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我正在计算给定音节组合字符串列表的可能单词数。音节组合列表如下所示：

syllable_combinations <- c("C", "CC", "CCCV-CCV", "CCCV-CCV-CV", "CCCV-CV-CCV", "CCCV-CCV-CCV-CV", "CCCV-CC-CV", "CCCV-CCV-C", "CCCV-CV", "CV-C-CCCV")

在这个列表的基础上，我想计算给定音位规则的英语中可能单词的数量。为此，我需要遍历音节组合列表中的各个项目，并计算给定音节音节组合的可能单词数。

要为给定的音节组合生成可能的单词数，我需要遍历音节组合并根据其环境依次查看每个字符。例如，对于第一个音节组合，我需要执行以下操作：

  1. 确定该单词以单个辅音 C 开头（而不是 2 或 3 个辅音）；
  2. 确定第一个单辅音后面跟着元音 V；
  3. 确定该单词以下一个音节继续（由连字符表示）；
  4. 确定第二个音节也以单个辅音 C 开头；
  5. 并以另一个元音 V 结尾。

此信息需要与可能出现在这些位置的声音信息相关联：

number_of_vowels <- 20
number_of_initial_consonants_length_1 <- 22
number_of_initial_consonants_length_2 <- 47
number_of_final_consonants_length_1 <- 24

为了计算英语中具有“CVCV”音节结构的可能单词的数量：

number_of_CVCV_words <- number_of_initial_consonants_length_1*number_of_vowels*number_of_initial_consonants_length_1*number_of_vowels

number_of_CVCV_words
193600

关于如何做到这一点的任何建议？

我对此有所了解，但遇到了一些问题。

首先，将音节组合拆分为单独的音节：

split_syllables <- c()

for(i in 1:length(syllable_combinations)){
strsplit(as.character(syllable_combinations[i]), split = "-") -> split_syllable
split_syllables <- append(split_syllables, split_syllable)
}

然后，一个可以匹配每个音节的函数（唯一的音节数量有限，所以这是可行的）（在给定特定音节结构的情况下，counter1 变量给出了英语中可能的声音组合的数量）：

detect_syllables <- function(syllable){
if(syllable == "C") {
counter1 <- 25
} else if(syllable == "CC") {
counter1 <- 528
} else if(syllable == "CCCV") {
counter1 <- 200 
} else if(syllable == "CCV") {
counter1 <- 940
} else if(syllable == "CV") {
counter1 <- 440
} else if(syllable == "CVC") {
counter1 <- 10560
} else 
print(syllable, "syllable not matched")
}

然后，对原始音节组合中的每个音节执行 detect_syllables 函数的函数：

one_syllable <- function(first_syllable){
lapply(split_syllables[[i]][1], FUN = detect_syllables)
counter1 -> first_syl
first_syl -> number1
print(number1)
}

two_syllables <- function(first_syllable, second_syllable){
lapply(split_syllables[[i]][1], FUN = detect_syllables)
counter1 -> first_syl
lapply(split_syllables[[i]][2], FUN = detect_syllables)
counter1 -> second_syl
first_syl*second_syl -> number2
print(number2) 
}

three_syllables <- function(first_syllable, second_syllable, third_syllable){
lapply(split_syllables[[i]][1], FUN = detect_syllables)
counter1 -> first_syl
lapply(split_syllables[[i]][2], FUN = detect_syllables)
counter1 -> second_syl
lapply(split_syllables[[i]][3], FUN = detect_syllables)
counter1 -> third_syl
first_syl*second_syl*third_syl -> number3
print(number3)
}

four_syllables <- function(first_syllable, second_syllable, third_syllable, fourth_syllable){
lapply(split_syllables[[i]][1], FUN = detect_syllables)
counter1 -> first_syl
lapply(split_syllables[[i]][2], FUN = detect_syllables)
counter1 -> second_syl
lapply(split_syllables[[i]][3], FUN = detect_syllables)
counter1 -> third_syl
lapply(split_syllables[[i]][4], FUN = detect_syllables)
counter1 -> fourth_syl
first_syl*second_syl*third_syl*fourth_syl -> number4
print(number4)
}

还有一个 for 循环，以确保正确使用 detect_syllables 函数：

for(i in 1:10){
if(length(split_syllables[[i]]) == 1) { 
lapply(split_syllables[[i]][1], FUN = one_syllable)
} else if(length(split_syllables[[i]]) == 2) {
lapply(split_syllables[[i]][1], split_syllables[[i]][2], FUN = two_syllables)
} else if(length(split_syllables[[i]]) == 3) {
lapply(split_syllables[[i]][1], split_syllables[[i]][2], split_syllables[[i]][3], FUN = three_syllables)
} else if(length(split_syllables[[i]]) == 4) {
lapply(split_syllables[[i]][1], split_syllables[[i]][2], split_syllables[[i]][3], split_syllables[[i]][4], FUN = four_syllables)
} else 
print("number of syllables is bigger than 4")
}

但是，当我尝试使用 for 循环时，我收到以下错误消息：

Error in four_syllables(split_syllables[[1]]) : object 'counter1' not found

我意识到这与评估“counter1”的环境有关，如此处所述： Using get inside lapply, inside a function，但我不知道如何解决它。如果我尝试将它们指向正确的环境（FUN 中的错误（“C”[[1L]]，...）：未使用的参数），那么两个 lapply 似乎都不喜欢它。

如果不使用 lapply()，则可能会非常不雅地获得所需的结果。如果有人有其他解决方案，我很乐意了解它。

for(i in 1:10){
if(length(split_syllables[[i]]) == 1) { 
detect_syllables(split_syllables[[i]][1]) -> counter1
counter1 -> first_syl
first_syl -> number1
print(number1)
} else if(length(split_syllables[[i]]) == 2) {
detect_syllables(split_syllables[[i]][1]) -> counter1
counter1 -> first_syl
detect_syllables(split_syllables[[i]][2]) -> counter1
counter1 -> second_syl
first_syl*second_syl -> number2
print(number2)
} else if(length(split_syllables[[i]]) == 3) {
detect_syllables(split_syllables[[i]][1]) -> counter1
counter1 -> first_syl
detect_syllables(split_syllables[[i]][2]) -> counter1
counter1 -> second_syl
detect_syllables(split_syllables[[i]][3]) -> counter1
counter1 -> third_syl
first_syl*second_syl*third_syl -> number3
print(number3)
} else if(length(split_syllables[[i]]) == 4) {
detect_syllables(split_syllables[[i]][1]) -> counter1
counter1 -> first_syl
detect_syllables(split_syllables[[i]][2]) -> counter1
counter1 -> second_syl
detect_syllables(split_syllables[[i]][3]) -> counter1
counter1 -> third_syl
detect_syllables(split_syllables[[i]][4]) -> counter1
counter1 -> fourth_syl
first_syl*second_syl*third_syl*fourth_syl -> number4
print(number4)
} else 
print("number of syllables is bigger than 4")
}
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1 回答 1

0

不确定我是否遵循您想做的所有事情，但这里有一些代码可能会帮助您入门。

# save first two syllables
split_combs <- strsplit(syllable_combinations, "-")
syl1 <- sapply(split_combs, "[", 1)
syl2 <- sapply(split_combs, "[", 2)

# function to look at how a string starts
check.start <- function(string, start) {
    # does the string start with this?
    tfn <- substring(string, 1, nchar(start))==start
    tfn[is.na(tfn)] <- FALSE
    tfn
    }

# show all syllable combinations with the first two syllables starting with CV
syllable_combinations[check.start(syl1, "CV") & check.start(syl2, "CV")]
于 2013-06-21T13:14:13.783 回答