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Anonymous's avatar
Anonymous
Not applicable
6 years ago
Solved

List.Unzip wanted

let suppose we have two lists for which I want to keep the correspondence in the order of the elements. So if I change the order of the first one of the lists, the elements of the second list also a...
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  • Mariusz's avatar
    Mariusz
    6 years ago

    Hi Anonymous 

     

    You can zip it again, this will reverse the zip, or am I missing something?

    = List.Zip( List.Zip( { { 1, 2 }, { 3, 4 }, { 5, 6 } } ) )

     

    Best Regards,
    Mariusz

    If this post helps, then please consider Accepting it as the solution.

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Mariusz's avatar
Mariusz
Community Champion
6 years ago

Hi Anonymous 

 

You can zip it again, this will reverse the zip, or am I missing something?

= List.Zip( List.Zip( { { 1, 2 }, { 3, 4 }, { 5, 6 } } ) )

 

Best Regards,
Mariusz

If this post helps, then please consider Accepting it as the solution.

Please feel free to connect with me.
LinkedIn

 

Anonymous's avatar
Anonymous
Not applicable
6 years ago

wow. this is the solution :-).

 

I was speculating on the InsertRanges

insertRanges 

 

function and I was thinking about situations of possible errors to manage.
In particular, I was thinking of cases where the different length of the vectors involved can lead to error.
Aside from these cases, a potential error situation occurred to me which also depends on the order of the elements in the vectors.
And on this I looked for ways to manage the situation.

 

la situazione si potrebbe presentare quando positions and ranges are, for instance,

{1, 5, 2} and  {22,55, {33,44}}.

 

 

 

 

I tryed to solve in the following way:

 

 

 

let
    Source = List.Sort(List.Zip({{1, 5, 2}, {22,55, {33,44}}}),each _{0})
in
    List.Accumulate(Source, {{},{}},(s,c)=>{s{0}&{c{0}},s{1}&{c{1}}})

 

 

But now there is not needs of news unzip function.

  • Anonymous's avatar
    Anonymous
    Not applicable
    6 years ago

    ...

    in some respects the transformation carried out by the list.zip function on a set of lists is analogous (isomorphic?!) to that made by the table.transpose function on the set of columns of a table.
    therefore like the transpose (transpose (T)) = T, should be also List.Zip (list.Zip (lists)) = lists

    this property indicates yet another way of realizing the list.zip function through other functions

     

     

     

     

    let
    ListZipUnZip = (lsts) =>  Table.ToColumns(Table.Transpose(Table.FromColumns(lsts)))
in
    ListZipUnZip

     

     

     

    of course also the dual situation is valid.

    if you wanted, for very personal reasons, to simulate the Table.Transpose function, you could use List.Zip

     

     

    let
    transpose = (tab) =>  Table.FromColumns(List.Zip(Table.ToColumns(tab)))
in
    transpose