如何根据一些参数生成字母串

Question

我有一组句子，每个句子中的单词数不同。我需要用一串字母替换每个单词，但字母串需要基于特定的标准。例如，字母“t”只能替换为字母“i”、“l”、“f”；对于字母表中的每个字母，字母“e”只能被“o”或“c”替换，依此类推。此外，单词之间的空格以及句号、撇号和其他标点符号需要保持完整。举个例子： 原文：他爱狗。 带有一串字母的句子：Fc tcwoz bcy。

有没有办法在 R 中自动化这个过程？谢谢。

添加：我需要替换大约 400 个句子。句子存储在数据框的变量（data$sentences）中。

Answer 1

更新 2：一些代码重构，添加了一个简单的后备策略来处理丢失的字符（因此我们可以对给定字符串中的所有字符进行编码，即使我们没有确切的字符一对一映射），并在字符串向量上添加了示例循环。

# we define two different strings to be encode
mystrings <- c('bye', 'BYE')

# the dictionary with the replacements for each letter
# for the lowercase letters we are defining the exact entries
replacements <- {}
replacements['a'] <- 'xy'
replacements['b'] <- 'zp'
replacements['c'] <- '91'
# ... 
replacements['e'] <- 'xyv'
replacements['y'] <- 'opj'

# then we define a generic "fallback" entry
# to be used when we have no clues on how to encode a 'new' character
replacements['fallback'] <- '2345678'


# string, named vector -> character
# returns a single character chosen at random from the dictionary
get_random_entry <- function(entry, dictionary) {

  value <- dictionary[entry]

  # if we don't know how to encode it, use the fallback
  if (is.na(value)) {
    value <- dictionary['fallback']
  }

  # possible replacement for the current character
  possible.replacements <- strsplit(value[[1]], '')[[1]]

  # the actual replacement
  result <- sample(possible.replacements, 1)

  return(result)
}

# string, named vector -> string
# encode the given string, using the given named vector as dictionary
encode <- function(s, dictionary) {

  # get the actual subsitutions 
  substitutions <- sapply (strsplit(s,'')[[1]], function(ch) {

    # for each char in the string 's'
    # we collect the respective encoded version
    return(get_random_entry(ch, dictionary))

  }, USE.NAMES = F,simplify = T);

  # paste the resulting vector into a single string
  result <- paste(substitutions, collapse = '')

  # and return it
  return(result);
}

# we can use sapply to process all the strings defined in mystrings
# for 'bye' we know how to translate
# for 'BYE' we don't know; we'll use the fallback entry
encoded_strings <- sapply(mystrings, function(s) {
                                        # encode a single string
                                        encode(s, replacements)
                                     }, USE.NAMES =  F)

encoded_strings