计算R中的方差和置信区间内和之间
https://stackoverflow.com/questions/1401894
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05-07-2019 - |
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Russian题
作为开发新的分析化学方法的一部分，我需要计算一些数据的运行差异和运行差异。我还需要使用R语言从这些数据中获得置信区间
我认为我需要使用anova或其他什么东西？
我的数据就像
> variance
Run Rep Value
1 1 1 9.85
2 1 2 9.95
3 1 3 10.00
4 2 1 9.90
5 2 2 8.80
6 2 3 9.50
7 3 1 11.20
8 3 2 11.10
9 3 3 9.80
10 4 1 9.70
11 4 2 10.10
12 4 3 10.00
解决方案 4
我一直在寻找类似的问题。我发现Burdick和Graybill提到了计算置信区间（Burdick，R。和Graybill，F。1992，Confidence Intervals on variance components，CRC Press）
使用一些代码我一直在尝试获取这些值
> kiaraov = aov(Value~Run+Error(Run),data=kiar)
> summary(kiaraov)
Error: Run
Df Sum Sq Mean Sq
Run 3 2.57583 0.85861
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
Residuals 8 1.93833 0.24229
> confint = 95
> a = (1-(confint/100))/2
> grandmean = as.vector(kiaraov#' avar Function
#'
#' Calculate thewithin, between and total %CV of a dataset by ANOVA, and the
#' associated confidence intervals
#'
#' @param dataf - The data frame to use, in long format
#' @param afactor Character string representing the column in dataf that contains the factor
#' @param aresponse Charactyer string representing the column in dataf that contains the response value
#' @param aconfidence What Confidence limits to use, default = 95%
#' @param digits Significant Digits to report to, default = 3
#' @param debug Boolean, Should debug messages be displayed, default=FALSE
#' @returnType dataframe containing the Mean, Within, Between and Total %CV and LCB and UCB for each
#' @return
#' @author Paul Hurley
#' @export
#' @examples
#' #Using the BGBottles data from Burdick and Graybill Page 62
#' assayvar(dataf=BGBottles, afactor="Machine", aresponse="weight")
avar<-function(dataf, afactor, aresponse, aconfidence=95, digits=3, debug=FALSE){
dataf<-subset(dataf,!is.na(with(dataf,get(aresponse))))
nmissing<-function(x) sum(!is.na(x))
n<-nrow(subset(dataf,is.numeric(with(dataf,get(aresponse)))))
datadesc<-ddply(dataf, afactor, colwise(nmissing,aresponse))
I<-nrow(datadesc)
if(debug){print(datadesc)}
if(min(datadesc[,2])==max(datadesc[,2])){
balance<-TRUE
J<-min(datadesc[,2])
if(debug){message(paste("Dataset is balanced, J=",J,"I is ",I,sep=""))}
} else {
balance<-FALSE
Jh<-I/(sum(1/datadesc[,2], na.rm = TRUE))
J<-Jh
m<-min(datadesc[,2])
M<-max(datadesc[,2])
if(debug){message(paste("Dataset is unbalanced, like me, I is ",I,sep=""))}
if(debug){message(paste("Jh is ",Jh, ", m is ",m, ", M is ",M, sep=""))}
}
if(debug){message(paste("Call afactor=",afactor,", aresponse=",aresponse,sep=""))}
formulatext<-paste(as.character(aresponse)," ~ 1 + Error(",as.character(afactor),")",sep="")
if(debug){message(paste("formula text is ",formulatext,sep=""))}
aovformula<-formula(formulatext)
if(debug){message(paste("Formula is ",as.character(aovformula),sep=""))}
assayaov<-aov(formula=aovformula,data=dataf)
if(debug){
print(assayaov)
print(summary(assayaov))
}
a<-1-((1-(aconfidence/100))/2)
if(debug){message(paste("confidence is ",aconfidence,", alpha is ",a,sep=""))}
grandmean<-as.vector(assayaov<*>quot;(Intercept)"[[1]][1]) # Grand Mean (I think)
if(debug){message(paste("n is",n,sep=""))}
#This line commented out, seems to choke with an aov object built from an external formula
#grandmean<-as.vector(model.tables(assayaov,type="means")[[1]]我一直在寻找类似的问题。我发现Burdick和Graybill提到了计算置信区间（Burdick，R。和Graybill，F。1992，Confidence Intervals on variance components，CRC Press）
使用一些代码我一直在尝试获取这些值
> kiaraov = aov(Value~Run+Error(Run),data=kiar)
> summary(kiaraov)
Error: Run
Df Sum Sq Mean Sq
Run 3 2.57583 0.85861
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
Residuals 8 1.93833 0.24229
> confint = 95
> a = (1-(confint/100))/2
> grandmean = as.vector(kiaraovGrand mean`) # Grand Mean (I think)
within<-summary(assayaov)[[2]][[1]]<*>quot;Mean Sq" # d2e, S2^2 Mean Square Value for Within Machine = 0.1819
dfRun<-summary(assayaov)[[1]][[1]]<*>quot;Df" # DF for within = 3
dfWithin<-summary(assayaov)[[2]][[1]]<*>quot;Df" # DF for within = 8
Run<-summary(assayaov)[[1]][[1]]<*>quot;Mean Sq" # S1^2Mean Square for Machine
if(debug){message(paste("mean square for Run ?",Run,sep=""))}
#Was between<-(Run-within)/((dfWithin/(dfRun+1))+1) but my comment suggests this should be just J, so I'll use J !
between<-(Run-within)/J # d2a (S1^2-S2^2)/J
if(debug){message(paste("S1^2 mean square machine is ",Run,", S2^2 mean square within is ",within))}
total<-between+within
between # Between Run Variance
within # Within Run Variance
total # Total Variance
if(debug){message(paste("between is ",between,", within is ",within,", Total is ",total,sep=""))}
betweenCV<-sqrt(between)/grandmean * 100 # Between Run CV%
withinCV<-sqrt(within)/grandmean * 100 # Within Run CV%
totalCV<-sqrt(total)/grandmean * 100 # Total CV%
n1<-dfRun
n2<-dfWithin
if(debug){message(paste("n1 is ",n1,", n2 is ",n2,sep=""))}
#within confidence intervals
if(balance){
withinLCB<-within/qf(a,n2,Inf) # Within LCB
withinUCB<-within/qf(1-a,n2,Inf) # Within UCB
} else {
withinLCB<-within/qf(a,n2,Inf) # Within LCB
withinUCB<-within/qf(1-a,n2,Inf) # Within UCB
}
#Mean Confidence Intervals
if(debug){message(paste(grandmean,"+/-(sqrt(",Run,"/",n,")*qt(",a,",df=",I-1,"))",sep=""))}
meanLCB<-grandmean+(sqrt(Run/n)*qt(1-a,df=I-1)) # wrong
meanUCB<-grandmean-(sqrt(Run/n)*qt(1-a,df=I-1)) # wrong
if(debug){message(paste("Grandmean is ",grandmean,", meanLCB = ",meanLCB,", meanUCB = ",meanUCB,aresponse,sep=""))}
if(debug){print(summary(assayaov))}
#Between Confidence Intervals
G1<-1-(1/qf(a,n1,Inf))
G2<-1-(1/qf(a,n2,Inf))
H1<-(1/qf(1-a,n1,Inf))-1
H2<-(1/qf(1-a,n2,Inf))-1
G12<-((qf(a,n1,n2)-1)^2-(G1^2*qf(a,n1,n2)^2)-(H2^2))/qf(a,n1,n2)
H12<-((1-qf(1-a,n1,n2))^2-H1^2*qf(1-a,n1,n2)^2-G2^2)/qf(1-a,n1,n2)
if(debug){message(paste("G1 is ",G1,", G2 is ",G2,sep=""))
message(paste("H1 is ",H1,", H2 is ",H2,sep=""))
message(paste("G12 is ",G12,", H12 is ",H12,sep=""))
}
if(balance){
Vu<-H1^2*Run^2+G2^2*within^2+H12*Run*within
Vl<-G1^2*Run^2+H2^2*within^2+G12*within*Run
betweenLCB<-(Run-within-sqrt(Vl))/J # Betwen LCB
betweenUCB<-(Run-within+sqrt(Vu))/J # Between UCB
} else {
#Burdick and Graybill seem to suggest calculating anova of mean values to find n1S12u/Jh
meandataf<-ddply(.data=dataf,.variable=afactor, .fun=function(df){mean(with(df, get(aresponse)), na.rm=TRUE)})
meandataaov<-aov(formula(paste("V1~",afactor,sep="")), data=meandataf)
sumsquare<-summary(meandataaov)[[1]]我一直在寻找类似的问题。我发现Burdick和Graybill提到了计算置信区间（Burdick，R。和Graybill，F。1992，Confidence Intervals on variance components，CRC Press）
使用一些代码我一直在尝试获取这些值
> kiaraov = aov(Value~Run+Error(Run),data=kiar)
> summary(kiaraov)
Error: Run
Df Sum Sq Mean Sq
Run 3 2.57583 0.85861
Error: Within
Df Sum Sq Mean Sq F value Pr(>F)
Residuals 8 1.93833 0.24229
> confint = 95
> a = (1-(confint/100))/2
> grandmean = as.vector(kiaraovSum Sq`
#so maybe S12u is just that bit ?
Runu<-(sumsquare*Jh)/n1
if(debug){message(paste("n1S12u/Jh is ",sumsquare,", so S12u is ",Runu,sep=""))}
Vu<-H1^2*Runu^2+G2^2*within^2+H12*Runu*within
Vl<-G1^2*Runu^2+H2^2*within^2+G12*within*Runu
betweenLCB<-(Runu-within-sqrt(Vl))/Jh # Betwen LCB
betweenUCB<-(Runu-within+sqrt(Vu))/Jh # Between UCB
if(debug){message(paste("betweenLCB is ",betweenLCB,", between UCB is ",betweenUCB,sep=""))}
}
#Total Confidence Intervals
if(balance){
y<-(Run+(J-1)*within)/J
if(debug){message(paste("y is ",y,sep=""))}
totalLCB<-y-(sqrt(G1^2*Run^2+G2^2*(J-1)^2*within^2)/J) # Total LCB
totalUCB<-y+(sqrt(H1^2*Run^2+H2^2*(J-1)^2*within^2)/J) # Total UCB
} else {
y<-(Runu+(Jh-1)*within)/Jh
if(debug){message(paste("y is ",y,sep=""))}
totalLCB<-y-(sqrt(G1^2*Runu^2+G2^2*(Jh-1)^2*within^2)/Jh) # Total LCB
totalUCB<-y+(sqrt(H1^2*Runu^2+H2^2*(Jh-1)^2*within^2)/Jh) # Total UCB
}
if(debug){message(paste("totalLCB is ",totalLCB,", total UCB is ",totalUCB,sep=""))}
# result<-data.frame(Name=c("within", "between", "total"),CV=c(withinCV,betweenCV,totalCV),
# LCB=c(sqrt(withinLCB)/grandmean*100,sqrt(betweenLCB)/grandmean*100,sqrt(totalLCB)/grandmean*100),
# UCB=c(sqrt(withinUCB)/grandmean*100,sqrt(betweenUCB)/grandmean*100,sqrt(totalUCB)/grandmean*100))
result<-data.frame(Mean=grandmean,MeanLCB=meanLCB, MeanUCB=meanUCB, Within=withinCV,WithinLCB=sqrt(withinLCB)/grandmean*100, WithinUCB=sqrt(withinUCB)/grandmean*100,
Between=betweenCV, BetweenLCB=sqrt(betweenLCB)/grandmean*100, BetweenUCB=sqrt(betweenUCB)/grandmean*100,
Total=totalCV, TotalLCB=sqrt(totalLCB)/grandmean*100, TotalUCB=sqrt(totalUCB)/grandmean*100)
if(!digits=="NA"){
result$Mean<-signif(result$Mean,digits=digits)
result$MeanLCB<-signif(result$MeanLCB,digits=digits)
result$MeanUCB<-signif(result$MeanUCB,digits=digits)
result$Within<-signif(result$Within,digits=digits)
result$WithinLCB<-signif(result$WithinLCB,digits=digits)
result$WithinUCB<-signif(result$WithinUCB,digits=digits)
result$Between<-signif(result$Between,digits=digits)
result$BetweenLCB<-signif(result$BetweenLCB,digits=digits)
result$BetweenUCB<-signif(result$BetweenUCB,digits=digits)
result$Total<-signif(result$Total,digits=digits)
result$TotalLCB<-signif(result$TotalLCB,digits=digits)
result$TotalUCB<-signif(result$TotalUCB,digits=digits)
}
return(result)
}
assayvar<-function(adata, aresponse, afactor, anominal, aconfidence=95, digits=3, debug=FALSE){
result<-ddply(adata,anominal,function(df){
resul<-avar(dataf=df,afactor=afactor,aresponse=aresponse,aconfidence=aconfidence, digits=digits, debug=debug)
resul$n<-nrow(subset(df, !is.na(with(df, get(aresponse)))))
return(resul)
})
return(result)
}
quot;(Intercept)"[[1]][1]) # Grand Mean (I think)
> within = summary(kiaraov)<*>quot;Error: Within"[[1]]<*>quot;Mean Sq" # S2^2Mean Square Value for Within Run
> dfRun = summary(kiaraov)<*>quot;Error: Run"[[1]]<*>quot;Df"
> dfWithin = summary(kiaraov)<*>quot;Error: Within"[[1]]<*>quot;Df"
> Run = summary(kiaraov)<*>quot;Error: Run"[[1]]<*>quot;Mean Sq" # S1^2Mean Square for between Run
> between = (Run-within)/((dfWithin/(dfRun+1))+1) # (S1^2-S2^2)/J
> total = between+within
> between # Between Run Variance
[1] 0.2054398
> within # Within Run Variance
[1] 0.2422917
> total # Total Variance
[1] 0.4477315
> betweenCV = sqrt(between)/grandmean * 100 # Between Run CV%
> withinCV = sqrt(within)/grandmean * 100 # Within Run CV%
> totalCV = sqrt(total)/grandmean * 100 # Total CV%
> #within confidence intervals
> withinLCB = within/qf(1-a,8,Inf) # Within LCB
> withinUCB = within/qf(a,8,Inf) # Within UCB
> #Between Confidence Intervals
> n1 = dfRun
> n2 = dfWithin
> G1 = 1-(1/qf(1-a,n1,Inf)) # According to Burdick and Graybill this should be a
> G2 = 1-(1/qf(1-a,n2,Inf))
> H1 = (1/qf(a,n1,Inf))-1 # and this should be 1-a, but my results don't agree
> H2 = (1/qf(a,n2,Inf))-1
> G12 = ((qf(1-a,n1,n2)-1)^2-(G1^2*qf(1-a,n1,n2)^2)-(H2^2))/qf(1-a,n1,n2) # again, should be a, not 1-a
> H12 = ((1-qf(a,n1,n2))^2-H1^2*qf(a,n1,n2)^2-G2^2)/qf(a,n1,n2) # again, should be 1-a, not a
> Vu = H1^2*Run^2+G2^2*within^2+H12*Run*within
> Vl = G1^2*Run^2+H2^2*within^2+G12*within*Run
> betweenLCB = (Run-within-sqrt(Vl))/J # Betwen LCB
> betweenUCB = (Run-within+sqrt(Vu))/J # Between UCB
> #Total Confidence Intervals
> y = (Run+(J-1)*within)/J
> totalLCB = y-(sqrt(G1^2*Run^2+G2^2*(J-1)^2*within^2)/J) # Total LCB
> totalUCB = y+(sqrt(H1^2*Run^2+H2^2*(J-1)^2*within^2)/J) # Total UCB
> result = data.frame(Name=c("within", "between", "total"),CV=c(withinCV,betweenCV,totalCV),LCB=c(sqrt(withinLCB)/grandmean*100,sqrt(betweenLCB)/grandmean*100,sqrt(totalLCB)/grandmean*100),UCB=c(sqrt(withinUCB)/grandmean*100,sqrt(betweenUCB)/grandmean*100,sqrt(totalUCB)/grandmean*100))
> result
Name CV LCB UCB
1 within 4.926418 3.327584 9.43789
2 between 4.536327 NaN 19.73568
3 total 6.696855 4.846030 20.42647
此处运行CV之间的置信区间小于零，因此报告为NaN。
我希望有更好的方法来做到这一点。如果我有时间，我可能会尝试创建一个函数来执行此操作。
保罗。
-
编辑：我最终写了一个函数，这里是（caveat emptor）
<*>
其他提示
你有四组三个观察结果：
> run1 = c(9.85, 9.95, 10.00)
> run2 = c(9.90, 8.80, 9.50)
> run3 = c(11.20, 11.10, 9.80)
> run4 = c(9.70, 10.10, 10.00)
> runs = c(run1, run2, run3, run4)
> runs
[1] 9.85 9.95 10.00 9.90 8.80 9.50 11.20 11.10 9.80 9.70 10.10 10.00
制作一些标签：
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
计算运行中的统计数据：
> withinRunStats = function(x) c(sum = sum(x), mean = mean(x), var = var(x), n = length(x))
> tapply(runs, group, withinRunStats)
你有四组三个观察结果：
> run1 = c(9.85, 9.95, 10.00)
> run2 = c(9.90, 8.80, 9.50)
> run3 = c(11.20, 11.10, 9.80)
> run4 = c(9.70, 10.10, 10.00)
> runs = c(run1, run2, run3, run4)
> runs
[1] 9.85 9.95 10.00 9.90 8.80 9.50 11.20 11.10 9.80 9.70 10.10 10.00
制作一些标签：
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
计算运行中的统计数据：
1`
sum mean var n
29.800000000 9.933333333 0.005833333 3.000000000
你有四组三个观察结果：
> run1 = c(9.85, 9.95, 10.00)
> run2 = c(9.90, 8.80, 9.50)
> run3 = c(11.20, 11.10, 9.80)
> run4 = c(9.70, 10.10, 10.00)
> runs = c(run1, run2, run3, run4)
> runs
[1] 9.85 9.95 10.00 9.90 8.80 9.50 11.20 11.10 9.80 9.70 10.10 10.00
制作一些标签：
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
计算运行中的统计数据：
2`
sum mean var n
28.20 9.40 0.31 3.00
你有四组三个观察结果：
> run1 = c(9.85, 9.95, 10.00)
> run2 = c(9.90, 8.80, 9.50)
> run3 = c(11.20, 11.10, 9.80)
> run4 = c(9.70, 10.10, 10.00)
> runs = c(run1, run2, run3, run4)
> runs
[1] 9.85 9.95 10.00 9.90 8.80 9.50 11.20 11.10 9.80 9.70 10.10 10.00
制作一些标签：
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
计算运行中的统计数据：
3`
sum mean var n
32.10 10.70 0.61 3.00
你有四组三个观察结果：
> run1 = c(9.85, 9.95, 10.00)
> run2 = c(9.90, 8.80, 9.50)
> run3 = c(11.20, 11.10, 9.80)
> run4 = c(9.70, 10.10, 10.00)
> runs = c(run1, run2, run3, run4)
> runs
[1] 9.85 9.95 10.00 9.90 8.80 9.50 11.20 11.10 9.80 9.70 10.10 10.00
制作一些标签：
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
计算运行中的统计数据：
4`
sum mean var n
29.80000000 9.93333333 0.04333333 3.00000000
你可以在这里做一些方差分析：
> data = data.frame(y = runs, group = factor(group))
> data
y group
1 9.85 1
2 9.95 1
3 10.00 1
4 9.90 2
5 8.80 2
6 9.50 2
7 11.20 3
8 11.10 3
9 9.80 3
10 9.70 4
11 10.10 4
12 10.00 4
> fit = lm(runs ~ group, data)
> fit
Call:
lm(formula = runs ~ group, data = data)
Coefficients:
(Intercept) group2 group3 group4
9.933e+00 -5.333e-01 7.667e-01 -2.448e-15
> anova(fit)
Analysis of Variance Table
Response: runs
Df Sum Sq Mean Sq F value Pr(>F)
group 3 2.57583 0.85861 3.5437 0.06769 .
Residuals 8 1.93833 0.24229
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> degreesOfFreedom = anova(fit)[, "Df"]
> names(degreesOfFreedom) = c("treatment", "error")
> degreesOfFreedom
treatment error
3 8
错误或组内差异：
> anova(fit)["Residuals", "Mean Sq"]
[1] 0.2422917
治疗或组间差异：
> anova(fit)["group", "Mean Sq"]
[1] 0.8586111
这应该让你有足够的信心去做置信区间。
如果要在诸如 Run 或 Rep 之类的因子中应用函数（例如 var ），则可以使用 tapply ：
> with(variance, tapply(Value, Run, var))
1 2 3 4
0.005833333 0.310000000 0.610000000 0.043333333
> with(variance, tapply(Value, Rep, var))
1 2 3
0.48562500 0.88729167 0.05583333
当我有更多时间时，我会对此采取行动，但同时，这是Kiar数据结构的 dput（）：
structure(list(Run = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4), Rep = c(1,
2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3), Value = c(9.85, 9.95, 10, 9.9,
8.8, 9.5, 11.2, 11.1, 9.8, 9.7, 10.1, 10)), .Names = c("Run",
"Rep", "Value"), row.names = c(NA, -12L), class = "data.frame")
...如果您想快速拍摄它。
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