我有一对数据框，包含多个模型参数的多对估计和方差，每个模型参数都在一个部分中。这是一个生成说明性样本的函数：
samplerats <- function(){
set.seed(310366)
d = data.frame(section=c(rep("S1",10),rep("S2",10),rep("S3",5)))
nr = nrow(d)
for(i in 1:5){
d[[paste0("est_v",i)]] = rnorm(nr)
d[[paste0("var_v",i)]] = runif(nr)
}
d
}
这是你得到的开始：
> d=samplerats()
> head(d)
section est_v1 var_v1 est_v2 var_v2 est_v3 var_v3
1 S1 0.3893008 0.1620882 -1.1915391 0.15439565 0.62022284 0.5487519
2 S1 0.8221099 0.3280630 0.7729817 0.14810283 -1.11337584 0.9947342
3 S1 0.8023230 0.1862810 -1.5285389 0.85648574 -1.74666907 0.4267944
4 S1 -0.2252865 0.5660111 -0.4348341 0.53013027 0.01823185 0.1379821
5 S1 -0.9475335 0.7904085 -1.0882961 0.40567780 1.69607397 0.3450983
6 S1 0.4415259 0.2969032 0.9200723 0.08754107 0.57010457 0.7579002
[with another two variables and 25 rows in total]
任务是计算每个参数的估计方差与每个参数的方差均值的比率，按部分分组。
例如，对于变量v1，粗略地只是为了得到数字：
> d %>% group_by(section) %>% summarise(var(est_v1)/mean(var_v1))
Source: local data frame [3 x 2]
section var(est_v1)/mean(var_v1)
1 S1 0.5874458
2 S2 2.4449153
3 S3 2.8621725
这给了我们v1的答案，我们只需要重复所有其他变量。请注意，列名称为est_或var_，后跟变量名称，可能是alpha或g2或其他一些字母。
当然，我有一个可怕的解决方案：
ratit <- function(d){
isVAR <- function(s){stringr::str_sub(s,1,4)=="var_"}
spreads = reshape2::melt(d) %>% mutate(isVAR=isVAR(variable), Variable = str_replace(variable,"^.*_",""))
vout = spreads %>% group_by(Variable, section, isVAR) %>% summarise(Z=if(isVAR(variable[1])){mean(value)}else{var(value)})
ratios = vout %>% group_by(section, Variable) %>% summarise(Vratio = Z[1]/Z[2]) %>% dcast(section ~ Variable)
ratios
}
给出：
> ratit(d)
Using section as id variables
Using Vratio as value column: use value.var to override.
section v1 v2 v3 v4 v5
1 S1 0.5874458 3.504169 3.676488 1.1716684 1.742021
2 S2 2.4449153 1.177326 1.106337 1.0700636 3.263149
3 S3 2.8621725 2.216099 3.846062 0.7777452 2.122726
您可以在其中看到第一列与v1相同 - 仅前面的示例。但是你好。
如果我可以融化，演员，dplyr或以其他方式将其整理成这种格式：
est var section variable
1 0.3893008 0.1620882 S1 v1
2 0.8221099 0.3280630 S1 v1
3 0.8023230 0.1862810 S1 v1
4 -0.2252865 0.5660111 S1 v1
5 -0.9475335 0.7904085 S1 v1
6 0.4415259 0.2969032 S1 v1
然后是琐碎的 - dd %>% group_by(section, variable) %>% summarise(rat=var(est)/mean(var)) %>% spread(variable, rat)
但那一步使我无法......
欢迎使用包括基础R，dplyr，tidyr，data.table等在内的任何解决方案。
答案 0 :(得分：11)
这应该可以解决问题：
dd <- reshape(d, varying = 2:11, direction = 'long', sep="_", timevar="variable")
head(dd)
# section variable est var id
# 1.v1 S1 v1 0.3893008 0.1620882 1
# 2.v1 S1 v1 0.8221099 0.3280630 2
# 3.v1 S1 v1 0.8023230 0.1862810 3
# 4.v1 S1 v1 -0.2252865 0.5660111 4
# 5.v1 S1 v1 -0.9475335 0.7904085 5
# 6.v1 S1 v1 0.4415259 0.2969032 6
答案 1 :(得分：9)
以下是base R使用mapply
est <- d[grep('^est|section', colnames(d))]
var1 <- d[grep('^var|section', colnames(d))]
lstest <- split(est[-1], est$section)
lstvar <- split(var1[-1], var1$section)
res <- t(mapply(function(x,y) mapply(function(.x, .y)
var(.x)/mean(.y), x, y), lstest, lstvar))
或使用dplyr
est1 <- est %>%
group_by(section) %>%
summarise_each(funs(var)) %>%
data.frame()
var2 <- var1 %>%
group_by(section) %>%
summarise_each(funs(mean)) %>%
data.frame()
est1[-1]/var2[-1]
数据
samplerats <- function(){
set.seed(310366)
d <- data.frame(section=sample(paste0("S", 1:20),
1e5, replace=TRUE))
nr <- nrow(d)
for(i in 1:20){
d[[paste0('est_v', i)]] <- rnorm(nr)
d[[paste0('var_v', i)]] <- runif(nr)
}
d
}
d <- samplerats()
<强>功能
akrun <- function(){
est1 <- d %>%
group_by(section) %>%
summarise_each(funs(var), starts_with('est'))
var1 <- d %>%
group_by(section) %>%
summarise_each(funs(mean), starts_with('var') )
cbind(unique(est1[1]), est1[-1]/var1[-1])
}
#Assuming that the `reshaped` dataset from @Josh O'Brien's method
#is further processed using `spread` from `tidyr`
josh <- function(){
dd <- reshape(d, varying = 2:ncol(d), direction = 'long',
sep="_", timevar="variable")
dd %>%
group_by(section, variable) %>%
summarise(rat=var(est)/mean(var)) %>%
spread(variable, rat)
}
#Using `data.table` for @Henriks' method as the output from
# `merged.stack is `data.table`
henrik <- function(){
dS <- merged.stack(data = getanID(d, "section"),
var.stubs = c("est", "var"), sep = "_")
dcast.data.table(dS[ , list(rat=var(est)/mean(var)),
.(section, .time_1)], section~.time_1, value.var='rat')
}
DMC <- function(){
d %>%
gather(key, value, -section) %>%
separate(key, into = c("type", "var")) %>%
group_by(section, var) %>%
summarise(result = var(value[type == "est"]) / mean(value[type == "var"]))%>%
spread(var, result)
}
基准
library(microbenchmark)
microbenchmark(akrun(), josh(), henrik(), DMC(), unit='relative', times=20L)
#Unit: relative
# expr min lq mean median uq max neval
#akrun() 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 20
# josh() 323.57129 335.51929 315.05115 312.02953 293.18614 308.30833 20
#henrik() 30.02737 33.95731 32.15254 34.72281 29.55944 35.26825 20
#DMC() 204.66445 211.82019 207.47286 201.33015 207.10875 231.24048 20
# cld
# a
# d
# b
# c
@ alexis_laz的解决方案来得有点迟了。这是system.time
system.time({cbind(levels(d$section),
aggregate(. ~ section, d[c(1, grep("^est_", names(d)))], var)[-1] /
aggregate(. ~ section, d[c(1, grep("^var_", names(d)))], mean)[-1])}
)
# user system elapsed
# 2.228 0.002 2.229
system.time(akrun())
# user system elapsed
# 0.034 0.000 0.035
答案 2 :(得分：7)
“等”解决方案：
library(splitstackshape)
Reshape(data = d, id.vars = "section", var.stubs = c("est", "var"), sep = "_")
# section .id time est var
# 1: S1 1 1 0.3893008 0.16208821
# 2: S1 2 1 0.8221099 0.32806300
# 3: S1 3 1 0.8023230 0.18628100
# 4: S1 4 1 -0.2252865 0.56601106
# 5: S1 5 1 -0.9475335 0.79040846
# ---
# 121: S3 1 5 0.4823552 0.57649912
# 122: S3 2 5 0.6624314 0.27159239
# 123: S3 3 5 -0.7515308 0.09077159
# 124: S3 4 5 -0.4426505 0.81389532
# 125: S3 5 5 1.3881995 0.01433167
# or
merged.stack(data = getanID(d, "section"), var.stubs = c("est", "var"), sep = "_")
# section .id .time_1 est var
# 1: S1 1 v1 0.38930083 0.16208821
# 2: S1 1 v2 -1.19153913 0.15439565
# 3: S1 1 v3 0.62022284 0.54875189
# 4: S1 1 v4 0.07671314 0.71301067
# 5: S1 1 v5 0.53539985 0.86674969
# ---
# 121: S3 5 v1 0.87184287 0.63119596
# 122: S3 5 v2 1.26976583 0.50432276
# 123: S3 5 v3 0.02390527 0.55614582
# 124: S3 5 v4 0.15269326 0.93073954
# 125: S3 5 v5 1.38819949 0.01433167
答案 3 :(得分：3)
此dplyr管道会跳过中间表。
library(dplyr)
library(tidyr)
d %>%
gather(key, value, est_v1:var_v5) %>%
separate(key, into = c("type", "var")) %>%
group_by(section, var) %>%
summarise(
result = var(value[type == "est"]) / mean(value[type == "var"])
)
答案 4 :(得分：2)
又一次尝试：
cbind(levels(d$section),
aggregate(. ~ section, d[c(1, grep("^est_", names(d)))], var)[-1] /
aggregate(. ~ section, d[c(1, grep("^var_", names(d)))], mean)[-1])
# levels(d$section) est_v1 est_v2 est_v3 est_v4 est_v5
#1 S1 0.5874458 3.504169 3.676488 1.1716684 1.742021
#2 S2 2.4449153 1.177326 1.106337 1.0700636 3.263149
#3 S3 2.8621725 2.216099 3.846062 0.7777452 2.122726