Question

  top5 <- ifelse(state = "PA", c("Oranges", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
             ifelse(state = "MA"), c("Oranges", "Bananas", "Peaches", "Leaves", "Cucumbers"),
             ifelse(state = "ME"), c("Oranges", "Bananas", "Avocados", "Tomatoes", "Watermelon"),
             ifelse(state = "NH"), c("Tea", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
             ifelse(state = "RI"), c("Oranges", "Coconuts", "Avocados", "Tomatoes", "Cucumbers"),
             ifelse(state = "VT"), c("Carrots", "Bananas", "Apples", "Tomatoes", "Cucumbers"))

以上是我现在所拥有的。我想根据输入将相应的向量分配到函数中。我把它放在一个函数中，虽然我甚至无法让它在函数之外工作。谢谢。

回答答案，这是很好的解决方案。但是，我希望将这些向量中的一个传递给基于标准的变量，该变量不起作用。

我需要这样做的原因是稍后调用向量来通过匹配状态的相应向量来过滤该状态的数据。

例如，PA的五大水果是：“橘子”，“香蕉”，“鳄梨”，“西红柿”，“黄瓜”。如果函数调用PA，我需要将其赋值给变量，以便函数稍后可以将数据过滤到那些结果。这是我的完整代码：

FruitAnalyses <- function(statefile,state) 
top5 <- ifelse(state = "PA", c("Oranges", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
         ifelse(state = "MA"), c("Oranges", "Bananas", "Peaches", "Leaves", "Cucumbers"),
         ifelse(state = "ME"), c("Oranges", "Bananas", "Avocados", "Tomatoes", "Watermelon"),
         ifelse(state = "NH"), c("Tea", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
         ifelse(state = "RI"), c("Oranges", "Coconuts", "Avocados", "Tomatoes", "Cucumbers"),
         ifelse(state = "VT"), c("Carrots", "Bananas", "Apples", "Tomatoes", "Cucumbers"))

  statefile <- fread(paste0(state,"_fruit.csv"), header = T, data.table = F, 
stringsAsFactors = F, na.strings = c(",,",""," ")) 
#Lower names and convert to tbl
names(statefile) <- tolower(names(statefile))
statefile <- as.tbl(statefile)
#Parse Month and Year into single string - This is for later use when 
dates <-mdy(statefile$fruitstartdate)
statefile <- statefile %>%
 mutate(Month = month(dates, label = T))
rm(dates) 
statefileGrouped <- statefile %>%
  filter(priorfruitgroup %in% top5) %>%
  group_by(priorfruitgroup, Month) %>%
   summarise(TotalSeriousFruit = sum(seriousfruit),
            TotalFruitCount = sum(fruit)) %>%
  arrange(Month)
fwrite(statefileGrouped,file=paste0(state,".csv"))
barplot <- ggplot(statefileGrouped, aes(x=Month, y=Totalfruitcount, 
   fill=TotalotherfruitCount)) +
  geom_bar(color="black",stat="identity") +
  facet_grid(.~priorfruitgroup)
return(barplot)
}

忽略变量名称中的差异，出于保密原因，我不得不更改一些名称。然而，这不是问题的主要议题。我想知道为什么top5不会根据标准正确分配给矢量。

Answer 1

我可能会选择switch函数：

，而不是使用ifelse

states <- c('PA', 'MA', 'ME', 'NH', 'RI', 'VT')

sapply(states, function(x){
  switch(x,
         PA = c("Oranges", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
         MA = c("Oranges", "Bananas", "Peaches", "Leaves", "Cucumbers"),
         ME = c("Oranges", "Bananas", "Avocados", "Tomatoes", "Watermelon"),
         NH = c("Tea", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
         RI = c("Oranges", "Coconuts", "Avocados", "Tomatoes", "Cucumbers"),
         VT = c("Carrots", "Bananas", "Apples", "Tomatoes", "Cucumbers"))
})

     PA          MA          ME           NH          RI          VT         
[1,] "Oranges"   "Oranges"   "Oranges"    "Tea"       "Oranges"   "Carrots"  
[2,] "Bananas"   "Bananas"   "Bananas"    "Bananas"   "Coconuts"  "Bananas"  
[3,] "Avocados"  "Peaches"   "Avocados"   "Avocados"  "Avocados"  "Apples"   
[4,] "Tomatoes"  "Leaves"    "Tomatoes"   "Tomatoes"  "Tomatoes"  "Tomatoes" 
[5,] "Cucumbers" "Cucumbers" "Watermelon" "Cucumbers" "Cucumbers" "Cucumbers"

可以按如下方式操纵：

xx <- sapply(states, function(x){
  switch(x,
         PA = c("Oranges", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
         MA = c("Oranges", "Bananas", "Peaches", "Leaves", "Cucumbers"),
         ME = c("Oranges", "Bananas", "Avocados", "Tomatoes", "Watermelon"),
         NH = c("Tea", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
         RI = c("Oranges", "Coconuts", "Avocados", "Tomatoes", "Cucumbers"),
         VT = c("Carrots", "Bananas", "Apples", "Tomatoes", "Cucumbers"))
})

apply(t(xx), 1, paste, collapse = ', ')

                                                PA 
 "Oranges, Bananas, Avocados, Tomatoes, Cucumbers" 
                                                MA 
    "Oranges, Bananas, Peaches, Leaves, Cucumbers" 
                                                ME 
"Oranges, Bananas, Avocados, Tomatoes, Watermelon" 
                                                NH 
     "Tea, Bananas, Avocados, Tomatoes, Cucumbers" 
                                                RI 
"Oranges, Coconuts, Avocados, Tomatoes, Cucumbers" 
                                                VT 
   "Carrots, Bananas, Apples, Tomatoes, Cucumbers"

Answer 2

或者，您可以使用列表。

my.list <- list("PA"=c("Oranges", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
                "MA"=c("Oranges", "Bananas", "Peaches", "Leaves", "Cucumbers"),
                "ME"= c("Oranges", "Bananas", "Avocados", "Tomatoes", "Watermelon"),
                "NH"=c("Tea", "Bananas", "Avocados", "Tomatoes", "Cucumbers"),
                "RI"=c("Oranges", "Coconuts", "Avocados", "Tomatoes", "Cucumbers"),
                "VT"=c("Carrots", "Bananas", "Apples", "Tomatoes", "Cucumbers"))

然后可以这样访问：

my.list$NH
[1] "Tea"       "Bananas"   "Avocados"  "Tomatoes"  "Cucumbers"

或：

x <- 'PA'
my.list[[x]]

使用变量时。

根据条件为矢量赋值

2 个答案: