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I'm trying to remove some items which both in two lists.But the remove command doesn't work.A='(0 1 2 3 4 5 6 7 8 9)B='(1 3 5 7 9)C=nilforeach(item A if(member(item B) then C=remove(item A) ) )why the result is C= (0 1 2 3 4 5 6 7 8) not (0 2 4 6 8)?how the remove works?Who can help to tell me the reason or share me a right code to make it true~~Thanks.
I got the right result by using [b] copy [/b] and [b] remd [/b]A='(0 1 2 3 4 5 6 7 8 9)B='(1 3 5 7 9)C=copy(A)foreach(item A if(member(item B) then remd(item C) ) )
Hi Leon,remove is a destructive function so you were correct to use copy.Another (faster) method is to use cons, to create a new list, or the slowish following methodC = setof(item, A, !member(item, B))Cheers, Dave
Thanks Dave,I havn't think of !member, At first, I used member(item B)=nil, it doesn't work. Haha~~
below is what I use, it is very fast and does not go through unnecessary loops if you have duplicates in list A (or if all items in A have already been removed)A='(1 2 1 3 1 4 1 5)B='(1 3 5)procedure(SubstractList(B A) let((item) while(B && A item=car(B) B=cdr(B) A=remove(item A) ) A ))Note that both lists are modified inside the procedure. This is ok since they are copies.If both lists are dbids, then creating another procedure that uses "req" instead of "remove" will be even faster.
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