遍歷r中的數據幀的列

Question

我有以下問題：

levelsvar <- c("arrears", "expenses", "warmhome", "telephone", "colorTV", "washer", "car", "meatfish", "holiday")

variables <- NULL

for (i in 1:length(levelsvar)) {

variables <- sapply(levelstest, function(x) (length(test$levelsvar[i][test$country==x & test$levelsvar[i]=="1"]) + length(test$levelsvar[i][test$country==x & test$levelsvar[i]=="2"])) / length(test$levelsvar[i][test$country==x]))

} 

我想使用一個for循環來執行在“ levelsvar”的所有級別上可以看到9次以上的功能。 我嘗試了很多次，但都失敗了。 我認為問題是r讀

test$"arrears"

代替

test$arrears

我已經嘗試使用noquote（）了，但是沒有幫助。

您有解決此問題的方法嗎？

先感謝您！

編輯：

舉個例子

levelstest <- c("AT", "BE")

levelsvar <- c("arrears", "expenses", "warmhome", "telephone", "colorTV", "washer", "car", "meatfish", "holiday")

structure(list(country = c("AT", "AT", "AT", "BE", "BE", "BE"
), arrears = c(1L, 1L, 1L, 2L, 1L, 1L), expenses = c(3L, 1L, 
3L, 1L, 1L, 2L), warmhome = c(1L, 2L, 2L, 1L, 1L, 1L), telephone = c(4L, 
1L, 4L, 4L, 3L, 3L), colorTV = c(2L, 1L, 3L, 4L, 3L, 1L), washer = c(4L, 
1L, 3L, 3L, 1L, 2L), car = c(4L, 4L, 4L, 4L, 3L, 2L), meatfish = c(2L, 
1L, 1L, 4L, 1L, 1L), holiday = c(2L, 2L, 1L, 3L, 4L, 2L)), .Names = c("country", 
"arrears", "expenses", "warmhome", "telephone", "colorTV", "washer", 
"car", "meatfish", "holiday"), row.names = c(NA, 6L), class = "data.frame")

現在我嘗試

variables <- NULL

for (i in 1:length(levelsvar)) {

variables <- sapply(levelstest, function(x) (length(test[levelsvar[i]][test$country==x & test[levelsvar[i]]=="1"]) + length(test[levelsvar[i]][test$country==x & test[levelsvar[i]]=="2"])) / length(test[levelsvar[i]][test$country==x]))

  }

但這不起作用。

Answer 1

我想要實現的是獲得百分比(length(test$arrears[test$country==x & test$arrears=="1"]) + length(test$arrears[test$country==x & test$arrears=="2"])) / length(test$arrears[test$country==x])) levelsvar (length(test$arrears[test$country==x & test$arrears=="1"]) + length(test$arrears[test$country==x & test$arrears=="2"])) / length(test$arrears[test$country==x]))的所有級別都為levelsvar （值1和2），並且在levelstest所有國家/ levelstest 。

解決我的問題的方法如下：

test <- (structure(list(country = c("AT", "AT", "AT", "BE", "BE", "BE"
), arrears = c(1L, 1L, 1L, 2L, 1L, 1L), expenses = c(3L, 1L, 
                                                 3L, 1L, 1L, 2L), warmhome = c(1L, 2L, 2L, 1L, 1L, 1L), telephone = c(4L, 
                                                                                                                      1L, 4L, 4L, 3L, 3L), colorTV = c(2L, 1L, 3L, 4L, 3L, 1L), washer = c(4L, 
                                                                                                                                                                                           1L, 3L, 3L, 1L, 2L), car = c(4L, 4L, 4L, 4L, 3L, 2L), meatfish = c(2L, 
                                                                                                                                                                                                                                                              1L, 1L, 4L, 1L, 1L), holiday = c(2L, 2L, 1L, 3L, 4L, 2L)), .Names = c("country", 
                                                                                                                                                                                                                                                                                                                                    "arrears", "expenses", "warmhome", "telephone", "colorTV", "washer", 
                                                                                                                                                                                                                                                                                                                                    "car", "meatfish", "holiday"), row.names = c(NA, 6L), class = "data.frame"))

levelsvar <- c("arrears", "expenses", "warmhome", "telephone", "colorTV", "washer", "car", "meatfish", "holiday")

levelstest <- c("AT", "BE")

variables <- NULL

for (i in 1:length(levelsvar)) {

variables <- cbind(variables, sapply(levelstest, function(x) (length(test[levelsvar[i]][test[1]==x & test[levelsvar[i]]=="1"]) + length(test[levelsvar[i]][test[1]==x & test[levelsvar[i]]=="2"])) / length(test[levelsvar[i]][test[1]==x])))

  } 

Answer 2

您需要做的只是測試，這是：

apply(test[-1],MARGIN = 2,function(x){
  tapply(x,test$country,function(y){
    sum(y %in% c(1,2))/length(y)
  })
})

帶有margin = 2的apply（）將沿着您的列前進，而tapply（）將基於分組（國家/地區）計算自定義函數。 它甚至保留您的變量名。 test [-1]將跳過國家/地區列。