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I have two giant array which looks like:

A = [11, 11, 12, 3, 3, 4, 4, 4 ];
B = [ 12, 4; 3, 11; 11, 1; 4, 13 ];

I want to create an array which takes values from B and column 1 from A to look like:

C = [ 11, 1; 11, 1; 12, 4; 3, 11; 3, 11; 4, 13; 4, 13; 4, 13 ];

I don't want to use for or any other kind of loop to optimize the process.

Sorry for being terse.

I will search each element from column 1 of A in column 1 of B and pick the corresponding column 2 elements from B and create a new array with column 1 elements of A and discovered column 2 elements from B.

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  • 2
    this needs EVEN MORE explaining! Commented Aug 19, 2015 at 16:01
  • Done. Check my edit. Commented Aug 19, 2015 at 16:15

2 Answers 2

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What you are doing in this problem is using A and searching the first column of B to see if there's a match. Once there's a match, extract out the row that corresponds to this matched location in B. Repeat this for the rest of the values in A.

Assuming that all values of A can be found in B and that the first column of B is distinct and that there are no duplicates, you can a unique call and sortrows call. The unique call is on A so that you can assign each value in A to be a unique label in sorted order. You would then use these labels to index into the sorted version of B to get your desired result:

[~,~,id] = unique(A);
Bs = sortrows(B);
C = Bs(id,:);

We get for C:

C =

    11     1
    11     1
    12     4
     3    11
     3    11
     4    13
     4    13
     4    13
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4 Comments

Your answer doesn't require an intermediate variable. On the other hand, it needs to sort a larger matrix
@LuisMendo - Correct! I suspect that your answer will be more efficient in the long run.
Sorry the answer was great but my problem was not well formulated I suppose. I have modified my problem statement.
@user1243255 you're very welcome. Sorry for being terse with you earlier... just that you new edit didn't make any sense until you corrected yourself.
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Thanks to @rayryeng for clarifying the question to me.

  • Assuming each element from A is present in column 1 of B:

    [~, ind] = max(bsxfun(@eq, A(:).', B(:,1)), [], 1);
C = B(ind,:);

  • If that assumption doesn't necessarily hold:

    [val, ind] = max(bsxfun(@eq, A(:).', B(:,1)), [], 1);
C = B(ind(val),:);

    So for example A = [11, 20, 12, 3, 3, 4, 4, 4 ]; would produce

    C =
    11     1
    12     4
     3    11
     3    11
     4    13
     4    13
     4    13
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1 Comment

Sorry the answer was great but my problem was not well formulated I suppose. I have modified my problem statement.

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