Hi Stephen,
As per my understanding, you would like the match the elements of B with values given in a vector A. There can be repetition and in such cases, we look for the next unmatched data. But as per the output provided by you, NaN needs to come before the matches for a given element in A. That is if A has [ 1 1 1] and B has [1 100; 1 200] then output should be [NaN NaN; 1 100; 2 100].
The following code assumes that A and B are in sorted order like in the example you provided. Hope this helps solve the issue you are facing.
a=[0 0 1 1 2 2 3 3 4 4 5 5]'; % a(:,1) in question
b=[0 110;0 210; 1 300; 1 405; 2 606; 3 707; 3 801;4 100; 5 204];
btmp=b;
final=zeros(length(a),2);%assign memory, same size as A
for i=1:length(a)
idx=find(btmp(:,1)==a(i),1);
if(length(idx)==1)
%if match then add it to final and remove from b temp
value=btmp(idx,:);
final(i,:)=value;
btmp(idx,:)=[];
else
j=i;
%if not found in current b temp but existed in b temp before, then
%shift previous elements down and add NaN at the top. This code
%does assume that the elements are in order like in the example. if
%not, modify the below while loop to shift values accordingly. Or
%you can remove the while loop entirely and how NaN be added after
%values are exhausted.
%{
That is the output will look like
0 110
0 210
1 300
1 405
2 606
NaN NaN
3 707
3 801
4 100
NaN NaN
5 204
NaN NaN
instead of
0 110
0 210
1 300
1 405
NaN NaN
2 606
3 707
3 801
NaN NaN
4 100
NaN NaN
5 204
like you had asked for.
%}
while j>1 && final(j-1,1)==a(i)
final(j,:)=final(j-1,:);
j=j-1;
end
final(j,:)=[NaN, NaN];
end
end
b=final
