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Retrieving vector names by time

I wrote this loop to extract the names of each element of a vector that occurs during a time interval (bin). I was wondering if I lacked a faster way to do this ... I want to implement the randomization aspect for vectors of length 1000 and therefore do not want to rely on the loop.

mydata <- structure(c(1199.91666666667, 1200.5, 1204.63333333333, 1205.5, 
                      1206.3, 1208.73333333333, 1209.06666666667, 1209.93333333333, 
                      1210.98333333333, 1214.56666666667, 1216.06666666667, 1216.63333333333, 
                      1216.91666666667, 1219.13333333333, 1221.35, 1221.51666666667, 
                      1225.35, 1225.53333333333, 1225.96666666667, 1227.61666666667, 
                      1228.91666666667, 1230.31666666667, 1233.53333333333, 1235.8, 
                      1237.51666666667, 1239.41666666667, 1241.6, 1247.08333333333, 
                      1247.45, 1252.7, 1253.26666666667), .Names = c("B", "A", "B", 
                                                                     "E", "A", "A", "B", "G", "G", "C", "A", "D", "E", "B", "B", "E", 
                                                                     "E", "G", "F", "A", "C", "A", "F", "B", "A", "F", "F", "G", "F", 
                                                                     "G", "F"))


mydata

      B        A        B        E        A        A        B        G        G        C        A        D        E        B        B        E        E 
1199.917 1200.500 1204.633 1205.500 1206.300 1208.733 1209.067 1209.933 1210.983 1214.567 1216.067 1216.633 1216.917 1219.133 1221.350 1221.517 1225.350 
       G        F        A        C        A        F        B        A        F        F        G        F        G        F 
1225.533 1225.967 1227.617 1228.917 1230.317 1233.533 1235.800 1237.517 1239.417 1241.600 1247.083 1247.450 1252.700 1253.267

They represent consecutive times in seconds of events. Say we want to make our intervals long. My approach is to make the start vector of each interval, and then use a loop to find the names of the elements occurring in this interval:

N=5
ints <- seq(mydata[1], mydata[length(mydata)], N)

out<-list()
for(i in 1:length(ints)){
  out[[i]] <- names(mydata[mydata>=ints[i] & mydata<ints[i]+N])
}

out


[[1]]
[1] "B" "A" "B"

[[2]]
[1] "E" "A" "A" "B"

[[3]]
[1] "G" "G" "C"

[[4]]
[1] "A" "D" "E" "B"

[[5]]
[1] "B" "E"

[[6]]
[1] "E" "G" "F" "A" "C"

[[7]]
[1] "A" "F"

[[8]]
[1] "B" "A" "F"

[[9]]
[1] "F"

[[10]]
[1] "G" "F"

[[11]]
[1] "G" "F"

This is good for small samples, but I see that it will be slow when you are dealing with very large samples that are rebuilt 1000 times.

+2
2

- findInterval ( ):

mydata2 = c(-Inf, mydata)
ints <- seq(mydata[1], mydata[length(mydata)]+5, N)
idx = findInterval(ints-1e-10, mydata2)

out<-list()
for(i in 1:(length(ints)-1)){
  out[[i]] <- names(mydata2[(idx[i]+1):(idx[i+1])])
}

, ( , , , epsilon). , :

> out
[[1]]
[1] "B" "A" "B"

[[2]]
[1] "E" "A" "A" "B"

[[3]]
[1] "G" "G" "C"

[[4]]
[1] "A" "D" "E" "B"

[[5]]
[1] "B" "E"

[[6]]
[1] "E" "G" "F" "A" "C"

[[7]]
[1] "A" "F"

[[8]]
[1] "B" "A" "F"

[[9]]
[1] "F"

[[10]]
[1] "G" "F"

[[11]]
[1] "G" "F"

:

> microbenchmark( jalapic = {out<-list(); for(i in 1:length(ints)){out[[i]] <- names(mydata[mydata>=ints[i] & mydata<ints[i]+N])}},
+   mts = {idx = findInterval(ints2-1e-10, mydata2); out<-list(); for(i in 1:(length(ints)-1)){out[[i]] <- names(mydata2[(idx[i]+1):(idx[i+1])])}}, 
+   alexis = {split(names(mydata), findInterval(mydata, ints))},
+   R_Yoda = {dt[, groups := cut2(data,ints)]; result <- dt[, paste0(names, collapse=", "), by=groups]})
Unit: microseconds
    expr      min        lq       mean    median       uq      max neval
 jalapic   67.177   76.9725   85.73347   82.8035   95.866  119.890   100
     mts   43.851   52.7150   62.72116   58.3130   73.007   96.099   100
  alexis   75.573   86.5360   95.72593   91.4340  100.531  234.649   100
  R_Yoda 2032.066 2158.4870 2303.68887 2191.3750 2281.409 8719.314   100

( 2000) :

set.seed(123)
mydata = sort(runif(n = 2000, min = 0, max = 100))
names(mydata) = sample(LETTERS[1:7], size = 2000, replace = T)
mydata2 = c(-Inf, mydata)
ints2 <- seq(mydata[1], mydata[length(mydata)]+5, N)
dt <- data.table(data=mydata, names=names(mydata) )
> microbenchmark( jalapic = {out<-list(); for(i in 1:length(ints)){out[[i]] <- names(mydata[mydata>=ints[i] & mydata<ints[i]+N])}},
+                 mts = {idx = findInterval(ints2-1e-10, mydata2); out<-list(); for(i in 1:(length(ints)-1)){out[[i]] <- names(mydata2[(idx[i]+1):(idx[i+1])])}}, 
+                 alexis = {split(names(mydata), findInterval(mydata, ints))},
+                 R_Yoda = {dt[, groups := cut2(data,ints)]; result <- dt[, paste0(names, collapse=", "), by=groups]})
Unit: microseconds
    expr      min        lq      mean    median        uq       max neval
 jalapic  804.243  846.9275  993.9957  862.0890  883.3140  7140.218   100
     mts   77.439   88.8685  100.6148  100.0640  106.5955   188.466   100
  alexis  187.066  204.7930  220.1689  215.5225  225.3190   299.026   100
  R_Yoda 3831.348 4066.4640 4366.5382 4140.1700 4248.8635 11829.923   100
+3

data.table:

: , ( mts)

library(Hmisc)
library(data.table)

# assuming that your mydata vector from the question is loaded
N=5   # code from your question...
ints <- seq(mydata[1], mydata[length(mydata)], N)   # code from your question...

dt <- data.table(data=mydata, names=names(mydata) )
dt[, groups := cut2(data,ints)]  # attention: shall the interval ends be included in the group or not?
groups <- dt[ , .(result=list(names)), by=groups]    # the elements of a data.table can be a list itself!
# to get the result as list:
out <- groups[,result]
out

: cut2 findInterval , :

out <- dt[, .(result=list(names)), by = findInterval(data,ints) ]

:

[[1]]
[1] "B" "A" "B"

[[2]]
[1] "E" "A" "A" "B"

[[3]]
[1] "G" "G" "C"

[[4]]
[1] "A" "D" "E" "B"

[[5]]
[1] "B" "E"

[[6]]
[1] "E" "G" "F" "A" "C"

[[7]]
[1] "A" "F"

[[8]]
[1] "B" "A" "F"

[[9]]
[1] "F"

[[10]]
[1] "G" "F"

[[11]]
[1] "G" "F"
+1

Source: https://habr.com/ru/post/1619014/

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