CODEWITHSUNDEEP
rif-statementintervalsbins
determin
determin

Reputation: 69

R case when according to intervals

In R ;

I want to put that 'points' according to that intervals in my table

enter image description here

df <- data.frame(
  category=c("aa","aa","aa","bb","bb","bb"), 
  intervals = c("[-Inf,20)", "[20,212)", "[212,Inf)", "[-Inf,150)", "[150,260)","[260,Inf)"), 
  points = c(45, 15, 35, -15,10,35))

the case is :

if 'category'=aa then if 'intervals'=[20,212) then points=15

but it should not be manual like that it should work related to my below table columns!

enter image description here

Thanks!

Upvotes: 1

Views: 425

Answers (1)

GKi
GKi

Reputation: 39647

You can use findInterval to link points to a number.

y$points <- sapply(seq_len(nrow(y)), function(i) {
  x[[y$category[i]]]$points[findInterval(y$intervals[i], x[[y$category[i]]]$intervalls)]
})
y
#  category intervals points
#1       aa       -50     45
#2       aa        25     15
#3       aa        55     15
#4       aa       250     35
#5       bb         5    -15
#6       bb       170     10
#7       bb       290     35

Data:

x <- list(aa=list(intervalls=c(-Inf, 20, 212, Inf), points=c(45,15,35))
         ,bb=list(intervalls=c(-Inf, 150, 260, Inf), points=c(-15,10,35)))

y <- data.frame(category=c("aa","aa","aa","aa","bb","bb","bb")
  , intervals=c(-50,25,55,250,5,170,290))

Create x from df given in the question:

tt <- lapply(strsplit(gsub("[[)]", "", df$intervals), ","), as.numeric)
x <- lapply(unique(df$category), function(i) {
  list(intervalls=sort(unique(unlist(tt[which(df$category == i)])))
     , points=df$points[df$category==i][order(unlist(lapply(tt[which(df$category == i)], "[[", 2)))])
})
names(x)  <- unique(df$category)

Upvotes: 1

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