Forum Discussion
Lambda To Rearrange Data
SergeiBaklan PeterBartholomew1 @lori_m @Chris_Gross
Using a single formula, involving use of Dynamic Array Formulas, I am trying to Convert data
---from this format
----To This Format
Seeking help from the learned friends
I owe you an apology. I rushed the previous formula before taking the dogs out and introduced an error. I think this is correct.
= LAMBDA(Centre,Transaction,[field], LET( k, SEQUENCE(ROWS(TblOriginal)), blockNum?, ISNUMBER(centreSrNo), blockHdrRow, FILTER(k,blockNum?), fld, IF(ISOMITTED(field), SEQUENCE(1,7), field), idx, INDEX(blockHdrRow,Centre), IF(fld<=3, INDEX(Original, idx, fld), INDEX(Original, idx+1+Transaction,fld-2)) ) )(3,4)As SergeiBaklan suggested, PQ is purpose written for this type of data manipulation and should be the first strategy to be considered. What I have tried to do in the attached is to take the restructuring in two steps. The first is to write a Lambda function capable of reading any given data item from the original table, treating it as an array in 3 dimensions (Centre, Unit, Transaction).
To normalise the data from there requires one to build a set of dimension indices corresponding to records within the intended normalised table. 'CentrNo', 'UnitNo' and 'FieldNo' do this but they are not that easy to create for irregular lists.
Something that is possible, having created the first Lambda function, is to omit the step of creating a normalised form of the dataset altogether, and simply work with the original. One loses the flexibility of the pivot table but, in some instances, generating the desired results requires less work than normalising the data.
Another formula of note within the solution is the FillDownλ function used to calculate the 'CentreNo' array
= LAMBDA(values, SCAN(0, values, LAMBDA(prev,curr, IF(ISNUMBER(curr), curr, prev)) ) )As yet, I would not claim to know what constitutes good practice when it comes to using Lambda functions. There is too much 'learning as one goes' for that and, as always, multiple strategies are frequently available..
9 Replies
A key part of the solution is to develop a Lambda function that will return a record given the centre and transaction numbers
= LAMBDA(centre,transaction, LET( blkSrNos, INDEX(Original,,1), blockHdrRow, XMATCH({1;2;3}, blkSrNos), field, SEQUENCE(1,7), idx, XLOOKUP(centre, blockHdrRow,blockHdrRow,,1), IF(field<=3, INDEX(Original, idx, field), INDEX(Original, idx+1+transaction,field-2)) ) )(2,2)which can be called using
= NewRecordλ(2,2)- Thanks a Lot Peter ! MAY God Bless you with Long and Healthy Life !
I have surely become a LAMBDA lover recently and I am sure that this is going to prove a Life long relationship. Thanks to people like you and SergeiBaklan I am never stuck, whenever their is an issue. Let me try to understand the above and apply it to my situation. Should the help be needed, I know where to knock ! ThanksI owe you an apology. I rushed the previous formula before taking the dogs out and introduced an error. I think this is correct.
= LAMBDA(Centre,Transaction,[field], LET( k, SEQUENCE(ROWS(TblOriginal)), blockNum?, ISNUMBER(centreSrNo), blockHdrRow, FILTER(k,blockNum?), fld, IF(ISOMITTED(field), SEQUENCE(1,7), field), idx, INDEX(blockHdrRow,Centre), IF(fld<=3, INDEX(Original, idx, fld), INDEX(Original, idx+1+Transaction,fld-2)) ) )(3,4)As SergeiBaklan suggested, PQ is purpose written for this type of data manipulation and should be the first strategy to be considered. What I have tried to do in the attached is to take the restructuring in two steps. The first is to write a Lambda function capable of reading any given data item from the original table, treating it as an array in 3 dimensions (Centre, Unit, Transaction).
To normalise the data from there requires one to build a set of dimension indices corresponding to records within the intended normalised table. 'CentrNo', 'UnitNo' and 'FieldNo' do this but they are not that easy to create for irregular lists.
Something that is possible, having created the first Lambda function, is to omit the step of creating a normalised form of the dataset altogether, and simply work with the original. One loses the flexibility of the pivot table but, in some instances, generating the desired results requires less work than normalising the data.
Another formula of note within the solution is the FillDownλ function used to calculate the 'CentreNo' array
= LAMBDA(values, SCAN(0, values, LAMBDA(prev,curr, IF(ISNUMBER(curr), curr, prev)) ) )As yet, I would not claim to know what constitutes good practice when it comes to using Lambda functions. There is too much 'learning as one goes' for that and, as always, multiple strategies are frequently available..
As for me that's much easier to do with Power Query as in attached. What's the reason to do that with lambdas?
- Thanks Sergei !
That is bang on ! There are other parts of the problem too, but I choose to do it one by one. I will try the initial portion also. I thought it would be easier using LAMBDA and so I asked it. In any case, I was sure that you people are magicians.
PeterBartholomew1 has provided a LAMBDA solution and I must admit, I am enchanted by the power of this function. So trying hard to get into the intricacies ! You all of you experts here ! MAY God Bless you with Long and Healthy Life !
