这是可行的:
testmodel=glm(breaks~wool,data=warpbreaks)
emmeans::emmeans(testmodel,"wool")这是可行的:
warpbreaks %>%
group_by(tension) %>%
do(models=glm(breaks~wool,data=.)) %>%
ungroup() %>%
mutate(means=map(models,~emmeans::emmeans(.x,"wool")))这不是:
warpbreaks %>%
group_by(tension) %>% nest() %>%
mutate(models=map(data,~glm(breaks~wool,data=.x))) %>%
mutate(means=map(models,~emmeans::emmeans(.x,"wool")))
Error in is.data.frame(data) : object '.x' not found
Error in mutate_impl(.data, dots) :
Evaluation error: Perhaps a 'data' or 'params' argument is needed.知道这是什么原因吗?
发布于 2018-02-03 08:18:42
我想通了。问题在于emmeans尝试从lm/glm对象恢复数据的方式:它尝试运行存储在对象中的调用,如果emmeans()在与原始glm()调用不同的环境中调用,则会失败:
emmeans:::recover_data.lm下面是一个简单的例子:
wb=warpbreaks
model=glm(breaks~wool,data=wb)
emmeans(model,"wool")
rm(wb)
emmeans(model,"wool")下面是使用map()实现emmeans()的方法:
warpbreaks %>%
group_by(tension) %>% nest() %>%
mutate(models=map(data,~glm(breaks~wool,data=.x))) %>%
mutate(means=map(models,~emmeans::emmeans(.x,"wool",data=.x$data)))奇怪的是,recover_data()并没有自动使用lm/glm对象的数据属性,而是假设调用将在当前环境中运行……
发布于 2018-02-02 15:06:22
我们可以分两步完成
df1 <- warpbreaks %>%
group_by(tension) %>%
nest() %>%
mutate(models = map(data,~glm(breaks~wool,data=.x)))
warpbreaks %>%
split(.$tension) %>%
map( ~glm(breaks ~ wool, data = .x) %>%
emmeans(., "wool")) %>%
mutate(df1, Means = .)
# A tibble: 3 x 4
# tension data models Means
# <fctr> <list> <list> <list>
#1 L <tibble [18 x 2]> <S3: glm> <S4: emmGrid>
#2 M <tibble [18 x 2]> <S3: glm> <S4: emmGrid>
#3 H <tibble [18 x 2]> <S3: glm> <S4: emmGrid>https://stackoverflow.com/questions/48576160
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