Question

我在R中编写一个函数来消除城市名称的向量歧义。基本思路是编写一个函数，只要它们与查找表匹配就返回原始值，否则尝试以各种方式清理数据（例如模糊匹配，删除标点符号等）。

我试图总结一下这个例子中的逻辑：

x <- "sun fish"
s <- function(x) {
  if (x=='animal') {                #condition A
    return(paste(x,"is an animal"))
  } else if (x=='fish') {           #condition B
    return(paste(x,"is a fish"))
  } else {                          #condition C (does some cleaning)
    x <- sapply(strsplit(x," "),'[[',2)
    return(paste(x, "is something else"))
  }
}
s(x)

如果输入条件C，通过条件A和条件B再次传递x的最佳方法是什么？

Answer 1

您可以使用递归再次应用测试：

x <- "sun fish"
s <- function(x) {
        if (x=='animal') {                #condition A
                return(paste(x,"is an animal"))
        } else if (x=='fish') {           #condition B
                return(paste(x,"is a fish"))
        } else {                          #condition C (does some cleaning)
                y <- sapply(strsplit(x," "),'[[',2)
                if(x!=y) return(s(y))
                return(paste(x, "is something else"))
        }
}
s(x)
[1] "fish is a fish"

编辑：

上面的代码不适用于 elses 。这应该在保留动物全名的同时修复它：

x <- c("animal", "sun fish", "an other bug")
s <- function(x) {
        ifelse(x=='animal', 
               paste(x,"is an animal"),
               ifelse(x=='fish',
                      paste(x,"is a fish"), 
                      ifelse(lengths(strsplit(x, " "))>1, 
                             paste(sub("([a-z]*) .*", "\\1", x), 
                                   s(sub("[a-z]* (.+)", "\\1", x))),
                             paste(x, "is something else"))))
}
s(x)
[1] "animal is an animal"            "sun fish is a fish"             "an other bug is something else"

Answer 2

尝试使用switch()代替多个if()来电：

x <- "sun fish"
s <- function(x) {
  z <- switch(x, 
              animal = "is an animal",
              fish = "is a fish",
              "is something else"
  )
  paste(x, z)
}

结果：

s(x)
[1] "sun fish is something else"

如何迭代嵌套的if语句？

2 个答案:

编辑：