Logical Indexing a spefic column in a matrix

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Hi,
I have a matrix with 5 columns and a variable number of rows (usually around 100). Column 4 of the matrix is a set of angles that can range anywhere from -180 to 180. I want to be able to create a set of smaller matrixes that splits up the larger matrix based on whether the angles fall in a 30 degree segment or not. For example a matrix that has 0 to 30 degrees, 30 to 60 degrees, 60 to 90 degrees, etc.
Can I use something like below. I don't think that will work though.
Zero_to_Thirty = A(4,:) < 30
Any help is appreciated!
Thanks,
Pat

采纳的回答

Guillaume
Guillaume 2014-12-17
编辑:Guillaume 2014-12-17
Don't store these submatrices in individual variables with different names.Rather store them in a cell array.
Use histcounts to find the distribution of the indices of the rows you want to split and arrayfun or a for loop to do the splitting:
m = rand(150, 6); m(:, 4) = randi([-180 180], 150, 1); %random matrix for demo, size 150*6
[~, ~, bins] = histcounts(m(:, 4), -180:30:180);
subm = arrayfun(@(bin) m(bins == bin, :), 1:max(bins), 'UniformOutput', false)
That last line with the arrayfun is equivalent to
subm = cell(1, max(bins));
for bin = 1:max(bins)
subm{bin} = m(bins == bin, :);
end
  5 个评论
Patrick
Patrick 2014-12-18
Ok this works fantastic. Except I forgot to mention that I actually needed the values in column 1 of the matrix in the subm cell array, that correspond to the largest values in column 5. Is there a way to keep them together?
Guillaume
Guillaume 2014-12-20
All the columns are present in the matrices of the subm cell array, so I'm not sure what you're asking.
If you want the value of column 1 that correspond to the maximum, use the second return value of max:
max_subm = zeros(numel(subm), 2)
lastmax = -Inf;
lastval = -Inf;
for idx = 1:numel(subm)
if ~isempty(subm{idx})
[lastmax row] = max(subm{idx}(:, 5));
lastval = subm{idx}(row, 1);
end
max_subm(idx, :) = [lastmax lastval];
end

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更多回答(1 个)

Roger Stafford
Roger Stafford 2014-12-17
If M is the original matrix,
c4 = M(:,4);
M1 = M(0<=c4 & c4<30,:); % <-- Having the angles between 0 and 30
M2 = M(30<=c4 & c4<60,:); % <-- Having the angles between 30 and 60
etc.
  2 个评论
Stephen23
Stephen23 2014-12-18
Surely it would be best not to split the array up into numbered variables. Doing this seems to invite the next question "how do I evaluate sequentially-numbered variable names?" which inevitably leads on to eval...
Roger Stafford
Roger Stafford 2014-12-18
Not necessarily. It depends on how many there are and how they are to be used.

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