File double_metaphone.rb has 445 lines of code (exceeds 250 allowed). Consider refactoring.

Open

require 'phonetic/algorithm'
require 'phonetic/double_metaphone/code'

module Phonetic
  # The Double Metaphone phonetic encoding algorithm is the second generation

Class DoubleMetaphone has 36 methods (exceeds 20 allowed). Consider refactoring.

Open

  class DoubleMetaphone < Algorithm
    START_OF_WORD_MAP = {
      # skip these when at start of word
      /^([GKP]N|WR|PS)/ => ['', '', 1],
      # initial 'X' is pronounced 'Z' e.g. 'Xavier'

Method encode_word has a Cognitive Complexity of 20 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_word(word, options = { size: 4 })
      code_size = options[:size] || 4
      w = word.upcase.gsub(/[^A-Z]/, '')
      return if w.empty?
      two = w[0, 2]

• Read up
  Read up

Method encode_j has a Cognitive Complexity of 16 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_j(w, i, len, code)
      r = 1
      last = len - 1
      # obvious spanish, 'jose', 'san jacinto'
      if w[i, 4] == 'JOSE' || w[0, 4] =~ /SAN\s/

• Read up
  Read up

Method encode_word has 45 lines of code (exceeds 25 allowed). Consider refactoring.

Open

    def self.encode_word(word, options = { size: 4 })
      code_size = options[:size] || 4
      w = word.upcase.gsub(/[^A-Z]/, '')
      return if w.empty?
      two = w[0, 2]

Method encode_word has 41 lines of code (exceeds 25 allowed). Consider refactoring.

Open

    def self.encode_word(word, options = { size: 4 })
      code_size = options[:size] || 4
      w = word.strip.upcase
      code = Code.new
      i = 0

Method encode_word has a Cognitive Complexity of 12 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_word(word, options = {})
      w = word.strip.upcase.gsub(/[^A-Z]+/, '')
      i = 0
      code = Code.new
      while i < w.size

• Read up
  Read up

Method encode_c has 33 lines of code (exceeds 25 allowed). Consider refactoring.

Open

    def self.encode_c(w, i, len, code)
      r = 1
      case
      # various germanic
      when c_germanic?(w, i)

Method encode_gh has a Cognitive Complexity of 10 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_gh(w, i, code)
      if i > 0 && !vowel?(w[i - 1])
        code.add 'K', 'K'
      # Parker's rule (with some further refinements)
      elsif !(i > 1 && w[i - 2] =~ /[BHD]/ || # e.g., 'hugh'

• Read up
  Read up

Method encode_s has a Cognitive Complexity of 10 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_s(w, i, len, code)
      r = 1
      last = len - 1
      case
      # special cases 'island', 'isle', 'carlisle', 'carlysle'

• Read up
  Read up

Method encode_s has 27 lines of code (exceeds 25 allowed). Consider refactoring.

Open

    def self.encode_s(w, i, len, code)
      r = 1
      last = len - 1
      case
      # special cases 'island', 'isle', 'carlisle', 'carlysle'

Method encode_sc has a Cognitive Complexity of 9 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_sc(w, i, code)
      # Schlesinger's rule
      if w[i + 2] == 'H'
        # dutch origin, e.g. 'school', 'schooner'
        if w[i + 3, 2] =~ /OO|UY|E[DM]/

• Read up
  Read up

Method encode_g has a Cognitive Complexity of 9 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_g(w, i, len, code)
      r = 2
      case
      when w[i + 1] == 'H'
        encode_gh(w, i, code)

• Read up
  Read up

Method encode_t has a Cognitive Complexity of 9 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_t(w, i, len, code)
      r = 1
      if w[i, 4] =~ /^(TION|TIA|TCH)/
        code.add 'X', 'X'
        r += 2

• Read up
  Read up

Method encode_c has a Cognitive Complexity of 9 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_c(w, n)
      metaph = ''
      if n == 0 || w[n - 1] != 'S' || !front_vowel?(w[n + 1])
        case
        when w[n + 1, 2] == 'IA'

• Read up
  Read up

Method encode_z has a Cognitive Complexity of 9 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_z(w, i, len, code)
      r = 1
      # chinese pinyin e.g. 'zhao'
      if w[i + 1] == 'H'
        code.add 'J', 'J'

• Read up
  Read up

Method encode_g has a Cognitive Complexity of 8 (exceeds 5 allowed). Consider refactoring.

Open

    def self.encode_g(w, n)
      metaph = ''
      silent = (w[n + 1] == 'H' && !vowel?(w[n + 2]))
      silent = true if n > 0 && w[n + 1] == 'N'
      silent = true if n > 0 && w[n - 1] == 'D' && front_vowel?(w[n + 1])

• Read up
  Read up

Consider simplifying this complex logical expression.

Open

        if i == last && i > 0 && vowel?(w[i - 1]) ||
           i > 0 && w[i - 1, 5] =~ /[EO]WSK[IY]/ ||
           w[0, 3] == 'SCH'
          code.add '', 'F'
        elsif w[i, 4] =~ /WICZ|WITZ/

Consider simplifying this complex logical expression.

Open

      elsif !(i > 1 && w[i - 2] =~ /[BHD]/ || # e.g., 'hugh'
              i > 2 && w[i - 3] =~ /[BHD]/ || # e.g., 'bough'
              i > 3 && w[i - 4] =~ /[BH]/)    # e.g., 'broughton'
        # e.g., 'laugh', 'McLaughlin', 'cough', 'gough', 'rough', 'tough'
        if i > 2 && w[i - 3, 3] =~ /[CGLRT].U/

Method find_code has 5 arguments (exceeds 4 allowed). Consider refactoring.

Open

    def self.find_code(map, w, i, last = nil, count = 0)