Create a binning variable in r to roll data up on? - r

Question

I'd like to create a variable that bins values from another variable based on a binwidth
The data would look something like this if I wanted to create a bin variable based on counts where:
1 to 5 = 1
6 to 10 = 2
11 to 15 = 3
Without hand recoding each bin is there a function that can do something like this in R?

Answer 1

Since it looks like you want to get a numeric rather than a factor result, try something like trunc((mydata$count-1)/5)+1
e.g.
mydata$bucket = trunc((mydata$count-1)/5)+1
There's also the ceiling function, which is a little simpler:
mydata$bucket = ceiling(mydata$count/5)
see ?round
So on your data:
mydata = data.frame(spend=c(21,32,34,43,36,39,33,47,47,47,25,50,44,44) ,
count=c(3L,1L,2L,15L,1L,8L,1L,11L,15L,11L,3L,12L,11L,4L) )
mydata$bucket = ceiling(mydata$count/5)
Which gives:
> mydata
spend count bucket
1 21 3 1
2 32 1 1
3 34 2 1
4 43 15 3
5 36 1 1
6 39 8 2
7 33 1 1
8 47 11 3
9 47 15 3
10 47 11 3
11 25 3 1
12 50 12 3
13 44 11 3
14 44 4 1

Answer 2

Yeah its called the cut function
? cut

Answer 3

You can use the generic cut() function. For a numeric vector x, the method has these arguments:
> args(cut.default)
function (x, breaks, labels = NULL, include.lowest = FALSE, right = TRUE,
dig.lab = 3L, ordered_result = FALSE, ...)
The argument breaks is central here. It is either a number of intervals or a vector of “breakpoints” defining your intervals. Note that all intervals are by default right-open (right = TRUE), so by creating an object x, containing the numbers from 1 to 100 and defining a vector of breakpoints (brk) {1, 20, 50, 100}, you will get these results (after using table() on the result):
> x <- 1:100
> brk <- c(1,20,50,100)
> table(cut(x = x, breaks = brk))
(1,20] (20,50] (50,100]
19 30 50
You can see that the first interval is $(1,\,20]$, so 1 is not part of it and the first observation will become a missing value NA (as all other observations outside the defined intervals).
By setting include.lowest = TRUE, R includes the lowest value (i.e., the first interval will be closed), so I think this will produce what you want:
> x <- 1:100
> brk <- c(1,20,50,100)
> table(cut(x = x, breaks = brk, include.lowest = TRUE))
[1,20] (20,50] (50,100]
20 30 50
The argument right reverses the whole process, so intervals are left-open by default and include.lowest will close the last interval (i.e., include the highest value in the last category).
As the resulting object will be of class "factor", you might consider setting ordered_result to TRUE, producing an ordered factor object (classes "ordered" and "factor").
Labelling, etc. is optional (see ?cut).

Answer 4

The cut function can actually accomplish binning a variable while keeping it as a continuous variable you just need to use the labels parameter:
myData$bucket <- cut(myData$counts, breaks = 30, labels = rep(1:30))