英文:
Can't get arcsin back-transformation with emmeans to work
问题
I am doing a GAM with my response variable as a percentage (0-100). I have used an arcsin transformation to improve model fit (asin(sqrt(myvariable/100))). I now want to evaluate contrasts between levels of my explanatory factor variable on the original scale. I have been trying to use emmeans and followed the steps in the Transformations and link functions vignette to set up my model with the transformation in a format emmeans can read. However, when I run the emmeans function I get the following error: Error in link$mu.eta(object@bhat[estble]) : attempt to apply non-function.
I set up the transformation object like this:
tran <- make.tran("asin.sqrt", 100)
I am confident that bit works, because when I tried it on a linear model with emmeans, it worked:
warp.t <- with(tran, lm(linkfun(breaks)~wool*tension, warpbreaks))
emmeans(warp.t, ~wool|tension, type="response")
tension = L:
wool response SE df lower.CL upper.CL
A 44.2 4.00 48 36.3 52.3
B 27.7 3.61 48 20.8 35.3
tension = M:
wool response SE df lower.CL upper.CL
A 23.5 3.41 48 17.0 30.7
B 28.4 3.63 48 21.4 35.9
tension = H:
wool response SE df lower.CL upper.CL
A 23.9 3.43 48 17.4 31.1
B 18.6 3.13 48 12.7 25.3
Confidence level used: 0.95
Intervals are back-transformed from the asin(sqrt(mu/100)) scale
However, if I try it in a gam (both straight into emmeans() or using regrid()), it doesn't work:
dat <- data.frame("x" = rep(1:3, times=12),
"y" = rep(4:6, times=12),
"z" = runif(36, 0, 100),
"m" = rep(1:12, times=3))
gam.t <- with(tran, gam(linkfun(z) ~ x * y + s(m), data=dat))
emmeans(gam.t, ~x|y, type="response")
Error in linkinv(result[[cnm[1]]]) : could not find function "linkinv"
#or
regrid(emmeans(gam.t, ~x|y), transform="response")
Error in flink$mu.eta(object@bhat[estble]) :
attempt to apply non-function
It's like it's looking for the inverse link function in the gam, but it's not where emmeans expects it to be. Can I assign it to the gam somehow? Does it not work with gams? Am I doing something wrong?
英文:
I am doing a GAM with my response variable as a percentage (0-100). I have used an arcsin transformation to improve model fit (asin(sqrt(myvariable/100))). I now want to evaluate contrasts between levels of my explanatory factor variable on the original scale. I have been trying to use emmeans and followed the steps in the Transformations and link functions vignette to set up my model with the transformation in a format emmeans can read. However, when I run the emmeans function I get the following error: Error in link$mu.eta(object@bhat[estble]) : attempt to apply non-function.
I set up the transformation object like this:
tran <- make.tran("asin.sqrt", 100)
I am confident that bit works, because when I tried it on a linear model with emmeans, it worked:
warp.t <- with(tran, lm(linkfun(breaks)~wool*tension, warpbreaks))
emmeans(warp.t, ~wool|tension, type="response")
tension = L:
wool response SE df lower.CL upper.CL
A 44.2 4.00 48 36.3 52.3
B 27.7 3.61 48 20.8 35.3
tension = M:
wool response SE df lower.CL upper.CL
A 23.5 3.41 48 17.0 30.7
B 28.4 3.63 48 21.4 35.9
tension = H:
wool response SE df lower.CL upper.CL
A 23.9 3.43 48 17.4 31.1
B 18.6 3.13 48 12.7 25.3
Confidence level used: 0.95
Intervals are back-transformed from the asin(sqrt(mu/100)) scale
However, if I try it in a gam (both straight into emmeans() or using regrid()), it doesn't work:
dat <- data.frame("x" = rep(1:3, times=12),
"y" = rep(4:6, times=12),
"z" = runif(36, 0, 100),
"m" = rep(1:12, times=3))
gam.t <- with(tran, gam(linkfun(z) ~ x * y + s(m), data=dat))
emmeans(gam.t, ~x|y, type="response")
Error in linkinv(result[[cnm[1]]]) : could not find function "linkinv"
#or
regrid(emmeans(gam.t, ~x|y), transform="response")
Error in flink$mu.eta(object@bhat[estble]) :
attempt to apply non-function
It's like it's looking for the inverse link function in the gam, but it's not where emmeans expects it to be. Can I assign it to the gam somehow? Does it not work with gams? Am I doing something wrong?
答案1
得分: 2
按照文中在标题“事后指定变换”下的说明,我认为以下代码会给你想要的结果:
gam.t2 <- gam(asin(sqrt(z/100)) ~ x * y + s(m), data = dat)
refgrid_gam.t2 <- update(ref_grid(gam.t2), tran = tran)
emmeans(refgrid_gam.t2, ~ x | y, type = "response")
在`gam()`调用中,响应变量被转换,然后我们在参考网格上调用`update()`来指定使用了什么变换。我得到的输出再次以百分比为单位:
y = 5:
x response SE df lower.CL upper.CL
2 52 10.3 32 31.5 72.2
置信水平使用: 0.95
区间从asin(sqrt(mu/100))的比例重新转换
英文:
Following the instructions in the vignette under the heading "Specifying a transformation after the fact" I think gives you the result you are looking for:
gam.t2 <- gam(asin(sqrt(z/100)) ~ x * y + s(m), data = dat)
refgrid_gam.t2 <- update(ref_grid(gam.t2), tran = tran)
emmeans(refgrid_gam.t2, ~ x | y, type = "response")
The response variable is transformed in the call to gam()
, then we call update()
on the reference grid to specify what transformation was used. The output I get is on the percent scale again:
y = 5:
x response SE df lower.CL upper.CL
2 52 10.3 32 31.5 72.2
Confidence level used: 0.95
Intervals are back-transformed from the asin(sqrt(mu/100)) scale
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