我编写了这个循环来提取在一个时间间隔（bin）内发生的向量的每个元素的名称。我想知道我是否错过了一种更快的方法来执行此操作...我想对长度为1000s的向量实现随机化方面，因此不希望依赖循环。
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
这些代表连续几秒的事件。假设我们想要间隔5s。我的方法是创建每个区间开头的向量，然后使用循环来查找在该区间内发生的元素的名称：
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"
对于小样本来说这很好 - 但我可以看到，当处理非常大的样本时，这会变得很慢。
答案 0 :(得分：3)
我的建议是使用findInterval（基于对this earlier question of mine的回答）：
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])])
}
正如你所看到的，我必须对开头做一点修改（添加一个小于第一个断点的第一个值，添加一个epsilon）。这是结果，它与你的完全相同：
> 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"
就示例的速度而言，有一些改进：
> 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
对于较大的载体（我选择长度2000），这更清楚：
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
答案 1 :(得分：1)
出于性能原因，我使用的是data.table：
编辑：这个解决方案有效，但速度不是很快（mts答案证明）
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
编辑：你可以用findInterval替换cut2并在一行中完成所有操作，但它仍然较慢：
out <- dt[, .(result=list(names)), by = findInterval(data,ints) ]
结果如下：
[[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"