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abhargavc0002
abhargavc0002
Participating Frequently
October 4, 2022
Solved

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

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}

    harshitb6843
    harshitb6843Answer
    Brainy
    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. 

    View in thread

    6 replies

    harshitb6843
    harshitb6843Answer
    Brainy
    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
      Participating Frequently
      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
        Brainy
        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
          Participating Frequently
          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

          sureshkumarj8140
          Known Participant
          December 20, 2024

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

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