r - 字符串中的字符频率

Question

我想创建一个带有两个参数的函数，用于显示给定单词中某个字符的出现频率:x <- 单词，y <- 字母。因此，我创建了以下函数:

frequency <- function(x,y)
{
    word <- strsplit(x,"")
    counter <- 0
    for (i in 1:length(word)){
        if (word[i] == y) counter=counter+1
    }
    print(counter)
}

这个函数的基本思想是拆分给定单词的字符，迭代它们并在满足条件时增加计数器的值。然而，此函数始终返回值 0。这是什么原因？

Answer 1

另一个版本是将您的“单词”翻译成 raw() 向量，并与作为 raw() 向量的“字母”进行比较。

frequency = function(word, letter)
    sum(charToRaw(word) == charToRaw(letter))

这里有四种不同的解决方案

f0 <- function(word, letter)
    sum(strsplit(word, "")[[1]] == letter)

f1 <- function(word, letter)
    sum(charToRaw(word) == charToRaw(letter))

f2a <- function(word, letter)
    length(unlist(gregexpr(letter, word)))

f2b <- function(word, letter)
    length(unlist(gregexpr(letter, word, fixed=TRUE)))

有一些正确性和性能比较

> word <- "foo"
> letter <- "o"
> identical(f0(word, letter), f1(word, letter))
[1] TRUE
> identical(f0(word, letter), f2a(word, letter))
[1] TRUE
> identical(f0(word, letter), f2b(word, letter))
[1] TRUE
> letter <- "a"
> identical(f0(word, letter), f1(word, letter))
[1] TRUE
> identical(f0(word, letter), f2a(word, letter))
[1] FALSE
> identical(f0(word, letter), f2b(word, letter))
[1] FALSE
> word <- paste(sample(letters, 10000, TRUE), collapse="")
> letter <- "a"
> microbenchmark(
+     f0(word, letter), f1(word, letter),
+     f2a(word, letter), f2b(word, letter)
+ )
Unit: microseconds
              expr     min       lq      mean   median       uq      max neval
  f0(word, letter) 558.433 562.4755 579.03451 583.5590 584.8920  628.946   100
  f1(word, letter)  71.482  78.7100 100.85787  80.0275  81.7035 2195.366   100
 f2a(word, letter) 277.618 278.7280 280.94280 279.4870 280.4270  302.683   100
 f2b(word, letter)  66.888  68.1800  69.07205  68.6205  69.3100   84.300   100

f2b() 是最快的，但也不正确； f1() 目前看来既快又正确(尽管速度对于手头的任务可能并不重要)。