如何将长格式的数据框转换为适当格式的列表？

Question

我有以下长格式的数据框：

我需要将它转换成一个看起来像这样的列表：

其中，列表的每个主要元素都是“实例编号”。及其子元素应包含所有对应的参数和值对 - 格式为“参数 X”=“abc”，如您在第二张图片中看到的那样，一个接一个地列出。

是否有任何现有的功能可以做到这一点？我真的找不到任何东西。任何帮助将不胜感激。

谢谢。

Answer 1

dplyr 解决方案

require(dplyr)
df_original <- data.frame("Instance No." = c(3,3,3,3,5,5,5,2,2,2,2),
                      "Parameter" = c("age", "workclass", "education", "occupation", 
                                      "age", "workclass", "education", 
                                      "age", "workclass", "education", "income"),
                      "Value" = c("Senior", "Private", "HS-grad", "Sales",
                                  "Middle-aged", "Gov", "Hs-grad",
                                  "Middle-aged", "Private", "Masters", "Large"),
                      check.names = FALSE)
    
# the split function requires a factor to use as the grouping variable.
# Param_Value will be the properly formated vector
df_modified <- mutate(df_original,
                      Param_Value = paste0(Parameter, "=", Value))
# drop the parameter and value columns now that the data is contained in Param_Value
df_modified <- select(df_modified,
                      `Instance No.`,
                      Param_Value)

# there is now a list containing dataframes with rows grouped by Instance No.
list_format <- split(df_modified, 
                     df_modified$`Instance No.`)

# The Instance No. is still in each dataframe. Loop through each and strip the column.
list_simplified <- lapply(list_format, 
                          select, -`Instance No.`)

# unlist the remaining Param_Value column and drop the names.                      
list_out <- lapply(list_simplified , 
                   unlist, use.names = F)
                     

现在应该有一个按要求格式化的向量列表。

$`2`
[1] "age=Middle-aged"   "workclass=Private" "education=Masters" "income=Large"     

$`3`
[1] "age=Senior"        "workclass=Private" "education=HS-grad" "occupation=Sales" 

$`5`
[1] "age=Middle-aged"   "workclass=Gov"     "education=Hs-grad"

贴出来的data.table解决方案比较快，不过我觉得这样比较好理解。

Answer 2

require(data.table)
your_dt <- data.table(your_df)

dt_long <- melt.data.table(your_dt, id.vars='Instance No.')
class(dt_long) # for debugging
dt_long[, strVal:=paste(variable,value, sep = '=')]

result_list <- list()

for (i in unique(dt_long[['Instance No.']])){
  result_list[[as.character(i)]] <- dt_long[`Instance No.`==i, strVal]
}

Answer 3

仅供参考。这是执行此操作的 R base oneliner。 df 是您的数据框。

l <- lapply(split(df, list(df["Instance No."])), 
            function(x) paste0(x$Parameter, "=", x$Value))