# finding the maximum value table A for values in table B bellow benchmark

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Alexandra 2016 年 2 月 15 日
コメント済み: Guillaume 2016 年 2 月 15 日
So, here's the situation: I have three tables: A = [50 100 30 4]; B = [100 150 90 50]; C = [50 100 150];
Value in C are benchmarks and we need a table that gives for each value of C, the maximum value in A such that all values in B are bellow/equal the benchmark. That is: D = [4 50 100]; In B only 50 is bellow/equal to 50 so 4 the máximum in A. But for 100, 100/90/50 are all bellow equal to 100 so 50 is the máximum in A.
I am trying with a loop but it's not working. Thanks a lot.
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Guillaume 2016 年 2 月 15 日
Note that A, B, C, D are not tables in the matlab sense. They're just vectors, a particular type of matrices.
Alexandra 2016 年 2 月 15 日
You're right

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### 採用された回答

Guillaume 2016 年 2 月 15 日
One possible way, assuming that values in A are always positive:
AA = repmat(A', 1, numel(C));
D = max(AA .* bsxfun(@le, B', C))
If you want it to work even with negative values in A, it's only slightly more complicated
AA = repmat(A', 1, numel(C));
D = max(AA .* (0./bsxfun(@le, B', C) + 1)) %0./x+1 change a logical matrix of [0, 1] into a matrix of [Nan, 1]
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Guillaume 2016 年 2 月 15 日
For the first version of the code:
• the transpose of A is replicated in columns as many times as there are elements in C
• bsxfun forms a 2d matrix by comparing B to C using the le function (i.e. <=). Each row corresponds to an element of B, each column to an element of C. Each element (i, j) of the matrix is either 0 (for Bi > Cj) or 1 (for Bi <= Cj)
• This matrix of 0 and 1 is then multiplied to the replicated and transpose version of A. This result in a matrix where for each column you have the original elements of A when B is smaller than the correspond element of C for that column, or 0 when B is greater.
• The max function then find the maximum of each column, elements in A that are not to be taken into account where changed to 0 in the previous step, so are never the maximum
The second version of the code is only modified so that instead of a matrix of 0 and 1 you get a matrix of NaN and 1. An element of A multiplied by NaN still give NaN and max ignores NaN.
Guillaume 2016 年 2 月 15 日
"Got this error: Error using bsxfun Non-singleton dimensions of the two input arrays must match each other."
Then, B or C are not vectors as in your question. At least one of them is 2D (or ND).

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