Binning data in R

Question

I have a vector with around 4000 values. I would just need to bin it into 60 equal intervals for which I would then have to calculate the median (for each of the bins).

v<-c(1:4000)

V is really just a vector. I read about cut but that needs me to specify the breakpoints. I just want 60 equal intervals

Answer 1

Use cut and tapply:

> tapply(v, cut(v, 60), median)
          (-3,67.7]          (67.7,134]           (134,201]           (201,268] 
               34.0               101.0               167.5               234.0 
          (268,334]           (334,401]           (401,468]           (468,534] 
              301.0               367.5               434.0               501.0 
          (534,601]           (601,668]           (668,734]           (734,801] 
              567.5               634.0               701.0               767.5 
          (801,867]           (867,934]         (934,1e+03]    (1e+03,1.07e+03] 
              834.0               901.0               967.5              1034.0 
(1.07e+03,1.13e+03]  (1.13e+03,1.2e+03]  (1.2e+03,1.27e+03] (1.27e+03,1.33e+03] 
             1101.0              1167.5              1234.0              1301.0 
 (1.33e+03,1.4e+03]  (1.4e+03,1.47e+03] (1.47e+03,1.53e+03]  (1.53e+03,1.6e+03] 
             1367.5              1434.0              1500.5              1567.0 
 (1.6e+03,1.67e+03] (1.67e+03,1.73e+03]  (1.73e+03,1.8e+03]  (1.8e+03,1.87e+03] 
             1634.0              1700.5              1767.0              1834.0 
(1.87e+03,1.93e+03]    (1.93e+03,2e+03]    (2e+03,2.07e+03] (2.07e+03,2.13e+03] 
             1900.5              1967.0              2034.0              2100.5 
 (2.13e+03,2.2e+03]  (2.2e+03,2.27e+03] (2.27e+03,2.33e+03]  (2.33e+03,2.4e+03] 
             2167.0              2234.0              2300.5              2367.0 
 (2.4e+03,2.47e+03] (2.47e+03,2.53e+03]  (2.53e+03,2.6e+03]  (2.6e+03,2.67e+03] 
             2434.0              2500.5              2567.0              2634.0 
(2.67e+03,2.73e+03]  (2.73e+03,2.8e+03]  (2.8e+03,2.87e+03] (2.87e+03,2.93e+03] 
             2700.5              2767.0              2833.5              2900.0 
   (2.93e+03,3e+03]    (3e+03,3.07e+03] (3.07e+03,3.13e+03]  (3.13e+03,3.2e+03] 
             2967.0              3033.5              3100.0              3167.0 
 (3.2e+03,3.27e+03] (3.27e+03,3.33e+03]  (3.33e+03,3.4e+03]  (3.4e+03,3.47e+03] 
             3233.5              3300.0              3367.0              3433.5 
(3.47e+03,3.53e+03]  (3.53e+03,3.6e+03]  (3.6e+03,3.67e+03] (3.67e+03,3.73e+03] 
             3500.0              3567.0              3633.5              3700.0 
 (3.73e+03,3.8e+03]  (3.8e+03,3.87e+03] (3.87e+03,3.93e+03]    (3.93e+03,4e+03] 
             3767.0              3833.5              3900.0              3967.0

Answer 2

In the past, i've used this function

evenbins <- function(x, bin.count=10, order=T) {
    bin.size <- rep(length(x) %/% bin.count, bin.count)
    bin.size <- bin.size + ifelse(1:bin.count <= length(x) %% bin.count, 1, 0)
    bin <- rep(1:bin.count, bin.size)
    if(order) {    
        bin <- bin[rank(x,ties.method="random")]
    }
    return(factor(bin, levels=1:bin.count, ordered=order))
}

and then i can run it with

v.bin <- evenbins(v, 60)

and check the sizes with

table(v.bin)

and see they all contain 66 or 67 elements. By default this will order the values just like cut will so each of the factor levels will have increasing values. If you want to bin them based on their original order,

v.bin <- evenbins(v, 60, order=F)

instead. This just split the data up in the order it appears