Calcular dentro y entre varianzas e intervalos de confianza en R
https://stackoverflow.com/questions/1401894
-
05-07-2019 - |
italiano
english
français
española
中国
日本の
العربية
Deutsch
한국어
Português
RussianPregunta
Necesito calcular las variaciones dentro y entre ejecuciones de algunos datos como parte del desarrollo de un nuevo método de química analítica. También necesito intervalos de confianza de estos datos usando el lenguaje R
¿Supongo que necesito usar anova o algo?
Mis datos son como
> 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
Solución 4
He estado viendo un problema similar. He encontrado referencias a los intervalos de confianza calculados por Burdick y Graybill (Burdick, R. y Graybill, F. 1992, Intervalos de confianza en los componentes de la varianza, CRC Press)
Al usar un código que he estado intentando, obtengo estos valores
> 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]] He estado viendo un problema similar. He encontrado referencias a los intervalos de confianza calculados por Burdick y Graybill (Burdick, R. y Graybill, F. 1992, Intervalos de confianza en los componentes de la varianza, CRC Press)
Al usar un código que he estado intentando, obtengo estos valores
> 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]] He estado viendo un problema similar. He encontrado referencias a los intervalos de confianza calculados por Burdick y Graybill (Burdick, R. y Graybill, F. 1992, Intervalos de confianza en los componentes de la varianza, CRC Press)
Al usar un código que he estado intentando, obtengo estos valores
> 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
Aquí, el intervalo de confianza más bajo entre CV de ejecución es menor que cero, por lo que se informa como NaN.
Me encantaría tener una mejor manera de hacer esto. Si tengo tiempo, podría intentar crear una función para hacer esto.
Paul.
-
Edición: eventualmente escribí una función, aquí está (caveat emptor)
<*>
Otros consejos
Tiene cuatro grupos de tres observaciones:
> 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
Crea algunas etiquetas:
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
Calcular estadísticas dentro de la ejecución:
> withinRunStats = function(x) c(sum = sum(x), mean = mean(x), var = var(x), n = length(x))
> tapply(runs, group, withinRunStats)
Tiene cuatro grupos de tres observaciones:
> 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
Crea algunas etiquetas:
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
Calcular estadísticas dentro de la ejecución:
1`
sum mean var n
29.800000000 9.933333333 0.005833333 3.000000000
Tiene cuatro grupos de tres observaciones:
> 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
Crea algunas etiquetas:
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
Calcular estadísticas dentro de la ejecución:
2`
sum mean var n
28.20 9.40 0.31 3.00
Tiene cuatro grupos de tres observaciones:
> 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
Crea algunas etiquetas:
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
Calcular estadísticas dentro de la ejecución:
3`
sum mean var n
32.10 10.70 0.61 3.00
Tiene cuatro grupos de tres observaciones:
> 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
Crea algunas etiquetas:
> n = rep(3, 4)
> group = rep(1:4, n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4
Calcular estadísticas dentro de la ejecución:
4`
sum mean var n
29.80000000 9.93333333 0.04333333 3.00000000
Puedes hacer algo de ANOVA aquí:
> 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
Error o variación dentro del grupo:
> anova(fit)["Residuals", "Mean Sq"]
[1] 0.2422917
Tratamiento o varianza entre grupos:
> anova(fit)["group", "Mean Sq"]
[1] 0.8586111
Esto debería darle suficiente confianza para hacer intervalos de confianza.
Si desea aplicar una función (como var ) a través de un factor como Run o Rep , puede usar 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
Tomaré una grieta en esto cuando tenga más tiempo, pero mientras tanto, aquí está el dput () para la estructura de datos de Kiar:
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")
... en caso de que te gustaría tomar una foto rápida.
Licenciado bajo: CC-BY-SA con atribución
No afiliado a StackOverflow
