Trying to make function which takes string as input and returns no. of words in whole string

Question

**It takes Input as a string such as this - \'Nice one\' And Output gives - 4,3 (which is no. Of words in sentence or string) **

function countx(str)
   local cou         


        

Answer 1

def countLenWords(s):
   s=s.split(" ")
   s=map(len,s)
   s=map(str,s)
   s=list(s)
   return s

The above functions returns a list containing number of characters in each word

s=s.split(" ") splits string with delimiter " " (space) s=map(len,s) maps the words into length of the words in int s=map(str,s) maps the values into string s=list(s) converts map object to list

Short version of above function (all in one line)

def countLenWords(s):
   return list(map(str,map(len,s.split(" "))))

Answer 2

-- Localise for performance.
local insert = table.insert

local text = 'I am a poor boy straight. I do not need sympathy'

local function word_lengths (text)
    local lengths = {}
    for word in text:gmatch '[%l%u]+' do
        insert (lengths, word:len())
    end
    return lengths
end

print ('{' .. table.concat (word_lengths (text), ', ') .. '}')

• gmatch returns an iterator over matches of a pattern in a string.
• [%l%u]+ is a Lua regular expression (see http://lua-users.org/wiki/PatternsTutorial) matching at least one lowercase or uppercase letter:
  • [] is a character class: a set of characters. It matches anything inside brackets, e.g. [ab] will match both a and b,
  • %l is any lowercase Latin letter,
  • %u is any uppercase Latin letter,
  • + means one or more repeats.

Therefore, text:gmatch '[%l%u]+' will return an iterator that will produce words, consisting of Latin letters, one by one, until text is over. This iterator is used in generic for (see https://www.lua.org/pil/4.3.5.html); and on any iteration word will contain a full match of the regular expression.

Answer 3

You can find a simple alternative with your new requirements:

function CountWordLength (String)
  local Results  = { }
  local Continue = true
  local Position = 1
  local SpacePosition
  
  while Continue do
    SpacePosition = string.find(String, " ", Position)
    if SpacePosition then
      Results[#Results + 1] = SpacePosition - Position
      Position = SpacePosition + 1
      -- if needed to print the string
      -- local SubString = String:sub(Position, SpacePosition)
      -- print(SubString)
    else
      Continue = false
    end    
  end

  Results[#Results + 1] = #String - Position + 1
  
  return Results  
end

Results = CountWordLength('I am a boy')

for Index, Value in ipairs(Results) do
  print(Value)
end

Which gives the following results:

1
2
1
3