How to assign "cut" range midpoints in R?

Question

I am using cut to divide my data into bins, which gives the resulting bin as something like (x1,x2]. Can anyone tell me how I might make a new column that expresses these bins as the midpoint of the bin? For example, with the following dataframe:

structure(list(x = c(1L, 4L, 6L, 7L, 8L, 9L, 12L, 18L, 19L), 
    y = 1:9), .Names = c("x", "y"), class = "data.frame", row.names = c(NA, 
-9L))

I can use

test$xRange <- cut(test$x, breaks=seq(0, 20, 5))

to give

    x   y   xRange
1   1   1   (0,5]
2   4   2   (0,5]
3   6   3   (5,10]
4   7   4   (5,10]
5   8   5   (5,10]
6   9   6   (5,10]
7   12  7   (10,15]
8   18  8   (15,20]
9   19  9   (15,20]

But the result I need should instead look like:

    x   y   xRange        xMidpoint
1   1   1   (0,5]         2.5
2   4   2   (0,5]         2.5
3   6   3   (5,10]        7.5
4   7   4   (5,10]        7.5
5   8   5   (5,10]        7.5
6   9   6   (5,10]        7.5
7   12  7   (10,15]       12.5
8   18  8   (15,20]       17.5
9   19  9   (15,20]       17.5

I've done some searching, and came upon a similar question at divide a range of values in bins of equal length: cut vs cut2, which gives a solution as

cut2 <- function(x, breaks) {
  r <- range(x)
  b <- seq(r[1], r[2], length=2*breaks+1)
  brk <- b[0:breaks*2+1]
  mid <- b[1:breaks*2]
  brk[1] <- brk[1]-0.01
  k <- cut(x, breaks=brk, labels=FALSE)
  mid[k]
}

But when I try this on my case, using

test$xMidpoint <- cut2(test$x, 5)

it does not return the correct midpoint. Perhaps I am entering the breaks incorrectly in cut2? Can anyone tell me what I'm doing incorrectly?

Feminine

Answer 1

Unless I miss something, something like this looks valid:

brks = seq(0, 20, 5)
ints = findInterval(test$x, brks, all.inside = T)
#mapply(function(x, y) (x + y) / 2, brks[ints], brks[ints + 1])  #which is ridiculous
#[1]  2.5  2.5  7.5  7.5  7.5  7.5 12.5 17.5 17.5
(brks[ints] + brks[ints + 1]) / 2  #as sgibb noted
#[1]  2.5  2.5  7.5  7.5  7.5  7.5 12.5 17.5 17.5
(head(brks, -1) + diff(brks) / 2)[ints] #or using thelatemail's idea from the comments
#[1]  2.5  2.5  7.5  7.5  7.5  7.5 12.5 17.5 17.5

Womenfolk

That worked beautifully, alexis_lax! Thank you so much for your help! – Feminine

@alexis_laz: You could simply replace your mapply by (brks[ints]+brks[ints+1])/2. – Plague

@Plague : Right! I'm embarassed! – Womenfolk

+1! I must become friend with findInterval...And yes, I was too tired when trying to answer. Thanks for pointing out my mistake! – Aigrette

@Aigrette : Your approach with hist was indeed elegant, though! If only it hadn't restrictions using it – Womenfolk

Similar, I was going to suggest: (head(brks,-1) + diff(brks)/2)[cut(test$x, breaks=brks)] – Culminant

Answer 2

I know this is a really old question, but this may help future googlers. I wrote a function that I called midcut that cuts the data and provides me with the midpoint of the bin.

midcut<-function(x,from,to,by){
   ## cut the data into bins...
   x=cut(x,seq(from,to,by),include.lowest=T)
   ## make a named vector of the midpoints, names=binnames
   vec=seq(from+by/2,to-by/2,by)
   names(vec)=levels(x)
   ## use the vector to map the names of the bins to the midpoint values
   unname(vec[x])
}

example

test$midpoint=midcut(test$x,0,20,5)
> test
   x y  xRange midpoint
1  1 1   (0,5]      2.5
2  4 2   (0,5]      2.5
3  6 3  (5,10]      7.5
4  7 4  (5,10]      7.5
5  8 5  (5,10]      7.5
6  9 6  (5,10]      7.5
7 12 7 (10,15]     12.5
8 18 8 (15,20]     17.5
9 19 9 (15,20]     17.5

Crystalloid

Answer 3

An alternative way of calculating midpoints regardless of how you specify the breaks in "cut" function (i.e. regardless of wether you supply a vector of breakpoints or a number of bins) is using the label text that the cut function supplies.

get_midpoint <- function(cut_label) {
  mean(as.numeric(unlist(strsplit(gsub("\\(|\\)|\\[|\\]", "", as.character(cut_label)), ","))))
}

test$xMidpoint <- sapply(test$xRange, get_midpoint)

Note that this requires the "labels" argument in the cut function to be set to TRUE.

Alfreda