Forum Discussion
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?
5 Replies
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.
...
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)) = liststhis 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 ListZipUnZipof 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
Hey Anonymous ,
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 UnZippedMariusz' answer is better and certainly easier. What I came up with just gives one list.
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 ZipContrary 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
