Count the characters matching alphabets

Certified Senior Developer

Hello,

how to build an expression rule  which has 2 inputs - 1 is string, 2nd is alphabet. print the count of characters matching alphabet with the string

(Umbrella to expected Result -u1m1b1r1e1l2a1)

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  • Try this expression, it should work for you assuming you want to only return the first instance (with count) of each character.  E.g., "Caterpillar" = "c1a2t1e1r2p1i1l2".

    a!localVariables(
      local!alphabet: "abcdefghijklmnopqrstuvwxyz",
      local!input: "Umbrella",
      
      local!inputArray: a!forEach(
        items: 1+enumerate(len(local!input)),
        expression: lower(charat(local!input,fv!index))
      ),
      local!chars: union(local!inputArray,local!inputArray),
      
      joinarray(
        reject(
          fn!isnull,
          a!forEach(
            items: local!chars,
            expression: if(
              find(fv!item,local!alphabet,1)>0,
              concat(
                fv!item,
                count(wherecontains(fv!item,local!inputArray))
              ),
              null
            )
          )
        )
      )
    )

Reply
  • Try this expression, it should work for you assuming you want to only return the first instance (with count) of each character.  E.g., "Caterpillar" = "c1a2t1e1r2p1i1l2".

    a!localVariables(
      local!alphabet: "abcdefghijklmnopqrstuvwxyz",
      local!input: "Umbrella",
      
      local!inputArray: a!forEach(
        items: 1+enumerate(len(local!input)),
        expression: lower(charat(local!input,fv!index))
      ),
      local!chars: union(local!inputArray,local!inputArray),
      
      joinarray(
        reject(
          fn!isnull,
          a!forEach(
            items: local!chars,
            expression: if(
              find(fv!item,local!alphabet,1)>0,
              concat(
                fv!item,
                count(wherecontains(fv!item,local!inputArray))
              ),
              null
            )
          )
        )
      )
    )

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