1

R で、ある文字列が別の文字列の省略形であるかどうかを判断する効率的な方法を探しています。私が取っている基本的なアプローチは、短い文字列の文字が長い文字列で同じ順序で表示されるかどうかを確認することです。たとえば、短い文字列が「abv」で長い文字列が「abbreviation」の場合、肯定的な結果が必要になりますが、短い文字列が「avb」の場合、否定的な結果が必要になります。機能する機能をまとめましたが、それはかなり洗練されていないソリューションのようで、正規表現の魔法が欠けているのではないかと思いました。R の「stringdist」関数も調べましたが、特にこれを行っているように見えるものは見つかりませんでした。これが私の機能です：

# 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)

}

助けてくれてありがとう！

4

2 に答える 2

1

基本的に各文字を取り、各文字の間に0回以上任意の文字に一致するオプションを追加するこのようなものはどうですか ( [a-z]*?)

f <- Vectorize(function(x, y) {
  xx <- strsplit(tolower(x), '')[[1]]
  grepl(paste0(xx, collapse = '[a-z]*?'), y)
  ## add this if you only want to consider letters in y
  # grepl(paste0(xx, collapse = sprintf('[%s]*?', tolower(y))), y)
}, vectorize.args = 'x')

f(c('ohb','hello','ob','ohc'), 'ohbother')
#  ohb hello    ob   ohc 
# TRUE FALSE  TRUE FALSE 

f(c('abbrev','abb','abv', 'avb'), 'abbreviation')
# abbrev    abb    abv    avb 
#   TRUE   TRUE   TRUE  FALSE
于 2015-11-02T17:36:07.280 に答える
0

答えが短いわけではありませんが、再帰を使用します（再帰はエレガントですよね？: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
于 2015-11-02T18:34:13.113 に答える