题
我想在存储在两个单独的列表中的对象之间执行ANOVA,但是然后在这个
的时间一次执行它们> anova(output.02[[1]], output.03[[1]])
Model df AIC BIC logLik Test L.Ratio p-value
output.02[[1]] 1 9 11221.77 11279.72 -5601.884
output.03[[1]] 2 13 11222.90 11306.60 -5598.450 1 vs 2 6.868822 0.143
> anova(output.02[[2]], output.03[[2]])
Model df AIC BIC logLik Test L.Ratio p-value
output.02[[2]] 1 9 10976.36 11034.31 -5479.182
output.03[[2]] 2 13 10974.90 11058.60 -5474.449 1 vs 2 9.465378 0.0505
.
我想使用循环来在每个列表中的对象之间执行ANOVA。我尝试使用Mapply函数,但输出没有产生我期待的结果。
> mapply(anova, output.02, output.03)
zimmrec zdelrec zdigiback zspotword zsdmt zglobcog zmmse
call factor,2 factor,2 factor,2 factor,2 factor,2 factor,2 factor,2
Model Integer,2 Integer,2 Integer,2 Integer,2 Integer,2 Integer,2 Integer,2
df Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2
AIC Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2
BIC Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2
logLik Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2
Test factor,2 factor,2 factor,2 factor,2 factor,2 factor,2 factor,2
L.Ratio Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2
p-value Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2 Numeric,2
.
对我如何执行此操作的任何建议?
感谢
编辑:可重复示例
attach(Orthodont)
set.seed(1234)
#example response variables
Orthodont$v1 <- rnorm(n=108, mean=20, sd=1)
Orthodont$v2 <- rnorm(n=108, mean=31, sd=2.8)
Orthodont$v3 <- rnorm(n=108, mean=15, sd=1.5)
head(Orthodont)
#function to loop the response variables through a lme function
#produces first batch of models
myfunc <- function(X){
lapply(X, function(.col){
y <- .col
out <- with(Orthodont, lme(y ~ age, random = ~ age | Subject, method = "ML", na.action = na.exclude, control = lmeControl(opt = "optim")))
out
})
}
output.02 <- myfunc(Orthodont[5:7]) #first list of models
myfunc2 <- function(X){
lapply(X, function(.col){
y <- .col
out <- with(Orthodont, lme(y ~ age + Sex, random = ~ age | Subject, method = "ML", na.action = na.exclude, control = lmeControl(opt = "optim")))
out
})
}
output.03 <- myfunc2(Orthodont[5:7])# second list of models
#anova for each pair of models
anova(output.02[[1]], output.03[[1]])
anova(output.02[[2]], output.03[[2]])
anova(output.02[[3]], output.03[[3]])
#mapply function
mapply(anova, output.02, output.03)
. 解决方案
使用SIMPLIFY
参数:
mapply(anova, output.02, output.03, SIMPLIFY=FALSE)
#$v1
# Model df AIC BIC logLik Test L.Ratio p-value
#dots[[1L]][[1L]] 1 6 324.4204 340.5132 -156.2102
#dots[[2L]][[1L]] 2 7 326.2229 344.9978 -156.1115 1 vs 2 0.1974693 0.6568
#
#$v2
# Model df AIC BIC logLik Test L.Ratio p-value
#dots[[1L]][[2L]] 1 6 524.0956 540.1884 -256.0478
#dots[[2L]][[2L]] 2 7 525.7577 544.5326 -255.8788 1 vs 2 0.337934 0.561
#
#$v3
# Model df AIC BIC logLik Test L.Ratio p-value
#dots[[1L]][[3L]] 1 6 387.4002 403.4930 -187.7001
#dots[[2L]][[3L]] 2 7 389.1333 407.9082 -187.5667 1 vs 2 0.266947 0.6054
. 其他提示
您可以使用sapply
MyRes <- sapply(1:length(output.02), function(x) {
anova(output.02[[x]], output.03[[x]])})
. 不隶属于 StackOverflow