Question

我正在寻找一种有效的方法来判断一个字符串是否可能是另一个字符串的缩写。我正在采取的基本方法是查看较短字符串中的字母是否在较长字符串中以相同顺序出现。例如，如果我的短字符串是“abv”而我的长字符串是“缩写”，我想要一个正面的结果，而如果我的短字符串是“avb”，我会想要一个否定的结果。我有一个功能，我把它放在一起工作，但它似乎是一个非常不优雅的解决方案，我想我可能会错过一些正则表达式的魔法。我也看过R的'stringdist'函数，但是我没有发现任何看起来特别喜欢它的东西。这是我的功能：

# This function computes whether one of the input strings (input strings x and y) could be an abbreviation of the other
# input strings should be all the same case, and probably devoid of anything but letters and numbers
abbrevFind = function(x, y) {

  # Compute the number of characters in each string
  len.x = nchar(x)
  len.y = nchar(y)

  # Find out which string is shorter, and therefore a possible abbreviation
  # split each string into its component characters
  if (len.x < len.y) {

    # Designate the abbreviation and the full string
    abv = substring(x, 1:len.x, 1:len.x)
    full = substring(y, 1:len.y, 1:len.y)

  } else if (len.x >= len.y) {

    abv = substring(y, 1:len.y, 1:len.y)
    full = substring(x, 1:len.x, 1:len.x)

  }

  # Get the number of letters in the abbreviation
  small = length(abv)

  # Set up old position, which will be a comparison criteria
  pos.old = 0

  # set up an empty vector which will hold the letter positions of already used letters
  letters = c()

  # Loop through each letter in the abbreviation
  for (i in 1:small) {

    # Get the position in the full string of the ith letter in the abbreviation
    pos = grep(abv[i], full)
    # Exclude positions which have already been used
    pos = pos[!pos %in% letters]
    # Get the earliest position (note that if the grep found no matches, the min function will return 'Inf' here)
    pos = min(pos)
    # Store that position
    letters[i] = pos

    # If there are no matches to the current letter, or the current letter's only match is earlier in the string than the last match
    # it is not a possible abbreviation. The loop breaks, and the function returns False
    # If the function makes it all the way through without breaking out of the loop, the function will return true
    if (is.infinite(pos) | pos <= pos.old) {abbreviation = F; break} else {abbreviation = T}

    # Set old position equal to the current position
    pos.old = pos

  }

  return(abbreviation)

}

感谢您的帮助！

Answer 1

这样的事情，你基本上采取每个字符并添加一个选项，以匹配任何字母0（或{1}}之间的任何字母）

[a-z]*?

Answer 2

不是答案，但使用递归（递归很优雅，对吧？：p）

#Just a library I prefer to use for regular expressions
library(stringr)

#recursive function
checkAbbr <- function(abbr,word){
  #Go through each letter in the abbr vector and trim the word string if found
  word <- substring(word,(str_locate(word,abbr[1])[,1]+1))
  abbr <- abbr[-1]

  #as long as abbr still has characters, continue to loop recursively 
  if(!is.na(word) && length(abbr)>0){
    checkAbbr(abbr,word)
  }else{
    #if a character from abbr was not found in word, it will return NA, which determines whether the abbr vector is an abbreviation of the word string
    return(!is.na(word))
  }
}


#Testing cases for abbreviation or not
checkAbbr(strsplit("abv","")[[1]],"abbreviation") #FALSE
checkAbbr(strsplit("avb","")[[1]],"abbreviation") #FALSE
checkAbbr(strsplit("z","")[[1]],"abbreviation") #FALSE

R在完整字符串中查找缩写

2 个答案: