JaeHyoLee's solution to Word Count on the Lua track

Instructions

Test suite

Solution

Note:

This solution was written on an old version of Exercism. The tests below might not correspond to the solution code, and the exercise may have changed since this code was written.

Given a phrase, count the occurrences of each word in that phrase.

For example for the input "olly olly in come free"

olly: 2
in: 1
come: 1
free: 1

Running the tests

To run the tests, run the command busted from within the exercise directory.

Further information

For more detailed information about the Lua track, including how to get help if you're having trouble, please visit the exercism.io Lua language page.

Source

This is a classic toy problem, but we were reminded of it by seeing it in the Go Tour.

Submitting Incomplete Solutions

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word-count_spec.lua

local word_count = require('word-count').word_count

describe('word-count', function()
  it('counts one word', function()
    local result = word_count('word')
    local expected = { word = 1 }
    assert.are.same(expected, result)
  end)

  it('counts one of each', function()
    local result = word_count('one of each')
    local expected = { one = 1, of = 1, each = 1 }
    assert.are.same(expected, result)
  end)

  it('counts multiple occurrences', function()
    local result = word_count('one fish two fish red fish blue fish')
    local expected = { one = 1, fish = 4, two = 1, red = 1, blue = 1 }
    assert.are.same(expected, result)
  end)

  it('ignores punctuation', function()
    local result = word_count('car : carpet as java : javascript!!&@$%^&')
    local expected = { car = 1, carpet = 1, as = 1, java = 1, javascript = 1 }
    assert.are.same(expected, result)
  end)

  it('includes numbers', function()
    local result = word_count('testing, 1, 2 testing')
    local expected = { testing = 2, ['1'] = 1, ['2'] = 1 }
    assert.are.same(expected, result)
  end)

  it('normalizes case', function()
    local result = word_count('go Go GO')
    local expected = { go = 3 }
    assert.are.same(expected, result)
  end)
end)

local function update_result(word_table, word_normalized_table, new_word)
  if type(word_table[new_word]) ~= 'nil' then   -- if original word exists either in the table or normalized metatable
    if type(word_table[new_word]) == 'number' then  -- increase count if the word already exists
      word_table[new_word] = word_table[new_word] + 1
    else -- if it does not exist then look into metatable and get the key (word) and increase that word count
      word_table[word_normalized_table[new_word:lower()]] = word_table[word_normalized_table[new_word:lower()]] + 1
    end
  elseif type(word_table[new_word:lower()]) ~= 'nil' then  -- normalized word exists either in the table or normalized metatable
      word_table[word_normalized_table[new_word:lower()]] = word_table[word_normalized_table[new_word:lower()]] + 1
  else -- register word in both original and normalized metatable if it is the first time
    word_table[new_word] = 1
    word_normalized_table[new_word:lower()] = new_word
  end
end

return {
  word_count = function (source)
  local word_table = {}
  local word_normalized_table = {}
  setmetatable(word_table, {__index = word_normalized_table})

  for new_word in source:gmatch("%w+") do
    update_result(word_table, word_normalized_table, new_word)
  end
  return word_table
end
}

Community comments

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JaeHyoLee

Solution Author

commented about 5 years ago

I may misunderstood the last test in spec file. 'normalizes case' I thought it supposed to treat all mixed lower and upper case letters as same and it should be the first word in the test string. so in my solution it will give different answer for the following case. 'go Go GO' -> go = 3 'Go go GO' -> Go = 3 (since the Go is the first 'go' in the test array) Now I figured out by looking at others' solution.. that 'normalizes case' means all lower case.. so 'Go GO GO' should be go = 3 (at least.. I figured out how to use metatable and method.. though :-))

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JaeHyoLee's solution

to Word Count in the Lua Track