如何使用R将分类数据转换为相对计数

Question

Minimal example

I have: input<-data.frame(id=c(1,1,1,2,2,2),A=as.factor(c(1,1,2,2,1,3)),B=as.factor(c(0,1,1,1,0,0)))

I want: output<-data.frame(id=c(1,2), A1=c(2/3,1/3), A2=c(1/3,1/3), A3=c(0/3,1/3), B0=c(1/3,2/3), B1=c(2/3,1/3))

Motivation

I have a data frame with categorical data. I would like to turn this into a dataframe with proportianal counts of each category occuring. In the "output" dataframe I would like to have a column for each variable-category combination (A1,A2, etc.). Each row gives the relative counts for a "id". For instance, id=1 has three entries in "input" with two times category 1 under variable "A". The column "A1" should show 2/3 in that row. Divided by three, because id=1 has three entries in "input".

What I started

function(input){

#create empty dataframe
distcID<-duplicated(input$id)
output<-data.frame(id=integer(0),A1=integer(0),A2=integer(0),A3=integer(0),
                 B0=integer(0),B1=integer(0))

count<-0

for (i in input$id[distcID]){
df.cID<-input[input$customer_ID==i]
m<- NROW(df.cID)
count<-count+1
output$customer_ID[count]<-i
output$A1[count]<-1/m*NROW(df.cID$A==1)
output$A2[count]<-1/m*NROW(df.cID$A==2)
output$A3[count]<-1/m*NROW(df.cID$A==3)
output$B0[count]<-1/m*NROW(df.cID$B==0)
output$B1[count]<-1/m*NROW(df.cID$B==1)
}
return(output)

}

What is wrong? - it is ugly. Given functions like apply and aggregate or a package like plyr, there should be nicer (ie shorter) solutions to this problem.

• R does not accept the initialization of output with empty columns.

• the column names of output are not created automatically, but by hand.

Thank you! Please tell me if my question lacks clarity. This is my first question here.

Answer 1

This expression creates a table for each of the non-ID columns (here, 2:3 ):

individuals <- lapply(2:3, function(i) {
  # Table of counts, by "id"
  x <- table(input[,c(1,i)])

  # Scale to proportions
  x <- x / rowSums(x)

  # Fix the names
  colnames(x) <- paste0(colnames(input)[i], colnames(x))

  return(x)
}
)

individuals
## [[1]]
##    A
## id         A1        A2        A3
##   1 0.6666667 0.3333333 0.0000000
##   2 0.3333333 0.3333333 0.3333333
## 
## [[2]]
##    B
## id         B0        B1
##   1 0.3333333 0.6666667
##   2 0.6666667 0.3333333

Now put them together with cbind :

do.call(cbind, individuals)
##          A1        A2        A3        B0        B1
## 1 0.6666667 0.3333333 0.0000000 0.3333333 0.6666667
## 2 0.3333333 0.3333333 0.3333333 0.6666667 0.3333333

The id column is not present, but the row names can be used for this purpose.

Answer 2

This isn't a complete answer, but should help you along the way (with a bit of resphape[2] -ing:

ct <- count(input, "id")
A <- data.frame(table(input[,c(1,2)])/ct[ct$id==1,]$freq)
B <- data.frame(table(input[,c(1,3)])/ct[ct$id==2,]$freq)

print(A)
  id A      Freq
1  1 1 0.6666667
2  2 1 0.3333333
3  1 2 0.3333333
4  2 2 0.3333333
5  1 3 0.0000000
6  2 3 0.3333333

print(B)
  id B      Freq
1  1 0 0.3333333
2  2 0 0.6666667
3  1 1 0.6666667
4  2 1 0.3333333

Answer 3

Here's on possible solution:

library(reshape2)
library(qdap)

x <- prop.table(ftable(melt(input, id="id")))
x2 <- colpaste2df(data.frame(x), 2:3, keep.orig = FALSE, sep="", name.sep = "")
x3 <- dcast(x2, id  ~ variablevalue, value.var = "Freq")
x3[, c(TRUE, colSums(x3[, -1]) != 0)]

##   id         A1         A2         A3         B0         B1
## 1  1 0.16666667 0.08333333 0.00000000 0.08333333 0.16666667
## 2  2 0.08333333 0.08333333 0.08333333 0.16666667 0.08333333

Answer 4

Can be seen as a pivot table (or two pivot tables):

>install.packages('reshape')
>library(reshape)
>ct <-count(input, "id")
>DF1<-cast(input, id ~ A, value='B')
>DF2<-cast(input, id ~ B, value="A")
>DF3<-cbind(DF1$id, DF1[names(DF1)!='id']/ct[1,]$freq, DF2[names(DF2)!='id']/ct[2,]$freq)
>names(DF3)<-c('id', paste('A', names(DF1)[-1], sep=''), paste('B', names(DF2)[-1], sep=''))
> DF3
  id        A1        A2        A3        B0        B1
1  1 0.6666667 0.3333333 0.0000000 0.3333333 0.6666667
2  2 0.3333333 0.3333333 0.3333333 0.6666667 0.3333333

Answer 5

This is what I think you wanted. Just add row or column names to suit your tastes.

 tbB <- with(input, table(B, id))
 tbA <- with(input, table(A, id))
 cbind( t( tbA/rowSums(tbA)), t(tbB/rowSums(tbB)) )
          1   2 3         0         1
1 0.6666667 0.5 0 0.3333333 0.6666667
2 0.3333333 0.5 1 0.6666667 0.3333333