既存の列の値から新しい列のリストを動的に作成する方法R

Question

Rプログラミング言語では新しいので、Rongでmongodbデータ分析を行う必要があります。

注：ここでは、プロセスとcustnameの列をグループ化して "値"列の新しいリストを作成しています。参照データフレームを参照してください。

マイデータフレーム

project   Process  custname  column1 column2 

    analytics   view  jackson    ZZ  2              
    analytics   Read  jackson    KK  3 
    analytics   Read  jackson    FF  4 
    analytics   Read  jackson    KK  8              
    analytics   Read  ander    MM  9              
    analytics   Write  jackson    UU  5 
    analytics   Write  jackson    UU  6 

出力データフレーム。

Domain   Process  custname  Fields    values    

    analytics   view  jackson  column1    list(colfield ="ZZ", "Totalcount"="1") 
    analytics   view  jackson  column2    list(colfield ="2", "Totalcount"="1") 
    analytics   Read  jackson  column1    list(colfield ="KK", "Totalcount"="2",list(colfield ="FF","Totalcount"="1")) 
    analytics   Read  jackson  column2    list(colfield ="3", "Totalcount"="1",list(colfield ="4", "Totalcount"="1"), list(colfield ="8", "Totalcount"="1")) 
    analytics   Read  ander   column1    list(colfield ="MM", "Totalcount"="1") 
    analytics   Read  ander   column2    list(colfield ="9", "Totalcount"="1") 
    analytics   Write  jackson  column1    list(colfield ="UU", "Totalcount"="2") 
    analytics   Write  jackson  column2    list(colfield ="5", "Totalcount"="1",list(colfield ="6", "Totalcount"="1")) 

Dput

structure(list(project = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L), .Label = "analytics", class = "factor"), Process = structure(c(2L, 
1L, 1L, 1L, 1L, 3L, 3L), .Label = c("Read", "view", "Write"), class = "factor"), 
    custname = structure(c(2L, 2L, 2L, 2L, 1L, 2L, 2L), .Label = c("ander", 
    "jackson"), class = "factor"), column1 = structure(c(5L, 
    2L, 1L, 2L, 3L, 4L, 4L), .Label = c("FF", "KK", "MM", "UU", 
    "ZZ"), class = "factor"), column2 = structure(c(1L, 2L, 3L, 
    6L, 7L, 4L, 5L), .Label = c("2", "3", "4", "5", "6", "8", 
    "9"), class = "factor")), .Names = c("project", "Process", 
"custname", "column1", "column2"), row.names = c(NA, -7L), class = "data.frame") 

Answer 1

のは、あなたの応答のために、この