R：熔化和合并数据

Question

这是我的数据集的一个示例：

ID = c(1, 2, 3, 4) 
Allegation = c("A::B::C::V", "A::C", "A::D", "D::E::D") 
Disposition = c("Open::Closed::Open", "Closed::Closed", "Open::Open", "Closed::Open") 
df <- data.frame(ID,Allegation, Disposition)  

  ID Allegation        Disposition
  1 A::B::C::V Open::Closed::Open
  2       A::C     Closed::Closed
  3       A::D         Open::Open
  4    D::E::D       Closed::Open

我想要以下结果：

ID  Allegation  Disposition Allegation_detail   Dispostion_detail
1   A::B::C::V  Open::Closed::Open  A       Open
1   A::B::C::V  Open::Closed::Open  B       Closed
1   A::B::C::V  Open::Closed::Open  C       Open  
1   A::B::C::V  Open::Closed::Open  V       NA
2     A::C      Closed::Closed      A       Closed

我曾尝试将数据融合，然后将其合并，但我没有获得所需的输出

这是我目前的方法：

#Create column to see num of allegations
df$num_allegations <- (str_count(as.character(df$Allegation), "::") +1) 

#Looking max allegations
max(df$num_allegations)

#Expanding allegations
df$Allegation1 <- sapply(strsplit(as.character(df$Allegation), "::", fixed= TRUE), `[`, 1)
df$Allegation2 <- sapply(strsplit(as.character(df$Allegation), "::", fixed= TRUE), `[`, 2)
df$Allegation3 <- sapply(strsplit(as.character(df$Allegation), "::", fixed= TRUE), `[`, 3)
df$Allegation4 <- sapply(strsplit(as.character(df$Allegation), "::", fixed= TRUE), `[`, 4)

#Expanding Disposition
df$Disposition1 <- sapply(strsplit(as.character(df$Disposition), "::", fixed= TRUE), `[`, 1)
df$Disposition2 <- sapply(strsplit(as.character(df$Disposition), "::", fixed= TRUE), `[`, 2)
df$Disposition3 <- sapply(strsplit(as.character(df$Disposition), "::", fixed= TRUE), `[`, 3)
df$Disposition4 <- sapply(strsplit(as.character(df$Disposition), "::", fixed= TRUE), `[`, 4)

#melting data
dfmelt1 <- melt(df[,c(1:8)], id=c("ID", "Allegation", "Disposition", "num_allegations"))
dfmelt2 <- melt(df[,c(1,2,3,4,9,10,11,12)], id=c("ID", "Allegation", "Disposition", "num_allegations"))
colnames(dfmelt2) <- c("ID" ,"Allegation" ,"Disposition","num_allegations", "variable2",
                   "value2")

但是当我合并数据时，我得到了这个结果，这不是我想要的：

merge(dfmelt1, dfmelt2, by = c("ID", "Allegation", "Disposition", "num_allegations"))

ID Allegation        Disposition num_allegations    variable value       variable2 value2
 1 A::B::C::V Open::Closed::Open               4 Allegation1     A Disposition1   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation1     A Disposition2 Closed
 1 A::B::C::V Open::Closed::Open               4 Allegation1     A Disposition3   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation1     A Disposition4   <NA>
 1 A::B::C::V Open::Closed::Open               4 Allegation2     B Disposition1   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation2     B Disposition2 Closed
 1 A::B::C::V Open::Closed::Open               4 Allegation2     B Disposition3   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation2     B Disposition4   <NA>
 1 A::B::C::V Open::Closed::Open               4 Allegation3     C Disposition1   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation3     C Disposition2 Closed
 1 A::B::C::V Open::Closed::Open               4 Allegation3     C Disposition3   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation3     C Disposition4   <NA>
 1 A::B::C::V Open::Closed::Open               4 Allegation4     V Disposition1   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation4     V Disposition2 Closed
 1 A::B::C::V Open::Closed::Open               4 Allegation4     V Disposition3   Open
 1 A::B::C::V Open::Closed::Open               4 Allegation4     V Disposition4   <NA>
 2       A::C     Closed::Closed               2 Allegation1     A Disposition1 Closed

我怎样才能合并，所以我得到处置 1，只有在它说指控 1 的地方？

谢谢

Answer 1

这是一个想法，

#get a vector with repeats for expanding the data.frame
ind <- stringr::str_count(df$Allegation, '\\w+') 
new_df <- df[rep(row.names(df), ind),]
#create vector with allegation details
v1 <- do.call(rbind, sapply(strsplit(as.character(df$Allegation), '::'), function(i)
                                                                  t(as.data.frame(t(i)))))
#create vector with Disposition details
v2 <- do.call(rbind, sapply(strsplit(as.character(df$Disposition), '::'), function(i)
                                                                  t(as.data.frame(t(i)))))
v2 <- v2[match(make.unique(rownames(v1)), make.unique(rownames(v2)))]

#construct final data frame
final_df <- data.frame(new_df, Allegation_detail=v1, Disposition_detail=v2, 
                                              stringsAsFactors = FALSE, row.names = NULL)

final_df
#    ID Allegation        Disposition Allegation_detail Disposition_detail
#1    1 A::B::C::V Open::Closed::Open                 A               Open
#2    1 A::B::C::V Open::Closed::Open                 B             Closed
#3    1 A::B::C::V Open::Closed::Open                 C               Open
#4    1 A::B::C::V Open::Closed::Open                 V               <NA>
#5    2       A::C     Closed::Closed                 A             Closed
#6    2       A::C     Closed::Closed                 C             Closed
#7    3       A::D         Open::Open                 A               Open
#8    3       A::D         Open::Open                 D               Open
#9    4    D::E::D       Closed::Open                 D             Closed
#10   4    D::E::D       Closed::Open                 E               Open
#11   4    D::E::D       Closed::Open                 D               <NA>

Answer 2

这是一个使用 data.table 的解决方案，但在逻辑上它类似于您的算法

library(data.table)
library(stringi)
setDT(df)
splitter <- function(x) as.vector(stri_list2matrix(stri_split_fixed(x, "::")))

#find the max parts for padding NA at the end
#http://stackoverflow.com/questions/17804389/pad-each-element-in-a-list-to-specific-length-in-r
df[, Len:=max(lengths(lapply(.SD, splitter))), by="ID"]

#split using ::
parsedDF <- df[, lapply(.SD, function(x) {
        ans <- splitter(x)
        length(ans) <- Len
        ans
    }), by="ID"][,
        Len:=NULL]
setnames(parsedDF, names(parsedDF), paste0(names(parsedDF),"_detail"))

#join back with original data.table
df[parsedDF, on=c("ID"="ID_detail")][,
    Len:=NULL]

## ID Allegation        Disposition Allegation_detail Disposition_detail
## 1:  1 A::B::C::V Open::Closed::Open                 A               Open
## 2:  1 A::B::C::V Open::Closed::Open                 B             Closed
## 3:  1 A::B::C::V Open::Closed::Open                 C               Open
## 4:  1 A::B::C::V Open::Closed::Open                 V                 NA
## 5:  2       A::C     Closed::Closed                 A             Closed
## 6:  2       A::C     Closed::Closed                 C             Closed
## 7:  3       A::D         Open::Open                 A               Open
## 8:  3       A::D         Open::Open                 D               Open
## 9:  4    D::E::D       Closed::Open                 D             Closed
## 10:  4    D::E::D       Closed::Open                 E               Open
## 11:  4    D::E::D       Closed::Open                 D                 NA