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
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
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
This result shows the 59 median values of the break-points. The 60 bin values are probably as close to equal as possible (but probably not exactly equal).
> sq <- seq(1, 4000, length = 60)
> sapply(2:length(sq), function(i) median(c(sq[i-1], sq[i])))
# [1] 34.88983 102.66949 170.44915 238.22881 306.00847 373.78814
# [7] 441.56780 509.34746 577.12712 644.90678 712.68644 780.46610
# ......
Actually, after checking, the bins are pretty darn close to being equal.
> unique(diff(sq))
# [1] 67.77966 67.77966 67.77966
来源：https://stackoverflow.com/questions/24359863/binning-data-in-r