r中的多重匹配函数

Question

我正在尝试使用以下变量 School（唯一）将两个数据集与需要教师的课程进行匹配。有的老师有一种专长，有的有不止一种。我一直在尝试使用 match() 和 which( %in% ) 基本函数，但我无法让它搜索所有可能的教师匹配项。它总是在第一场比赛后停止。以下是一些示例数据：

class<-c("english","history","art","art","math","history","art")
school<-c("C.H.S.","B.H.S.","D.H.S.","A.H.S.","Z.H.S.","M.H.S.","L.H.S.")
specialty<-c("math","history","English","history","literature","art","English")
teacher<-c("Jill","Jill","Sam","Liz","Liz","Liz","Rob")
teacher.skills<-data.frame(teacher, specialty)
school.needs<-data.frame(school,class)
teacher.match<-data.frame(Jill,Sam,Rob,Liz)

最终结果如下所示：

Jill<-c("No","Yes","No","No","Yes","Yes","No")
Sam<-c("Yes","No","No","No","No","No","No")
Liz<-c("No","Yes","Yes","Yes","No","Yes","Yes")
Rob<-c("Yes","No","No","No","No","No","No")
match.result<-data.frame(school.needs, teacher.match)
match.result

我什至尝试过使用这样的小功能，但仍然无法正确设置最终格式。

source.1<-school.needs
source.2<-teacher.skills
dist.name<-adist(source.1$class, source.2$specialty, partial = FALSE, ignore.case = TRUE)
min.name<-apply(dist.name, 1, min)
school.teacher.match<-NULL  
for(i in 1:nrow(dist.name))
{
    skills.ref<-match(min.name[i], dist.name[i,])
    school.ref<-i
    school.teacher.match<-rbind(data.frame(skills.ref=skills.ref, school.ref=school.ref, Teacher=source.2[skills.ref,]$teacher, Class=source.1[school.ref,]$class, School=source.1[school.ref,]$school, adist=min.name[i]), school.teacher.match)
    school.teacher.match<-subset(school.teacher.match, school.teacher.match$adist==0)
}
school.teacher.match

任何帮助将不胜感激，谢谢！

Answer 1

请注意，我必须修改您的输入数据以将每次匹配的 "English" 更改为 "english"。数据由：

school.needs <- structure(list(school = structure(c(3L, 2L, 4L, 1L, 7L, 6L, 5L
), .Label = c("A.H.S.", "B.H.S.", "C.H.S.", "D.H.S.", "L.H.S.", 
"M.H.S.", "Z.H.S."), class = "factor"), class = structure(c(2L, 
3L, 1L, 1L, 4L, 3L, 1L), .Label = c("art", "english", "history", 
"math"), class = "factor")), .Names = c("school", "class"), row.names = c(NA, 
-7L), class = "data.frame")

teacher.skills <- structure(list(teacher = structure(c(1L, 1L, 4L, 2L, 2L, 2L, 
3L), .Label = c("Jill", "Liz", "Rob", "Sam"), class = "factor"), 
specialty = structure(c(5L, 3L, 2L, 3L, 4L, 1L, 2L), .Label = c("art", 
"english", "history", "literature", "math"), class = "factor")), .Names = c("teacher", 
"specialty"), row.names = c(NA, -7L), class = "data.frame")

使用来自reshape2（或data.table）的merge 和dcast：

library(reshape2)
## use merge to match needs to skills
m <- merge(school.needs,teacher.skills,by.x="class",by.y="specialty")
m$val <- "Yes"   ## add a column for the "Yes"
## go to wide format for the final result filling NA with "No"
result <- dcast(m,school+class~teacher,value.var="val",fill="No")
##  school   class Jill Liz Rob Sam
##1 A.H.S.     art   No Yes  No  No
##2 B.H.S. history  Yes Yes  No  No
##3 C.H.S. english   No  No Yes Yes
##4 D.H.S.     art   No Yes  No  No
##5 L.H.S.     art   No Yes  No  No
##6 M.H.S. history  Yes Yes  No  No
##7 Z.H.S.    math  Yes  No  No  No

Answer 2

我会这样做：

（数据）

schools <- data.frame(
  school = c("C.H.S.", "B.H.S.", "D.H.S.", "A.H.S.","Z.H.S.", "M.H.S.", "L.H.S."),
  class  = c("english", "history", "art", "art", "math", "history", "art"),
  stringsAsFactors = F)

teachers <- data.frame(
  teacher   = c("Jill", "Jill", "Sam", "Liz", "Liz", "Liz", "Rob"),
  specialty = c("math", "history", "English", "history", "literature", "art", "English"),
  stringsAsFactors = F)

（关键概念）

# you can get the specialties of a given teacher like this:
subset(teachers, teacher == 'Jill')$specialty
# [1] "math"    "history"

# you can get the set of unique teachers like this:
unique(teachers$teacher)
# [1] "Jill" "Sam"  "Liz"  "Rob" 

（解决方案）

# for each teacher, do any of their specialties match the class need of each school?
matches <- 
  sapply(unique(teachers$teacher), function(this_t) {
    specs <- subset(teachers, teacher == this_t)$specialty
    schools$class %in% specs
    })

# combine with school data.frame
data.frame(schools, matches)

#   school   class  Jill   Sam   Liz   Rob
# 1 C.H.S. english FALSE FALSE FALSE FALSE
# 2 B.H.S. history  TRUE FALSE  TRUE FALSE
# 3 D.H.S.     art FALSE FALSE  TRUE FALSE
# 4 A.H.S.     art FALSE FALSE  TRUE FALSE
# 5 Z.H.S.    math  TRUE FALSE FALSE FALSE
# 6 M.H.S. history  TRUE FALSE  TRUE FALSE
# 7 L.H.S.     art FALSE FALSE  TRUE FALSE

一些注意事项：

1) 当您在代码中包含适当的间距时，它会更容易阅读（和思考）。此外，与其创建一堆向量然后组装成 data.frames，不如一步完成——它更短，它有助于显示向量如何相互关联，并且不会弄乱您的全局环境。

2) 我将匹配值保留为 FALSE/TRUE，因为这是布尔数据，因此使用适当的数据类型是有意义的。但是，如果您真的想要 No/Yes，您可以使用这些标签将这些值更改为因子

3) 结果与您的预期略有不同，因为'English' == 'english' 是FALSE。您可能想要清理您的起始数据。如果您知道大小写会混合并且您不区分大小写匹配，则可以在比较之前将所有值强制为小写：tolower(schools$class) %in% tolower(specs)