cancel
Showing results for
Did you mean:
Highlighted
Solution Sage

## 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 arrange themselves in corresponding positions.
This is easily done using the List.Zip function.

But to get back two separate lists  you need a sort of List.Unzip function.

I imagined one based on the use of List.accumulate

How else can you do the reverse operation of List.Zip?

1 ACCEPTED SOLUTION

Accepted Solutions
Highlighted
Super User IV

## Re: List.Unzip wanted

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.

5 REPLIES 5
Highlighted
Super User IV

## Re: List.Unzip wanted

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.

Highlighted
Solution Supplier

## Re: List.Unzip wanted

Hey @Rocco_sprmnt21 ,

I came up with this:

(ListGiven as list) as list =>
let
n = List.Count(ListGiven{0}),
Acc =
(ListGiven, k, n) =>
let
ListK =
if k < n then
List.Transform(ListGiven, each _{k})
else
{},
NewList =
if k < n then
ListK & @Acc(ListGiven, k + 1, n)
else
ListK
in
NewList,
UnZipped = Acc(ListGiven, 0, n)
in
UnZipped

@Mariusz' answer is better and certainly easier. What I came up with just gives one list.

Feel free to connect with me:

Highlighted
Solution Supplier

## Re: List.Unzip wanted

Well,

Now I've made a tweaked List.Zip:

(ListGiven as list) as list =>
let
n =
List.Max(
List.Transform(
ListGiven,
each try List.Count(_) otherwise 1
)
),
GetK =
(k) =>
List.RemoveFirstN(
List.RemoveLastN(
List.Transform(
ListGiven,
each try _{k} otherwise null
),
each _ = null
),
each _ = null
),
Zip =
List.Generate(
() =>
[
l =
List.RemoveFirstN(
List.RemoveLastN(
List.Transform(
ListGiven,
each try _{0} otherwise _
),
each _ = null
),
each _ = null
),
k = 1
],
each [k] <= n,
each
[
l = GetK([k]),
k = [k] + 1
],
each [l]
)
in
Zip

Contrary to List.Zip, it will also take single values, and cut leading and trailing nulls.

If you want to test it, you could name it fnListZip and use this:

let
l = {1,3,5,7,9},
k = {2,4,6},
#"ListZip(l)" = List.Zip({l}),
#"fnListZip(l)" = fnListZip({l}),
#"ListZip(k,l)" = List.Zip({k,l}),
#"fnListZip(k,l)" = fnListZip({k, l}),
#"ListZip(ListZip(k,l))" = List.Zip(#"ListZip(k,l)"),
#"fnListZip(fnListZip(k,l))" = fnListZip(#"fnListZip(k,l)"),
#"ListZip(l,k,l)" = List.Zip({l, k, l}),
#"fnListZip(l,k,l)" = fnListZip({l, k, l}),
#"ListZip(ListZip(l,k,l))" = List.Zip(#"ListZip(l,k,l)"),
#"fnListZip(fnListZip(l,k,l))" = fnListZip(#"fnListZip(l,k,l)"),
NotReallyAListZip = {{1,2},{3,4},{5,6},7,9},
#"ListZip(NotReallyAListZip)" = List.Zip(NotReallyAListZip),
#"fnListZip(NotReallyAListZip)" = fnListZip(NotReallyAListZip),
NotReallyAListZipB = {1,2,3,4,{5,"Here Is Something"},7,9},
#"ListZip(NotReallyAListZipB)" = List.Zip(NotReallyAListZipB),
#"fnListZip(NotReallyAListZipB)" = fnListZip(NotReallyAListZipB)
in
#"fnListZip(NotReallyAListZipB)"

Best,

Spyros

Feel free to connect with me:

Highlighted
Solution Sage

## Re: List.Unzip wanted

wow. this is the solution :-).

I was speculating on the 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.

Highlighted
Solution Sage

## Re: List.Unzip wanted

...

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

Announcements

#### August 2020 Community Challenge: Can You Solve These?

We're excited to announce our first cross-community 'Can You Solve These?' challenge!

#### Community Blog

Visit our Community Blog for articles, guides, and information created by fellow community members.

#### Community Summit Australia – Join Online!

Be a part of the leading Microsoft Business Applications digital event, curated for the APAC community.

Top Solution Authors
Top Kudoed Authors