dplyr::do() requires named function?

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梦毁少年i
梦毁少年i 2020-12-03 08:02

The following works fine:

library(dplyr) 
m <- function(df) {
  mod <- lm(Sepal.Length ~ Sepal.Width, data = df)
  pred <- predict(mod,newdata = df[         


        
2条回答
  •  慢半拍i
    慢半拍i (楼主)
    2020-12-03 08:23

    You don't need an anonymous function:

    library(dplyr)
    iris %>%
      group_by(Species) %>%
      do({
        mod <- lm(Sepal.Length ~ Sepal.Width, data = .)
        pred <- predict(mod, newdata = .["Sepal.Width"])
        data.frame(., pred)
      })
    

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