我在模拟研究中遇到lapply（）的应用。这些数据旨在帮助我们了解标准化公式如何影响提案评级结果。
评估者有三个条件：没有偏见，统一偏见（评估者的偏见增加）和双向偏见（偏见在评分者中均衡为正和负）。
假设提案的真实价值已知。
我们希望在每个偏差条件下生成一组复制数据集，以便数据集可以模拟单个提案评估期（面板）。然后，我们希望复制小组以模拟具有许多提案评估期。
以下是数据结构的示意图：
The data structure looks like this:
p = number of proposals
r = number of raters
n.panels = number of replicate panels
t.reps = list of several replicate panels
three bias conditions: n.bias - no bias
u.bias - uniform bias (raters higher than previous rater)
b.bias - bidirectional bias (balanced up and down bias)
-|
t 1 |..| --> 10*(n.bias(p*r)) + 10*(u.bias(p*r)) + 10*(b.bias(p*r) {panel replication 1}
. 2 |..| --> 10*(n.bias(p*r)) + 10*(u.bias(p*r)) + 10*(b.bias(p*r) {panel replication 2}
r : : : : :
e : : : : :
p n.panels |..| --> 10*(n.bias(p*r)) + 10*(u.bias(p*r)) + 10*(b.bias(p*r) {n. panels replications}
s
_|
以下R代码正确生成数据：
########## start of simulation parameters
set.seed(271828)
means <- matrix(c(rep(50,3), rep(60,3), rep(70,4) ), ncol = 1) # matrix of true proposal values
bias.u <- matrix(c(0,2,4,6,8), nrow=1) # unidirectional bias
bias.b <- matrix(c(0,3,-3, 5, -5), nrow=1) # bidirectional bias
ones.u <- matrix(rep(1,ncol(bias.u)), nrow = 1) # number of raters is the number of columns (r)
ones.b <- matrix(rep(1,ncol(bias.b)), nrow = 1)
ones.2 <- matrix(rep(1,nrow(means)), ncol = 1) # number of proposals is the number of rows (p)
true.ratings <- means%*%ones.u # gives matrix of true proposal value for each rater (p*r)
uni.bias <- ones.2%*%bias.u
bid.bias <- ones.2%*%bias.b # gives matrix of true rater bias for each proposal (p*r)
n.val <- nrow(means)*ncol(ones.u)
# true.ratings
# uni.bias
# bid.bias
library(MASS)
#####
##### generating replicate data...
#####
##########-------------------- analyzing mse of adjusted scores across replications
##########-------------------- developing random replicates of panel data
##########----- This means that there are (reps) replications in each of the bias conditions
##########----- to represent a plausible set of ratings in a particular collection
##########----- of panels. So for one proposal cycle (panel) , there are 3 * (reps) * nrow(means)
##########----- number of proposal ratings.
##########-----
##########----- There are (n.panels) replications of the total number of proposal ratings placed in a list
##########----- (t.reps).
n.panels <- 2 # put in the number of replicate panels that should be produced
reps <- 10 # put in the number of times each bias condition should be included in a panel
t.reps <- list()
n.bias <- list()
u.bias <- list()
b.bias <- list()
for (i in 1:n.panels)
{
{
for(j in 1:reps)
n.bias[[j]] <- true.ratings + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
for(j in 1:reps)
u.bias[[j]] <- true.ratings + uni.bias + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
for(j in 1:reps)
b.bias[[j]] <- true.ratings + bid.bias + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
}
t.reps[[i]] <- list(n.bias, u.bias, b.bias)
}
# t.reps
列表中的每个元素（t.reps）是一组审阅者的随机复制 对于整套提案。
我想应用以下功能来“调整”面板内的分数 使用整套提案分数的特征（涵盖所有评估者和提案） 在评估者中调整值。我们的想法是以某种方式纠正任何偏见 （例如，在对提案进行评级时过于苛刻或过于简单）。
应对每个（reps）数据集应用调整。
因此，对于一个面板，将有30个重复数据集（每个偏置条件10个） 每个复制数据集将有10个提案，由5个评级者评定，结果 共有300份提案。
因此，我们的想法是随机复制以了解调整后的分数与未调整分数的比较。
我曾尝试在（t.reps）列表中的列表中使用lapply（）函数，但它不起作用。
adj.scores <- function(x, tot.dat)
{
t.sd <- sd(array(tot.dat))
t.mn <- mean(array(tot.dat))
ones.t.mn <- diag(1,ncol(x))
p <- nrow(x)
r <- ncol(x)
ones.total <- matrix(1,p,r)
r.sd <- diag(apply(x,2, sd))
r.mn <- diag(apply(x,2, mean))
den.r.sd <- ginv(r.sd)
b.shift <- x%*%den.r.sd
a <- t.mn*ones.t.mn - den.r.sd%*%r.mn
a.shift <- ones.total%*%a
l.x <- b.shift + a.shift
return(l.x)
}
########## I would like to do something like this...
########## apply the function to each element in the list t.reps
dat.1 <- matrix(unlist(t.reps[[1]]), ncol=5)
adj.rep.1 <- lapply(t.reps[[1]], adj.scores, tot.dat = dat.1)
我对其他方法/解决方法持开放态度，允许使用整套评级中的统计数据在一组提案评级中进行评估。可能有一些R功能，我只是不知道或没有遇到过。
此外，如果任何人都可以推荐一本书来编写这样的数据结构（在R，Perl或Python中），那将是非常感激的。到目前为止，我发现的文本没有详细解决这些问题。
很多，非常感谢提前。
-Jon
答案 0 :(得分：1)
我不能说我完全理解整个问题（我，不是统计人员！），但你的lapply行失败的原因是adj.scores通过了x中的列表何时需要矩阵。
由于你有列表列表（列表！），rapply似乎更合适。以下似乎产生了一些合理的东西：
adj.rep.1 <- rapply(t.reps[[1]], adj.scores, how='replace', tot.dat = dat.1)
# comparing the structures
str(t.reps[[1]])
str(adj.rep.1)
希望这有帮助！
答案 1 :(得分：0)
我发布适用于我的解决方案的时间已经很晚了。我确信可以进行改进，所以请随意发表评论！
本练习的目的是了解提案评级的线性转换在多大程度上会对提案的选择产生影响。我们的想法是尝试将“提议质量”与“评估偏见”和“小组偏见”区分开来。
实现此目的的一种方法实质上是面板上所有评级的中心，然后使用所有评级的总体均值和sd对面板中心评级进行均值/ sd转换。此过程位于函数adj.scores。
这是非常重要的，因为提案是由人们评估的，并且可能会有大量的财务激励措施依赖于成功的提案评估（拨款，合同等）。
欢迎任何有关改进或竞争策略的想法。
####################
########## proposal ratings project
########## 17 June 2011
########## original code by: jjb with help from es
##########------ functions to be read in and called when desired
########## applying this function to a single matrix will give detailed output
########## calculating generalizability theory components
########## not a very robust formulation, but a good place to start
########## for future, put panel facet on this design
g.pxr.long = function(x)
{
m.raters <<- colMeans(x)
n.raters <<- length(m.raters)
m.props <<- rowMeans(x)
n.props <<- length(m.props)
m.total <<- mean(x)
n.total <<- nrow(x)*ncol(x)
m.raters.2 <<- m.raters^2
m.props.2 <<- m.props^2
sum.m.raters.2 <<- sum(m.raters.2)
sum.m.props.2 <<- sum(m.props.2)
ss.props <<- n.raters*(sum.m.props.2) - n.total*(m.total^2)
ss.raters <<- n.props*(sum.m.raters.2) - n.total*(m.total^2)
ss.pr <<- sum(x^2) - n.raters*(sum.m.props.2) - n.props*(sum.m.raters.2) + n.total*(m.total^2)
df.props <- n.props - 1
df.raters <- n.raters - 1
df.pr <- df.props*df.raters
ms.props <- ss.props / df.props
ms.raters <- ss.raters / df.raters
ms.pr <- ss.pr / df.pr
var.p <- (ms.props - ms.pr) / n.raters
var.r <- (ms.raters - ms.pr) / n.props
var.pr <- ms.pr
out.1 <- as.data.frame( matrix(c( df.props, df.raters, df.pr,
ss.props, ss.raters, ss.pr,
ms.props, ms.raters, ms.pr,
var.p, var.r, var.pr), ncol = 4))
out.2 <- as.data.frame(matrix(c("p", "r", "pr" ), ncol = 1))
g.out.1 <- as.data.frame(cbind(out.2, out.1))
colnames(g.out.1) <- c(" source", " df ", " ss ", " ms ", " variance")
var.abs <- (var.r / n.raters) + (var.pr / n.raters)
var.rel <- (var.pr / n.raters)
var.xbar <- (var.p / n.props) + (var.r / n.raters) + (var.pr / (n.raters*n.props) )
gen.coef <- var.p / (var.p + var.rel)
dep.coef <- var.p / (var.p + var.abs)
out.3 <- as.data.frame(matrix(c(var.rel, var.abs, var.xbar, gen.coef, dep.coef), ncol=1))
out.3 <- round(out.3, digits = 4)
out.4 <- as.data.frame(matrix(c("relative error variance", "absolute error variance", "xbar error variance", "E(rho^2)", "Phi"), ncol=1))
g.out.2 <- as.data.frame(cbind(out.4,out.3))
colnames(g.out.2) <- c(" estimate ", " value")
outs <- list(g.out.1, g.out.2)
names(outs) <- c("generalizability theory: G-study components", "G-study Indices")
return(outs)
}
##########----- calculating confidence intervals using Chebychev, Cramer, and Normal
##########----- use this to find alpha for desired k
factor.choose = function(k)
{
alpha <- 1 / k^2
return(alpha)
}
conf.intervals = function(dat, alpha)
{
k <- sqrt( 1 / alpha ) # specifying what the k factor is...
x.bar <- mean(dat)
x.sd <- sd(dat)
n.obs <- length(dat)
sem <- x.sd / sqrt(n.obs)
ub.cheb <- x.bar + k*sem
lb.cheb <- x.bar - k*sem
ub.crem <- x.bar + (4/9)*k*sem
lb.crem <- x.bar - (4/9)*k*sem
ub.norm <- x.bar + qnorm(1-alpha/2)*sem
lb.norm <- x.bar - qnorm(1-alpha/2)*sem
out.lb <- matrix(c(lb.cheb, lb.crem, lb.norm), ncol=1)
out.ub <- matrix(c(ub.cheb, ub.crem, ub.norm), ncol=1)
dat <- as.data.frame(dat)
mean.raters <- as.matrix(rowMeans(dat))
dat$z.values <- matrix((mean.raters - x.bar) / x.sd)
select.cheb <- dat[ which(abs(dat$z.values) >= k ) , ]
select.crem <- dat[ which(abs(dat$z.values) >= (4/9)*k ) , ]
select.norm <- dat[ which(abs(dat$z.values) >= qnorm(1-alpha/2)) , ]
count.cheb <- nrow(select.cheb)
count.crem <- nrow(select.crem)
count.norm <- nrow(select.norm)
out.dat <- list()
out.dat <- list(select.cheb, select.crem, select.norm)
names(out.dat) <- c("Selected cases: Chebychev", "Selected cases: Cramer's", "Selected cases: Normal")
out.sum <- matrix(c(x.bar, x.sd, n.obs), ncol=3)
colnames(out.sum) <- c("mean", "sd", "n")
out.crit <- matrix(c(alpha, k, (4/9)*k, qnorm(1-alpha/2)) ,ncol=4 )
colnames(out.crit) <- c("alpha", "k", "(4/9)*k", "z" )
out.count <- matrix(c(count.cheb, count.crem, count.norm) ,ncol=3 )
colnames(out.count) <- c("Chebychev", "Cramer" , "Normal")
rownames(out.count) <- c("count")
outs <- list(out.sum, out.crit, out.count, out.dat)
names(outs) <- c("Summary of data", "Critical values", "Count of Selected Cases", "Selected Cases")
return(outs)
}
rm(list = ls())
set.seed(271828)
means <- matrix(c( rep(50,19), rep(70,1) ), ncol = 1) # matrix of true proposal values
bias.u <- matrix(c(0,2,4), nrow=1) # unidirectional bias
bias.b <- matrix(c(0,5, -5), nrow=1) # bidirectional bias
ones.u <- matrix(rep(1,ncol(bias.u)), nrow = 1) # number of raters is the number of columns (r)
ones.b <- matrix(rep(1,ncol(bias.b)), nrow = 1)
ones.2 <- matrix(rep(1,nrow(means)), ncol = 1) # number of proposals is the number of rows (p)
true.ratings <- means%*%ones.u # gives matrix of true proposal value for each rater (p*r)
uni.bias <- ones.2%*%bias.u
bid.bias <- ones.2%*%bias.b # gives matrix of true rater bias for each proposal (p*r)
pan.bias.pos <- matrix(20,nrow=nrow(true.ratings), ncol=ncol(true.ratings)) # gives a matrix of bias for every member in a panel (p*r)
n.val <- nrow(true.ratings)*ncol(true.ratings)
# true.ratings
# uni.bias
# bid.bias
# pan.bias.pos
library(MASS)
#####
##### generating replicate data...
#####
n.panels <- 10 # put in the number of replicate panels that should be produced
reps <- 2 # put in the number of times each bias condition should be included in a panel
t.reps <- list()
n.bias <- list()
u.bias <- list()
b.bias <- list()
pan.bias <- list()
n.bias.out <- list()
u.bias.out <- list()
b.bias.out <- list()
pan.bias.out <- list()
for (i in 1:n.panels)
{
{
for(j in 1:reps)
n.bias[[j]] <- true.ratings + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
for(j in 1:reps)
u.bias[[j]] <- true.ratings + uni.bias + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
for(j in 1:reps)
b.bias[[j]] <- true.ratings + bid.bias + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
}
pan.bias[[i]] <- true.ratings + pan.bias.pos + matrix(round(rnorm(n.val,4,2), digits=0), nrow = nrow(means))
n.bias.out[[i]] <- n.bias
u.bias.out[[i]] <- u.bias
b.bias.out[[i]] <- b.bias
t.reps[[i]] <- c(n.bias, u.bias, b.bias, pan.bias[i])
}
# t.reps
# rm(list = ls())
##########----- this is the standardization formula to be applied to proposal ratings **WITHIN** a panel
adj.scores <- function(x, tot.dat)
{
t.sd <- sd(array(tot.dat))
t.mn <- mean(array(tot.dat))
ones.t.mn <- diag(1,ncol(x))
p <- nrow(x)
r <- ncol(x)
ones.total <- matrix(1,p,r)
r.sd <- diag(apply(x,2, sd))
r.mn <- diag(apply(x,2, mean))
den.r.sd <- ginv(r.sd)
b.shift <- x%*%den.r.sd
a <- t.mn*ones.t.mn - den.r.sd%*%r.mn
a.shift <- ones.total%*%a
l.x <- b.shift + a.shift
return(l.x)
}
##########----- applying standardization formula and getting
##########----- proposal means for adjusted and unadjusted ratings
adj.rep <- list()
m.adj <- list()
m.reg <- list()
for (i in 1:n.panels)
{
panel.data <- array(unlist(t.reps[[i]]))
adj.rep[[i]] <- lapply(t.reps[[i]], adj.scores, tot.dat = panel.data)
m.adj[[i]] <- lapply(adj.rep[[i]], rowMeans)
m.reg[[i]] <- lapply(t.reps[[i]], rowMeans)
}
##########-----
##########----- This function extracts the replicate proposal means for each set of reviews
##########----- across panels. So, if there are 8 sets of reviewers that are simulated 10 times,
##########----- this extracts the first set of means across the 10 replications and puts them together,
##########----- and then extracts the second set of means across replications and puts them together, etc..
##########----- The result will be 8 matrices made up of 10 columns with rows .
##########-----
##########----- first for unadjusted means
means.reg <- matrix(unlist(m.reg), nrow=nrow(means))
sets <- length(m.reg[[1]])
counter <- n.panels*length(m.reg[[1]])
m.reg.panels.set <- list()
for (j in 1:sets)
{
m.reg.panels.set[[j]] <- means.reg[ , c(seq( j, counter, sets))]
}
##########----- now for adjusted means
means.adj <- matrix(unlist(m.adj), nrow=nrow(means))
sets <- length(m.adj[[1]])
counter <- n.panels*length(m.adj[[1]])
m.adj.panels.set <- list()
for (j in 1:sets)
{
m.adj.panels.set[[j]] <- means.adj[ , c(seq( j, counter, sets))]
}
########## calculating msd as bias^2 and error^2
msd.calc <- function(x)
{
true.means <- means
calc.means <- as.matrix(rowMeans(x))
t.means.mat <- matrix(rep(true.means, n.panels), ncol=ncol(x))
c.means.mat <- matrix(rep(calc.means, n.panels), ncol=ncol(x))
msd <- matrix( rowSums( (x - t.means.mat )^2) / ncol(c.means.mat) )
bias.2 <- (calc.means - true.means)^2
e.var <- matrix( rowSums( (x - c.means.mat )^2) / ncol(c.means.mat ) )
msd <- matrix(c(msd, bias.2, e.var), ncol = 3)
colnames(msd) <- c("msd", "bias^2", "e.var")
return(msd)
}
########## checking work...
# msd.calc <- bias.2 + e.var
# all.equal(msd, msd.calc)
##########----- applying function to each set across panels
msd.reg.panels <- lapply(m.reg.panels.set, msd.calc)
msd.adj.panels <- lapply(m.adj.panels.set, msd.calc)
msd.reg.panels
msd.adj.panels
########## for the unadjusted scores, the msd is very large (as is expected)
########## for the adjusted scores, the msd is in line with the other panels
##########----- trying to evaluate impact of adjusting scores on proposal awards
reg.panel.1 <- matrix(unlist(m.reg[[1]]), nrow=nrow(means))
adj.panel.1 <- matrix(unlist(m.adj[[1]]), nrow=nrow(means))
reg <- matrix(array(reg.panel.1), ncol=1)
adj <- matrix(array(adj.panel.1), ncol=1)
panels.1 <- cbind(reg,adj)
colnames(panels.1) <- c("unadjusted", "adjusted")
cor(panels.1, method="spearman")
plot(panels.1)
######## identify(panels.1)
plot(panels.1, xlim = c(25,95), ylim = c(25,95) )
abline(0,1, col="red")
########## the adjustment greatly reduces variances of ratings
########## compare the projection of the panel means onto the respective margins
##########----- examining the selection of the top 10% of the proposals
pro.names <- matrix(seq(1,nrow(reg),1))
df.reg <- as.data.frame(cbind(pro.names, reg))
colnames(df.reg) <- c("pro", "rating")
df.reg$q.pro <- ecdf(df.reg$rating)(df.reg$rating)
df.adj <- as.data.frame(cbind(pro.names, adj))
colnames(df.adj) <- c("pro", "rating")
df.adj$q.pro <- ecdf(df.adj$rating)(df.adj$rating)
awards.reg <- df.reg[ which(df.reg$q.pro >= .90) , ]
awards.adj <- df.adj[ which(df.adj$q.pro >= .90) , ]
awards.reg[order(-awards.reg$q.pro) , ]
awards.adj[order(-awards.adj$q.pro) , ]
awards.reg[order(-awards.reg$pro) , ]
awards.adj[order(-awards.adj$pro) , ]
##### using unadjusted scores, the top 10% of proposals come mostly from one (biased) panel, and the rest are made up of the
##### highest scoring proposal (from the biased rater) from the remaining panels.
##### using the adjusted scores, the top 10% of proposals are made up of the highest scoring proposal (the biased rater) from the
##### several panels, and a few proposals from other panels. Doesn't seem to be a systematic explanation as to why...
#########----- treating rating data in an ANOVA model
ratings <- do.call("rbind", t.reps[[1]] )
panels <- factor(gl(7,20))
fit <- manova(ratings ~ panels)
summary(fit, test="Wilks")
adj.ratings <- do.call("rbind", adj.rep[[1]] )
adj.fit <- manova(adj.ratings ~ panels)
summary(adj.fit, test="Wilks")
##########----- thinking... extra ideas for later
##########----- calculating Mahalanobis distance to identify biased panels
mah.dist = function(dat)
{md.dat <- as.matrix(mahalanobis(dat, center = colMeans(dat) , cov = cov(dat) ))
md.pv <- as.matrix(pchisq(md.dat, df = ncol(dat), lower.tail = FALSE))
out <- cbind(md.dat, md.pv)
out.2 <- as.data.frame(out)
colnames(out.2) <- c("MD", "pMD")
return(out.2)
}
mah.dist(ratings)
mah.dist(adj.ratings)