将新列添加到df中，并按组[重复项]获得最大值

Question

此问题已经在这里有了答案：

Calculate group mean (or other summary stats) and assign to original data（4个答案）

关闭58分钟前。

我试图首先通过某些分组value找到最大值id，然后我需要使用该分组ID的提取值来创建一列。

类似：

df$maxdvalue <-aggregate(value ~ id, data = df, max)

我收到以下错误：

$<-.data.frame（*tmp*中的错误，最大值，值= list（id = 1：1763，：替换项有1763行，数据项有74619]

Answer 1

有很多方法可以完成，并且您是R / stackoverflow的新手，所以我希望我早就知道几种方法：

# Groupwise aggregation 
# (note: the aggregate function applied can be changed to whatever is required) 

  # 1st base R method using "ave" function, assigned as vector:

  df$maxmpg <- ave(df$mpg, df$car_brand, FUN = max)

  # 2nd base R method, transforming dataframe using  "ave" function, assigned as dataframe: 

  df <- transform(df, maxmpg = ave(mpg, car_brand, FUN = max))

  # 3rd method using with syntax in conjunction with "ave", assigned as vector:  

  df$maxmpg <- with(df, ave(mpg, car_brand, FUN = max))

  # 4th method using cbind in conjunction with "ave", assigned as dataframe:  

  df <- cbind(df, maxmpg = ave(df$mpg, df$car_brand, FUN = max))

  # 5th method using tapply, assigned as vector:  

  df$maxmpg <- tapply(df$mpg, df$car_brand, max)

  # 6th base R method using lapply with lambda function  
  # and row-binding the each list element back into a df (assigned as df): 

  df <- do.call("rbind", lapply(split(df, df$car_brand), 

                                function(x){

                                  x$maxmpg <- max(x$mpg)

                                  return(x)

                                }

                              )
                          )

  # 7th base R solution using aggregate and merge (assigned as df): 

  df <- merge(df, setNames(aggregate(mpg ~ car_brand, df, FUN = max),

                           c("car_brand", "max_mpg")), by = "car_brand", all.x = TRUE)


  # Using pacakges: 

  # Create a vector of the required packages: 

  necessary_packages <- c("dplyr", "data.table")

  # Create a vector containing the names of any packages requiring installation: 

  new_packages <- necessary_packages[!(necessary_packages %in% installed.packages()[,"Package"])]

  # If the vector has more than 0 elements, install the new pacakges
  # (and it's/their) associated dependencies: 

  if(length(new_packages) > 0){

    install.packages(new_packages, dependencies = TRUE)

  }

  # Initialise the packages in the session: 

  lapply(necessary_packages,

         require,

         character.only = TRUE)

  # 8th method using dplyr, assigned as vector:

  df <- 

    df %>% 

    group_by(car_brand) %>% 

    mutate(maxmpg = max(mpg)) %>%

    ungroup()

# 9th solution using data.table (set as vector): 

dt <- data.table(df)

dt[, maxmpg := max(mpg), by = car_brand]

使用的数据：

df <- data.frame(car_type = row.names(mtcars),

                 car_brand = gsub(" .*", "", row.names(mtcars)),

                 mtcars, 

                 row.names = NULL)