}", "{yogurt,rice} => {other vegetables}"
), class = "factor"), support = c(0.00193187595322827, 0.00101677681748856, 
0.00172852058973055, 0.00101677681748856, 0.00111845449923742, 
0.00132180986273513), confidence = c(0.904761904761905, 0.909090909090909, 
0.80952380952381, 0.833333333333333, 0.916666666666667, 0.8125
), lift = c(11.2352693602694, 3.55786275006331, 3.16819206791352, 
3.26137418755803, 3.58751160631383, 3.17983983286908)), .Names = c("rules", 
"support", "confidence", "lift"), row.names = c(NA, 6L), class = "data.frame")

我需要的是将这些规则组织成广泛的格式，其中，对于规则的每个LHS部分中的每个项目，将有一个值为1的指定列（表示规则在其LHD部分中有该项目），规则的RHS也是如此，例如，采用前两个规则：

{liquor,red/blush wine} => {bottled beer} 0.0019 0.90 11.2
{curd,cereals} => {whole milk} 0.0010 0.91 3.6

结果应该是如下所示的数据帧：

'rules_id' 'lhs_liquor' 'lhs_red/blush wine' 'lhs_curd' 'lhs_cereals' 'rhs_bottled beer' 'rhd_whole milk' 'support' 'confidence' 'lift'
1 1 1 0 0 1 0 0.0019 0.90 11.2
2 0 0 1 1 0 1 0.0010 0.91 3.6 

由于我是新的R和堆栈溢出，请让我知道，如果问题没有很好地定义

---

感谢您的任何帮助

您可以这样做

library(dplyr)
library(tidyr)
library(reshape2) 
rules %>% 
  mutate(id = seq_len(n())) %>% 
  separate(rules, c("lhs", "rhs"), "\\} => \\{") %>% 
  separate_rows(lhs) %>% filter(lhs!="") %>% 
  gather(value, var, lhs, rhs) %>% 
  mutate(var=paste(value, sub("}", "", var, fixed=T), sep="_")) %>%
  dcast(id+support+confidence+lift~var, fun.aggregate = function(x) (length(x)>0)+0L)
#   id     support confidence      lift lhs_beer lhs_blush lhs_bottled lhs_butter lhs_cereals
# 1  1 0.001931876  0.9047619 11.235269        0         1           0          0           0
# 2  2 0.001016777  0.9090909  3.557863        0         0           0          0           1
# 3  3 0.001728521  0.8095238  3.168192        0         0           0          0           1
# 4  4 0.001016777  0.8333333  3.261374        0         0           0          1           0
# 5  5 0.001118454  0.9166667  3.587512        1         0           1          0           0
# 6  6 0.001321810  0.8125000  3.179840        0         0           0          0           0
#   lhs_curd lhs_house lhs_jam lhs_keeping lhs_liquor lhs_napkins lhs_products lhs_red
# 1        0         0       0           0          1           0            0       1
# 2        1         0       0           0          0           0            0       0
# 3        0         0       0           0          0           0            0       0
# 4        0         0       1           0          0           0            0       0
# 5        0         0       0           0          0           0            0       0
# 6        0         1       0           1          0           1            1       0
#   lhs_soups lhs_wine lhs_yogurt rhs_bottled beer rhs_whole milk
# 1         0        1          0                1              0
# 2         0        0          0                0              1
# 3         0        0          1                0              1
# 4         0        0          0                0              1
# 5         1        0          0                0              1
# 6         0        0          0                0              1

请随意使用tidyr的

spread

而不是Reformae2的

dcast

-我仍然觉得dcast更不实用…

您可以这样做

dummies <- function(x, prefix) {
    x.names <- unique(unlist(strsplit(x, ',')))
    out <- array(0L, c(nrow(df), length(x.names)), list(NULL, x.names))
    mapply(function(i, val) out[i, val] <<- 1L, 1:nrow(out), strsplit(x, ','))
    if (!missing(prefix))
        colnames(out) <- paste0(prefix, colnames(out))
    out
}

pat <- '[{](.*)[}] => [{](.*)[}]'

cbind(as.data.frame(
    cbind(dummies(sub(pat, '\\1', df$rules), 'lhs.'),
          dummies(sub(pat, '\\2', df$rules), 'rhs.'))),
    df[c('support','confidence','lift')])

---

你的最后一段，因为它不是问题本身的一部分，通常会放在这里的注释中。长格式可能更有用：

df%>%separate（col=rules，into=c（'lhs'，'rhs'），sep='=>'））%%>%separate_行（col=lhs，into=lhs，sep='，'）%%>%gather（key=side，value=product，lhs，rhs）%%>%mutate（product=gsub（'[{}]'，''，product））

太棒了！非常好的解决方案。我不太可能理解这些说法，所以如果你能解释一下那里发生了什么，我将不胜感激，谢谢anyway@NirRegev实际上，这个解决方案有点老套。第一个问题：

dummies

对

df

有一个硬编码的引用模拟Python中的

enumerate（）

，迭代

1:nrow（out）

和

strsplit（x，，'）

的元素。它还依赖于

separate_rows（）is unknown函数。我想知道这个函数属于哪个包？
  lhs.liquor lhs.red/blush wine lhs.curd lhs.cereals lhs.yogurt lhs.butter
1          1                  1        0           0          0          0
2          0                  0        1           1          0          0
3          0                  0        0           1          1          0
4          0                  0        0           0          0          1
5          0                  0        0           0          0          0
6          0                  0        0           0          0          0
  lhs.jam lhs.soups lhs.bottled beer lhs.napkins lhs.house keeping products
1       0         0                0           0                          0
2       0         0                0           0                          0
3       0         0                0           0                          0
4       1         0                0           0                          0
5       0         1                1           0                          0
6       0         0                0           1                          1
  rhs.bottled beer rhs.whole milk     support confidence      lift
1                1              0 0.001931876  0.9047619 11.235269
2                0              1 0.001016777  0.9090909  3.557863
3                0              1 0.001728521  0.8095238  3.168192
4                0              1 0.001016777  0.8333333  3.261374
5                0              1 0.001118454  0.9166667  3.587512
6                0              1 0.001321810  0.8125000  3.179840