基于不同变量的交互连接r

Question

我有两个数据框架如下：

 df<-data.frame(
  id=c("1-1","2-2","3-3","4-4","5-5","6-6"),
  identifer=c(1,2,3,4,5,6),
  key=c("A","B","C","D","E","F"),
  product=c("productA","productB","productC","productD","productE","productF"),
  ingredient=c("ingredientA","ingredientB","ingredientC","ingredientD","ingredientE","ingredientF"),
  DF=c("Tablet","Powder","Suspension","System","Capsule","Capsule"))

  df_2<-data.frame(
  identifer=c(1,2,2,3,4,6),
  key=c("A","B","B","C","D","F"),
  product=c("productA","productB","productB","productCC","productDD","productFF"),
  ingredient=c("ingredientA","ingredientBB","ingredientB","ingredientC","ingredientDD","ingredeintFF"),
  DF=c("Tablet","Powder","Powder","Suspension","injection","tablet"),
  Route=c("ORAL","INHALATION","INHALATION","topical","injecatable","oral")
)

我想首先在以下变量上加入这两个数据集+创建一个名为"match“的新列来描述连接：

1) identifier,key, product, ingredient,DF
match="identifier,key, product, ingredient,DF"

然后，我想加入这些变量的其余行：

2)identifier, key, product, DF
match="identifier,key, product,DF"

然后是步骤2中关于这些变量的其余行，等等。

3) identifier, key, Ingredient, DF
4) identifier, key, DF 
5) identifer, key, product, ingredient
7) identifer, key, product
8) identifer, key, ingredient 
9) identifier, key 

我希望返回没有匹配的行。我知道如何逐步地做这件事，但我想知道是否有更简单的方法来做到这一点？

这是预期的产出：

df_out<-data.frame(
  identifer=c(1,2,3,4,5,6),
  key=c("A","B","C","D","E","F"),
  product_1=c("productA","productB","productC","productD","productE","productF"),
  ingredient_1=c("ingredientA","ingredientB","ingredientC","ingredientD","ingredientE","ingredientF"),
  DF_1=c("Tablet","Powder","Suspension","System","Capsule","Capsule"),
  product_2=c("productA","productB","productCC","productDD",NA,"productFF"),
  ingredient_2=c("ingredientA","ingredientB","ingredientC","ingredientDD",NA,"ingredeintFF"),
  DF_2=c("Tablet","Powder","Suspension","injection",NA,"tablet"),
  Route_2=c("ORAL","INHALATION",'topical',"injecatable",NA,"oral"),
  Match=c("identifer+key+product+ingredient+DF","identifier+key+product+ingredient+DF","identifier+key+ingredient+DF","identifer+key","None","identifer+key+product+ingredient"))

Answer 1

下面是一个使用data.table的选项

library(data.table)
setDT(df)
setDT(df_2)

keyord <- list(
    c("product", "ingredient", "DF"),
    c("product", "DF"),
    c("ingredient", "DF"),
    "DF",
    c("product", "ingredient"),
    "product",
    "ingredient",
    c()
)

cols <- c("product", "ingredient", "DF", "Route")
df[, Match := NA_character_]

for (v in keyord) {
    k <- c("identifier", "key", v)
    df[df_2, on=k, c(paste0(cols, "_2"), "check") := c(mget(paste0("i.", cols)), .(TRUE))]
    df[is.na(Match) & check, Match := toString(k)]
}
setnames(df, cols, paste0(cols, "_1"), skip_absent=TRUE)

产出：

    id identifier key product_1 ingredient_1       DF_1                                    Match product_2 ingredient_2       DF_2     Route_2 check
1: 1-1          1   A  productA  ingredientA     Tablet identifier, key, product, ingredient, DF  productA  ingredientA     Tablet        ORAL  TRUE
2: 2-2          2   B  productB  ingredientB     Powder identifier, key, product, ingredient, DF  productB  ingredientB     Powder  INHALATION  TRUE
3: 3-3          3   C  productC  ingredientC Suspension          identifier, key, ingredient, DF productCC  ingredientC Suspension     topical  TRUE
4: 4-4          4   D  productD  ingredientD     System                          identifier, key productDD ingredientDD  injection injecatable  TRUE
5: 5-5          5   E  productE  ingredientE    Capsule                                     <NA>      <NA>         <NA>       <NA>        <NA>    NA
6: 6-6          6   F  productF  ingredientF    Capsule     identifier, key, product, ingredient  productF  ingredientF     tablet        oral  TRUE

修正OP中某些排印后的数据：

df <- data.frame(
    id=c("1-1","2-2","3-3","4-4","5-5","6-6"),
    identifier=c(1,2,3,4,5,6),
    key=c("A","B","C","D","E","F"),
    product=c("productA","productB","productC","productD","productE","productF"),
    ingredient=c("ingredientA","ingredientB","ingredientC","ingredientD","ingredientE","ingredientF"),
    DF=c("Tablet","Powder","Suspension","System","Capsule","Capsule"))

df_2 <- data.frame(
    identifier=c(1,2,2,3,4,6),
    key=c("A","B","B","C","D","F"),
    product=c("productA","productB","productB","productCC","productDD","productF"),
    ingredient=c("ingredientA","ingredientBB","ingredientB","ingredientC","ingredientDD","ingredientF"),
    DF=c("Tablet","Powder","Powder","Suspension","injection","tablet"),
    Route=c("ORAL","INHALATION","INHALATION","topical","injecatable","oral")
)

编辑多个匹配：

df_2 <- data.frame( identifier=c(1,2,2,3,4,4,6), key=c("A","B","B","C","D","D","F"), product=c("productA","productB","productB","productCC","productDD","productDd","productF"), ingredient=c("ingredientA","ingredientBB","ingredientB","ingredientC","ingredientDD",NA,"ingredientF"), DF=c("Tablet","Powder","Powder","Suspension","injection",NA,"tablet"), Route=c("ORAL","INHALATION","INHALATION","topical","injecatable",NA,"oral") )
setDT(df_2)
df[, c("Match", "check") := .(NA_character_, FALSE)]

ocols <- unique(unlist(keyord))
rbindlist(lapply(keyord, function(v) {
    k <- c("identifier", "key", v)
    a <- df_2[df[(!check)], on=k, nomatch=0L, c(.(id=id),
        setNames(mget(paste0("i.", ocols)), paste0(ocols, "_1")), 
        setNames(mget(paste0("x.", c(ocols, "Route"))), paste0(c(ocols, "Route"), "_2"))) 
    ]
    df[id %chin% a$id, check := TRUE]
    a
}), use.names=TRUE)

产出：

    id product_1 ingredient_1       DF_1 product_2 ingredient_2       DF_2     Route_2
1: 1-1  productA  ingredientA     Tablet  productA  ingredientA     Tablet        ORAL
2: 2-2  productB  ingredientB     Powder  productB  ingredientB     Powder  INHALATION
3: 3-3  productC  ingredientC Suspension productCC  ingredientC Suspension     topical
4: 6-6  productF  ingredientF    Capsule  productF  ingredientF     tablet        oral
5: 4-4  productD  ingredientD     System productDD ingredientDD  injection injecatable
6: 4-4  productD  ingredientD     System productDd         <NA>       <NA>        <NA>

Answer 2

这里有一个解决方案，它可能会让人觉得有点过度设计，但却实现了预期的结果：

library(dplyr)
library(purrr)
library(stringr)

get_match=function(data, cols, keys){
  rtn = ifelse(rowSums(is.na(data[paste0(cols, "_1")]))==rowSums(is.na(data[paste0(cols, "_2")])), paste(keys, collapse="+"), "None")

  rtn2 = cols %>% 
    map(~{
      case_when(as.character(data[[paste0(.x, "_1")]])==as.character(data[[paste0(.x, "_2")]])~.x)
    }) %>% 
    reduce(paste, sep="+") %>% str_replace_all("\\+?NA\\+?", "")

  paste(rtn, rtn2, sep="+") %>% str_replace_all("\\+$", "")
}

df_out = left_join(df, df_2, by=c("identifer", "key"), suffix=c("_1", "_2")) %>% 
    mutate(Match = get_match(., cols=c("product", "ingredient", "DF"), keys=c("identifer", "key")), 
           match_strength = str_count(Match, "\\+")) %>% 
    group_by(id) %>% 
    filter(match_strength==max(match_strength, na.rm=TRUE))

dplyr::left_join移除by键，所以我发现添加它们的唯一方法是检查所有的_1或_2是否丢失。我可以使用keep=TRUE选项并在以后删除/重命名它们.