r语言 - purrr::map返回多行而不是一行

Question

我试图通过函数式编程运行线性回归。但是，我不能成功地得到输出。purrr:::map每个嵌套列表返回多行，而不是一行。

#perform linear regression for each cylinder
mtcars_result <- mtcars%>%
nest(-cyl)%>%
mutate(model=map(data,~ lm(as.formula("mpg~disp"),data=.)),
n=map(data,~nrow(.)))
#predict values
mtcars_result$predict <- 1:3
#helper function to obtain predict values
get_prediction <- function(m,varname,predict){
predictdata <- data.frame(predict)
names(predictdata) <- c(varname)
predict(m,newdata=predictdata,interval="confidence",level=0.95)
}
#prediction, notice it returns three rows per nested list
mtcars_result2 <- mtcars_result%>%mutate(predicted_values=map(model,get_prediction,"disp",predict))
mtcars_result2$predicted_values
[[1]]
fit      lwr      upr
1 19.08559 11.63407 26.53712
2 19.08920 11.67680 26.50160
3 19.09280 11.71952 26.46609
[[2]]
fit      lwr      upr
1 40.73681 32.68945 48.78418
2 40.60167 32.62715 48.57619
3 40.46653 32.56482 48.36824
[[3]]
fit      lwr      upr
1 22.01316 14.74447 29.28186
2 21.99353 14.74479 29.24227
3 21.97390 14.74511 29.20268

我的尝试:

我注意到主要问题可能是由于get_prediction()中的predict参数。当我运行这个版本的get_prediction()

get_prediction <- function(m,varname,predict){
predict_global<<-predict
predictdata <- data.frame(predict)
names(predictdata) <- c(varname)
predict(m,newdata=predictdata,interval="confidence",level=0.95)
}

> predict_global
[1] 1 2 3

因此，我的直觉是使用rowwise()，但它最终出现错误:

mtcars_result2 <- mtcars_result%>%rowwise()%>%mutate(predicted_values=map(model,get_prediction,"disp",predict))
Error in UseMethod("predict") : 
no applicable method for 'predict' applied to an object of class "c('double', 'numeric')" 

谁能给我点灯?也许我们可以用purrr::pmap代替purrr::map?

Answer 1

一种选择是使用imap并使用索引.y作为predict列的子集。

mtcars_result %>%
mutate(predicted_values = imap(model, ~ get_prediction(.x, "disp", predict[.y])))

或者我们可以使用rowwise()

mtcars_result %>% 
rowwise() %>% 
mutate(predicted_values = list(get_prediction(model, "disp", predict)))
#> # A tibble: 3 × 6
#> # Rowwise: 
#>     cyl data               model  n         predict predicted_values
#>   <dbl> <list>             <list> <list>      <int> <list>          
#> 1     6 <tibble [7 × 10]>  <lm>   <int [1]>       1 <dbl [1 × 3]>   
#> 2     4 <tibble [11 × 10]> <lm>   <int [1]>       2 <dbl [1 × 3]>   
#> 3     8 <tibble [14 × 10]> <lm>   <int [1]>       3 <dbl [1 × 3]>

^{创建于2022-12-01使用reprex v2.0.2}

Answer 2

与接受的答案非常相似，但我们也可以使用purrr::map2并在get_prediction

中切换参数的顺序

get_prediction <- function(m,predict,varname){
predictdata <- data.frame(predict)
names(predictdata) <- c(varname)
predict(m,newdata=predictdata,interval="confidence",level=0.95)
}
#prediction, notice there are duplicates
mtcars_result2 <- mtcars_result%>%mutate(predicted_values=map2(model,predict,get_prediction,"disp"))
mtcars_result2$predicted_values
[[1]]
fit      lwr      upr
1 19.08559 11.63407 26.53712
[[2]]
fit      lwr      upr
1 40.60167 32.62715 48.57619
[[3]]
fit      lwr      upr
1 21.9739 14.74511 29.20268

Answer 3

另一种可能性是通过cyl分割并使用map2:

library(tidyverse)
options(pillar.sigfig = 7)
mtcars %>%
split(f = .$cyl) %>% 
map2_dfr(c(2, 1, 3), 
~lm(mpg ~ disp, data = .) %>% 
get_prediction("disp", .y) %>% 
as_tibble(),
.id = "cyl")

返回预测值2, 1, 3

# A tibble: 3 × 4
cyl        fit      lwr      upr
<chr>    <dbl>    <dbl>    <dbl>
1 4     40.60167 32.62715 48.57619
2 6     19.08559 11.63407 26.53712
3 8     21.97390 14.74511 29.20268