Re: (Long post) Metaphone Algorithm In AWK

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Sujet : Re: (Long post) Metaphone Algorithm In AWK
De : porkchop (at) *nospam* invalid.foo (Mike Sanders)
Groupes : comp.lang.awk
Date : 20. Aug 2024, 12:33:33
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <va1uuc$3c4if$1@dont-email.me>
References : 1
User-Agent : tin/2.6.2-20221225 ("Pittyvaich") (NetBSD/9.3 (amd64))
Okay, essentially complete (enough for me at least).

- added sorely needed 'TH' digraph handling

- helper functions refactored

- misc code cleanup

# Metaphone Algorithm In AWK v4: Michael Sanders - 2024
#
# https://en.wikipedia.org/wiki/Metaphone
#
# example invocation: awk -f metaphone.awk -v find=butter < words.txt

BEGIN { find_code = metaphone(find) }

# -----------------------------------------------------------------

# emit metaphone codes only
# { for (x = 1; x <= NF; x++) print $x " : " metaphone($x) }

# tweek levenshtein distance to open/constrain results...
{

for (x = 1; x <= NF; x++)
   if (metaphone($x) == find_code && levenshtein($x, find) <= 2)
      print $x

}

# -----------------------------------------------------------------

function metaphone(w, m, c, n, z, i) {
  w = toupper(w)
  gsub(/[^A-Z]/, "", w)  # strip non-alphabetic characters
  z = length(w)

  # handle initial letters
  if (substr(w, 1, 2) ~ /^(KN|GN|PN|WR|PS)/) {
    w = substr(w, 2)
    z--
  }

  for (i = 1; i <= z; i++) {
    c = substr(w, i, 1)
    n = (i < z) ? substr(w, i + 1, 1) : ""

    # skip duplicate letters except for 'C'
    if (i > 1 && c == substr(w, i - 1, 1) && c != "C") continue

    # handle vowels: retain only if it's 1st letter
    if (isvowel(c)) {
      if (i == 1) m += c
    }
    # consonants
    else if (c == "B") {
      if (!(i == z && substr(w, i - 1, 1) == "M")) m += "B"
    }
    else if (c == "C") {
      if (substr(w, i, 2) == "CH") {
        m += "X"
        i++
      } else if (substr(w, i, 2) ~ /^(CI|CE|CY)/) {
        m += "S"
      } else {
        m += "K"
      }
    }
    else if (c == "D") {
      if (substr(w, i, 2) == "DG" && substr(w, i + 2, 1) ~ /[IEY]/) {
        m += "J"
        i += 2
      } else {
        m += "T"
      }
    }
    else if (c == "G") {
      if (substr(w, i, 2) == "GH" && (i == 1 || !isvowel(substr(w, i - 1, 1)))) {
        i++
      } else if (substr(w, i, 2) == "GN" || (i == z && c == "G")) {
        continue
      } else if (substr(w, i, 3) ~ /^(GIA|GIE|GEY)/) {
        m += "J"
      } else {
        m += "K"
      }
    }
    else if (c == "H") {
      if (i == 1 || substr(w, i - 1, 1) !~ /[CSPTG]/) {
        if (i < z && !isvowel(n)) {
          m += "H"
        }
      }
    }
    else if (c == "K") {
      if (i == 1 || substr(w, i - 1, 1) != "C") m += "K"
    }
    else if (c == "P") {
      if (substr(w, i, 2) == "PH") {
        m += "F"
        i++
      } else {
        m += "P"
      }
    }
    else if (c == "Q") {
      m += "K"
    }
    else if (c == "S") {
      if (substr(w, i, 2) == "SH") {
        m += "X"
        i++
      } else if (substr(w, i, 3) == "TIA" || substr(w, i, 3) == "TIO") {
        m += "X"
        i += 2
      } else {
        m += "S"
      }
    }
    else if (c == "T") {
      if (substr(w, i, 2) == "TH") {
        m += "0"  # add '0' for 'TH' digraph to distinguish from regular 'T'
        i++
      } else if (substr(w, i, 3) == "TIA" || substr(w, i, 3) == "TIO") {
        m += "X"
        i += 2
      } else {
        m += "T"
      }
    }
    else if (c == "V") {
      m += "F"
    }
    else if (c == "W" || c == "Y") {
      if (i < z && isvowel(n)) m += c
    }
    else if (c == "X") {
      m += "KS"
    }
    else if (c == "Z") {
      m += "S"
    }
    # ensure 'M', 'N', and 'L' are always retained
    else if (c == "M" || c == "N" || c == "L") {
      m += c
    }
  }

  return m
}

# -----------------------------------------------------------------

function levenshtein(w1, w2, l1, l2, i, j, cst, diz) {
  l1 = length(w1)
  l2 = length(w2)

  # initialize distance array
  for (i = 0; i <= l1; i++) diz[i, 0] = i
  for (j = 0; j <= l2; j++) diz[0, j] = j

  # compute distance
  for (i = 1; i <= l1; i++) {
    for (j = 1; j <= l2; j++) {
      cst = (substr(w1, i, 1) == substr(w2, j, 1)) ? 0 : 1
      diz[i, j] = min3(diz[i-1, j] + 1,     # deletion
                       diz[i, j-1] + 1,     # insertion
                       diz[i-1, j-1] + cst) # substitution
    }
  }

  return diz[l1, l2]
}

# -----------------------------------------------------------------

# metaphone helper function
function isvowel(char) { return char ~ /[AEIOU]/ }

# -----------------------------------------------------------------

# levenshtein helper function
function min3(a, b, c) {
  return (a < b ? (a < c ? a : c) : (b < c ? b : c))
}

# eof

--
:wq
Mike Sanders


Date Sujet#  Auteur
17 Aug 24 * (Long post) Metaphone Algorithm In AWK17Mike Sanders
19 Aug 24 +* Re: (Long post) Metaphone Algorithm In AWK10Ben Bacarisse
19 Aug 24 i+- Re: (Long post) Metaphone Algorithm In AWK1Ben Bacarisse
19 Aug 24 i+* Re: (Long post) Metaphone Algorithm In AWK2Mike Sanders
19 Aug 24 ii`- Re: (Long post) Metaphone Algorithm In AWK1Mike Sanders
20 Aug 24 i`* Re: (Long post) Metaphone Algorithm In AWK6Mike Sanders
21 Aug 24 i `* Re: (Long post) Metaphone Algorithm In AWK5Ben Bacarisse
21 Aug 24 i  `* Re: (Long post) Metaphone Algorithm In AWK4Mike Sanders
21 Aug 24 i   +- Re: (Long post) Metaphone Algorithm In AWK1Mike Sanders
21 Aug 24 i   `* Re: (Long post) Metaphone Algorithm In AWK2Ben Bacarisse
21 Aug 24 i    `- Re: (Long post) Metaphone Algorithm In AWK1Mike Sanders
20 Aug 24 +- Re: (Long post) Metaphone Algorithm In AWK1Mike Sanders
21 Aug 24 +* Re: (Long post) Metaphone Algorithm In AWK3Mike Sanders
21 Aug 24 i`* AWK language trivia (Was: (Long post) Metaphone Algorithm In AWK)2Kenny McCormack
21 Aug 24 i `- Re: AWK language trivia1Mike Sanders
21 Aug 24 +- Re: (Long post) Metaphone Algorithm In AWK1Mike Sanders
23 Aug 24 `- Re: (Long post) Metaphone Algorithm In AWK1Mike Sanders

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