How can I stack each 3 sets of 4 rows?

I have a report that vertically stacks yearly data made of two columns:

1. The first column shows the year and each of its 4 quarters (format: "YYYY Qx").
2. The second column shows a $ amount.

A year is made of 4 rows corresponding to 4 quarters. As the reports vertically stacks data every year, 4 new rows are added every new year.

• 4 quarters per year * 3 years (2022, 2023 and 2024) = 12 rows.
• When year 2025 will be added it becomes the 4th year, the report will have 16 rows (4 quarters per year * 4 years). The year after, 2026, there will be 20 rows and so on.

I would like to horizontally stack each 3 sets (2022, 2023 and 2024) of 4 rows with adjacent column #2 data. The result is a table of 4 rows and 3*2=6 columns.

Start table in cell A1:

Desired output:

I do not need to generate the header row. I thought it would be simple:

=wraprows(tocol(A2:B13);8)

I still cannot make it:

=wrapcols(torow(wrapcols(torow(A2:B13);2));4)

I would join the two columns and wrap the result but how do I split later on?

=let(
list1;byrow(A1:B12;lambda(x;join("|";x)));
wrapcols(list1;4)
)

One way is to use Makearray. You move down four rows in the input each time you move across to the next pair of columns in the output, and for odd columns of the output take input from the left-hand column, or for even columns of the output take input from the right-hand column.

=makearray(4,6,lambda(r,c,index(A1:B12,r+4*quotient(c-1,2),iseven(c)+1)))

A more general form of the formula where you could adjust rows and columns would be (for three columns of input and nine columns of output)

=makearray(4,9,lambda(r,c,index(A15:C26,r+4*quotient(c-1,3),mod(c-1,3)+1)))

If you wanted to make the formula even more general, but assuming for now that the data divides exactly by the required number of columns , you could try this:

=let(range,A2:B13,inCols,columns(range),inRows,rows(range),reps,3,outRows,inRows/reps,outCols,inCols*reps,makearray(outRows,outCols,lambda(r,c,index(range,r+inRows/reps*quotient(c-1,inCols),mod(c-1,inCols)+1))))

Or for the case where the data doesn't divide exactly into the required number of columns,

=let(range,A2:B12,inCols,columns(range),inRows,rows(range),reps,3,outRows,roundup(inRows/reps),outCols,inCols*reps,iferror(makearray(outRows,outCols,lambda(r,c,index(range,r+outRows*quotient(c-1,inCols),mod(c-1,inCols)+1)))))

Maybe this is nearer to OP's requirement, where the number of rows of output is fixed (four quarters per column) and the number of columns is adjusted to match the number of rows of data provided.

=let(range,A2:B13,inCols,columns(range),inRows,rows(range),outRows,4,outCols,roundup(inRows/outRows)*inCols,iferror(makearray(outRows,outCols,lambda(r,c,index(range,r+outRows*quotient(c-1,inCols),mod(c-1,inCols)+1)))))