Forum Discussion
Problem with INDEX : returns only first match as a column
You said "I did take that into account. Here's the formula I adapted from the one Patrick2788 suggested : =LET(r;LIGNES(Tableau1);INDEX(Tableau1;SEQUENCE(r);{3;1;2;4})) ..."
So did you try the formula with "3,1,2,4" FIRST? or immediately replace the commas with semi-colons? The rows must be in an array perpendicular to the array for columns (i.e. 1 must use commas and the other semi-colons if those are the 2 deliminators).EDIT: here is a visual example of how rows & columns affect the output of INDEX:
notice how if both input are commas, the output is 3 cells in a row {(1,2),(4,4),(5,3)}
and if both are using semi-colons the output is 3 in a column {(1,2);(4,4);(5,3)}, same pairs as above but in a column instead of a row.
then if the 2 are different and you swap them you end up with the 2 matrix that are transpose of each other.
You said "I did take that into account. Here's the formula I adapted from the one Patrick2788 suggested : =LET(r;LIGNES(Tableau1);INDEX(Tableau1;SEQUENCE(r);{3;1;2;4})) ..."
So did you try the formula with "3,1,2,4" FIRST? or immediately replace the commas with semi-colons? The rows must be in an array perpendicular to the array for columns (i.e. 1 must use commas and the other semi-colons if those are the 2 deliminators).
EDIT: here is a visual example of how rows & columns affect the output of INDEX:
notice how if both input are commas, the output is 3 cells in a row {(1,2),(4,4),(5,3)}
and if both are using semi-colons the output is 3 in a column {(1,2);(4,4);(5,3)}, same pairs as above but in a column instead of a row.
then if the 2 are different and you swap them you end up with the 2 matrix that are transpose of each other.