Создать таблицу данных, содержащую все возможные комбинации значений из четырех списков

Question

У меня есть следующие четыре списка.

varnames <- list("beefpork", "breakfast", "breakfast_yn", "diet_soda", "food_label", "fruit_and_veggie", "fruit_juice", "fruits", "milk",                      "min_foods","regular_soda", "ssb", "total_fruit", "vegetables",                      "asthma", "bmiclass3", "bmiclass4","bmiclass5", "dental_absence",                     "dental_appt", "diabetes", "food_allergies", "sore_teeth", "trying_weight",                     "count_pa60days", "count_vigpa20days", "gaming_bedroom", "other_organized_pa", "pa30outdoor","paguidelines", "pc_time", "school_transport", "sport_teams", "tv_bedroom", "tv_time_char", "video_games_char")
grades <- list("2", "4", "8", "11")
groups <- list("none", "ethnic", "bordercounty")
regions <- list("state", "hsr")

И следующая функция, которая возвращает целое число:

all_empty = function(outcome, groupvar, gradevar, regionvar){
  #How many observations?

  if (groupvar == "none") 
    fmla <- as.formula(paste0("~", outcome))
  else 
    fmla <- as.formula(paste0("~", outcome, "+", groupvar))

  if (regionvar == "hsr")
    mydata = span_phrwts
  else if (regionvar == "state" & groupvar %in% c("none", "ethnic"))
    mydata = span_statewts
  else if (regionvar == "state" & groupvar == "bordercounty")
    mydata = span_borderwts
  else mydata = span_statewts

  myrow = svytable(fmla, subset(mydata, grade==gradevar)) %>% nrow()
  return(myrow)
}

Я пытаюсь написать код, который будет запускатьработает со всеми 864 возможными комбинациями значений из моих списков и создает одну таблицу данных с 864 строками и 5 столбцами.

Я бы хотел, чтобы итоговая таблица выглядела примерно так, но не увенчалась успехом:

Variable          Grade          Group           Region     Obs
beefpork          2              none            state      5
beefpork          4              none            state      5
beefpork          8              none            state      3
beefpork          11             none            state      0

Это моя попытка выполнить это, но я не могу правильно рассчитать rownum .

output_all <- matrix(ncol = 5, nrow = length(varnames)*length(grades)*length(groups)*length(regions))
for(l in 1:length(regions)) {
  for (k in 1:length(grades)) {
    for(j in 1:length(groups)) {
      for(i in 1:length(varnames)){
        rownum = i + ((length(groups)*length(grades)*length(regions)) - 1)
        output_all[rownum, 1] = varnames[[i]]
        output_all[rownum, 2] = groups[[j]]
        output_all[rownum, 3] = grades[[k]]
        output_all[rownum, 4] = regions[[l]]
        output_all[rownum, 5] = all_empty(varnames[[i]], groups[[j]], grades [[k]], regions[[l]])

      }
    }
  } 
}
output_all %>% as_data_frame() %>% View()

Любая помощь / совет будет высоко ценится!

Answer 1

Если можно использовать векторы, а не списки, tidyr::crossing кажется простым подходом.

varnames <- c("beefpork", "breakfast", "breakfast_yn", "diet_soda", "food_label", "fruit_and_veggie", "fruit_juice", "fruits", "milk",                      "min_foods","regular_soda", "ssb", "total_fruit", "vegetables",                      "asthma", "bmiclass3", "bmiclass4","bmiclass5", "dental_absence",                     "dental_appt", "diabetes", "food_allergies", "sore_teeth", "trying_weight",                     "count_pa60days", "count_vigpa20days", "gaming_bedroom", "other_organized_pa", "pa30outdoor","paguidelines", "pc_time", "school_transport", "sport_teams", "tv_bedroom", "tv_time_char", "video_games_char")
grades <- c("2", "4", "8", "11")
groups <- c("none", "ethnic", "bordercounty")
regions <- c("state", "hsr")


tidyr::crossing(varnames, grades, groups, regions)


# A tibble: 864 x 4
   varnames grades groups       regions
   <chr>    <chr>  <chr>        <chr>  
 1 asthma   11     bordercounty hsr    
 2 asthma   11     bordercounty state  
 3 asthma   11     ethnic       hsr    
 4 asthma   11     ethnic       state  
 5 asthma   11     none         hsr    
 6 asthma   11     none         state  
 7 asthma   2      bordercounty hsr    
 8 asthma   2      bordercounty state  
 9 asthma   2      ethnic       hsr    
10 asthma   2      ethnic       state  

Answer 2

Используя data.table, у вас есть функция CJ для создания перекрестного соединения.Затем мы добавляем строку num (Idx) для выполнения построчного вызова функции.Мы наконец удалим столбец Idx

library(data.table)
dt <- CJ(varnames=varnames,grades=grades,groups=groups,regions=regions)
dt[,Idx:=.I]
dt[,by=Idx, Obs:=all_empty(outcome, groupvar, gradevar, regionvar)]
dt[,Idx:=NULL]

Answer 3

Рассмотрим expand.grid, затем вызовите вашу функцию с помощью mapply для поэлементной передачи значений столбца в пользовательский метод.

varnames <- c("beefpork", "breakfast", "breakfast_yn", "diet_soda", 
              "food_label", "fruit_and_veggie", "fruit_juice", 
              "fruits", "milk", "min_foods", "regular_soda", 
              "ssb", "total_fruit", "vegetables", "asthma", 
              "bmiclass3", "bmiclass4","bmiclass5", "dental_absence",
              "dental_appt", "diabetes", "food_allergies", 
              "sore_teeth", "trying_weight", "count_pa60days", 
              "count_vigpa20days", "gaming_bedroom", "other_organized_pa", 
              "pa30outdoor","paguidelines", "pc_time", "school_transport", 
              "sport_teams", "tv_bedroom", "tv_time_char", "video_games_char")
grades <- c("2", "4", "8", "11")
groups <- c("none", "ethnic", "bordercounty")
regions <- c("state", "hsr")

df <- expand.grid(varnames=varnames, grades=grades, groups=groups, regions=regions,
                  stringsAsFactors = FALSE)
str(df)
# 'data.frame': 864 obs. of  4 variables:
# $ varnames: chr  "beefpork" "breakfast" "breakfast_yn" "diet_soda" ...
# $ grades  : chr  "2" "2" "2" "2" ...
# $ groups  : chr  "none" "none" "none" "none" ...
# $ regions : chr  "state" "state" "state" "state" ...
# ...

df$fmla <- ifelse(df$groups == "none", paste0("~", outcome), paste0("~", outcome, "+", groupvar))

df$mydata <- ifelse(df$regions == "hsr", "span_phrwts",
                    ifelse(df$regions == "state" & df$groups %in% c("none", "ethnic"), "span_statewts",
                           ifelse(df$regions == "state" & df$groups == "bordercounty", "span_borderwts", 
                                  "span_statewts")))

Вызов функции

all_empty <- function(outcome, groupvar, gradevar, regionvar, fmla, mydata){
  # How many observations?
  myrow <- svytable(as.formula(fmla), subset(get(mydata), grade==gradevar))
  return(nrow(myrow))
}

df$Obs <- mapply(all_empty, df$varnames, df$groups, df$grades, 
                 df$regions, df$fmla, df$mydata)