Question

我的数据看起来像：

require(data.table)
DT <- data.table(x=c(19,19,19,21,21,19,19,22,22,22),
             y=c(53,54,55,32,44,45,49,56,57,58))

我想沿x搜索，并计算y的均值。但是，使用时。

DT[, .(my=mean(y)), by=.(x)]

我得到了x的重合值的总体方法。我想沿x搜索，每次x改变，我想计算一个新的均值。对于提供的示例，输出将为：

DTans <- data.table(x=c(19,21,19,22),
             my=c(54,38,47,57))

Answer 1

我们可以使用rleid创建另一个分组变量，获取＆＃39; y＆＃39;的mean，然后分配＆＃39; indx＆＃39;为NULL

library(data.table) # v 1.9.5+
DT[, .(my = mean(y)), by = .(indx = rleid(x), x)][, indx := NULL]
#    x my
#1: 19 54
#2: 21 38
#3: 19 47
#4: 22 57

基准

set.seed(24)
foo <- function(x) sample(x, 1e7L, replace = TRUE)
DT  <- data.table(x = foo(100L), y = foo(10000L))

josilber <- function() {
    new.group <- c(1, diff(DT$x) != 0)
    res <- data.table(x = DT$x[new.group == 1], 
              my = tapply(DT$y, cumsum(new.group), mean))
}

Roland <- function() {
    DT[, .(my = mean(y), x = x[1]), by = cumsum(c(1, diff(x) != 0))]
}

akrun <- function() { 
    DT[, .(my = mean(y)), by = .(indx = rleid(x), x)][,indx := NULL]
}

bgoldst <- function() {
    with(rle(DT$x), data.frame(x = values, 
       my = tapply(DT$y, rep(1:length(lengths), lengths), mean)))
}

system.time(josilber())
#   user  system elapsed 
#159.405   1.759 161.110 

system.time(bgoldst())
#   user  system elapsed 
#162.628   0.782 163.380 

system.time(Roland())
#   user  system elapsed 
# 18.633   0.052  18.678 

system.time(akrun())
#   user  system elapsed 
# 1.242   0.003   1.246

Answer 2

您可以识别连续元素组，然后确定每个元素的平均值和值：

(new.group <- c(1, diff(DT$x) != 0))
# [1] 1 0 0 1 0 1 0 1 0 0
DT[, list(x = x[1L], my = mean(y)), by = list(indx = cumsum(new.group))]
#    indx  x my
# 1:    1 19 54
# 2:    2 21 38
# 3:    3 19 47
# 4:    4 22 57

Answer 3

with(rle(DT$x),data.frame(x=values,my=tapply(DT$y,rep(1:length(lengths),lengths),mean)));
##    x my
## 1 19 54
## 2 21 38
## 3 19 47
## 4 22 57

r沿向量搜索并计算平均值

3 个答案:

基准