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abhargavc0002
abhargavc0002
October 4, 2022
Solved

How can we get 1st index elements of each array in different new array

  • October 4, 2022
  • 6 replies
  • 0 views

How can we get 1st index elements of each array in different new array. Like
{2,13,65}
{7,15,45}
{14,3,56}

So output of 1st elements should be {2,7,14}
And for 2nd and 3rd {13,15,3}
{65,45,56}

    Best answer by harshitb6843

    Appian doesn't really understand list of list and therefore it will flatten them into one big array. You might have to cast it in a dictionary so it can iterate on that. 

    a!localVariables(
  local!array:{
    {
      list: {2,13,65}
    },
    {
      list: {7,15,45}
    },
    {
      list: {14,3,56}
    }
    
  },
  a!forEach(
    items: local!array,
    expression: fv!item.list[1]
  )
)

    You can change the value of 1 to your desired index. 

    6 replies

    harshitb6843
    harshitb6843Answer
    October 4, 2022

    Appian doesn't really understand list of list and therefore it will flatten them into one big array. You might have to cast it in a dictionary so it can iterate on that. 

    a!localVariables(
  local!array:{
    {
      list: {2,13,65}
    },
    {
      list: {7,15,45}
    },
    {
      list: {14,3,56}
    }
    
  },
  a!forEach(
    items: local!array,
    expression: fv!item.list[1]
  )
)

    You can change the value of 1 to your desired index. 

      amans0009
      amans0009
      October 4, 2022

      for getting all the three arrays as desired/required

      a!localVariables(
  local!array:{
    {
      list: {2,13,65}
    },
    {
      list: {7,15,45}
    },
    {
      list: {14,3,56}
    },
    {
      list: {1,2,3}
    }

  },
  a!forEach(
    items:enumerate(3)+1,  /* three is the no. arrays you have to make/ output arrays */
    expression: a!localVariables(
      local!temp:fv!item,
      a!forEach(
    items: local!array,
    expression: fv!item.list[local!temp]
  )))
)
      This 3 should not pass the minimun no. of elements(fg-3) in any any list of the array .  

        harshitb6843
        harshitb6843
        October 4, 2022

        What is the point of copy-pasting the entire code but with increased complexity - one more a!forEach()? That just adds more executing time. 

          amans0009
          amans0009
          October 4, 2022

          Sir, as he said in the question that he wants n(3) arrays accordingly .. so I thought what if the arrays are {1,2,3,4,.....,n} ... So i've edited your code with one extra loop in which he can put the number of required arrays for output. 
            eg - [1,2,3] , [2,4,6], [4,8,9]  should be [1,2,4],[2,4,8],[3,6,9] 
               
          so i've added an extra loop for getting all the arrays as - if you put 2 in the first loop than you can get only [1,2,4],[2,4,6]. and same will work for the n times with a index function also.
          please correct me if I'm wrong

          S
          sureshkumarj8140
          December 20, 2024

          a!localVariables(
local!merger: merge({ 2, 13, 65 }, { 7, 15, 45 }, { 14, 3, 56 }),
local!merger
)

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