Here the solution will get slower and slower the more elements it has
because it will hold the entire list produce in memory and then iterates
over and adds to it for each new element for your solution and then that
is the end output. This current pr...
Dear Michel,The following function will do what you asked:q)update
(t[`b]!t`c)b from u where b in t`ba b-------0 "ert"1 "vbn"2 "ghr"3
"lkj"4 "poi"The dictionary inside the update statement between b and c
columns in t to replace the b values in u whe...