找到R中每个组中最常见的组合

Question

我有下面的数据集，显示了每个产品中包含的成分；

data <- data.frame("PRODUCT" = c("Creme","Creme","Creme","Creme","Medoc","Medoc","Medoc","Medoc","Medoc","Hububu","Hububu","Hububu","Hububu","Troll","Troll","Troll","Troll","Suzuki","Suzuki","Gluglu","Gluglu","Gluglu"), 
            "INGREDIENT" = c("zeze","zaza","zozo","zuzu","zaza","sasa","haha","zuzu","zemzem","zaza","zuzu","zizi","haha","zozo","zaza","zemzem","zuzu","sasa","zuzu","ozam","zaza","hayda"))

我想知道每个产品中最常见的配料组合；哪种配料与哪种其他配料相关？我应用了我在这个线程here中找到的代码：

combinaisons_par_PRODUCT = data %>% 
  full_join(data, by="PRODUCT") %>% 
  group_by(INGREDIENT.x, INGREDIENT.y) %>% 
  summarise(n = length(unique(PRODUCT))) %>% 
  filter(INGREDIENT.x!=INGREDIENT.y) %>%
  mutate(item = paste(INGREDIENT.x, INGREDIENT.y, sep=", "))

它可以工作，但有一个最后的缺陷；我希望忽略该命令。例如，这段代码会给我一个HAHA和SASA的关联，还有一个SASA和HAHA的关联。但对我来说，这些都是一样的事情。因此，我希望代码忽略配料的顺序，并给我一个独特的关联2哈哈&萨萨。

我试着在应用代码之前对成分进行排序，但也不起作用。有人能帮帮我吗？我怎么能在不考虑顺序的情况下拥有这些组合呢？

非常感谢!

Answer 1

这是你想要的吗？我仅限于组合词按字母顺序排列的情况，避免重复计数。

data %>% 
  full_join(data, by="PRODUCT") %>%
  filter(INGREDIENT.x < INGREDIENT.y) %>%
  count(combo = paste(INGREDIENT.x, INGREDIENT.y, sep = ", "))

Answer 2

使用graph_from_adjacency_matrix的igraph选项

library(igraph)

get.data.frame(
    graph_from_adjacency_matrix(
        crossprod(table(data)),
        mode = "undirected",
        weighted = TRUE
    )
)

给出

     from     to weight
1    haha   haha      2
2    haha   sasa      1
3    haha   zaza      2
4    haha zemzem      1
5    haha   zizi      1
6    haha   zuzu      2
7   hayda  hayda      1
8   hayda   ozam      1
9   hayda   zaza      1
10   ozam   ozam      1
11   ozam   zaza      1
12   sasa   sasa      2
13   sasa   zaza      1
14   sasa zemzem      1
15   sasa   zuzu      2
16   zaza   zaza      5
17   zaza zemzem      2
18   zaza   zeze      1
19   zaza   zizi      1
20   zaza   zozo      2
21   zaza   zuzu      4
22 zemzem zemzem      2
23 zemzem   zozo      1
24 zemzem   zuzu      2
25   zeze   zeze      1
26   zeze   zozo      1
27   zeze   zuzu      1
28   zizi   zizi      1
29   zizi   zuzu      1
30   zozo   zozo      2
31   zozo   zuzu      2
32   zuzu   zuzu      5

Answer 3

我们可以使用base R

m1 <- crossprod(table(data))
subset(as.data.frame.table(m1 * lower.tri(m1, diag = TRUE)), Freq != 0)

编辑:来自@ThomasIsCoding的评论