What is the most efficient way to find the position in the column of a matrix where the value drops below a given threshold (values are constantly decreasing down the columns)?

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J2A2B2 2017-12-13

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评论： Jos (10584) 2017-12-14

I have a large (up to 1000x1000) matrix which is the solution to a pde - the columns are the increments in time and the rows are the increments in space. The values down each column are decreasing and I want to find the row of each column where the value drops below a certain value (1 in the code below) and store these values in a vector where the value at each position is the row where it drops below the threshold. My method works perfectly well but is very slow, is there a better way?

My code:

timeivector = 2:state.Numberoftimesteps;  %starts at 2 since initial condition is zero everywhere
spaceivector = 1:state.Numberofspacesteps;
for ti = timeivector
  for xi = spaceivector
      if largematrix(xi,ti) <= 1
          continue 
      end
      outputvector(ti) = xi+1; 
  end
end

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Answer 1

Joel Miller 2017-12-13

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Similar to the previous answer, but without the meshgrid:

largelogical = largematrix <= 1;
[value, index] = max(largelogical);

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Rik 2017-12-13

value will by definition be 1, but I must admit this is more elegant (and faster) than my solution.

J2A2B2 2017-12-14

perfect, it takes practically no time at all to run. thanks

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Answer 2

Walter Roberson 2017-12-13

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first_row_below_threshold = sum(largematrix >= 1, 1) + 1;

Note: if none of the rows are below the threshold then the value will be 1 more than the number of rows in the matrix.

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Jos (10584) 2017-12-14

This makes excellent use of the given fact that the columns are in decreasing order. +1

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Answer 3

Birdman 2017-12-13

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编辑：Birdman 2017-12-13

在 MATLAB Online 中打开

One approach:

outputvector=largematrix(largematrix<=1);

To find specific rows and columns for the values:

[r,c]=find(largematrix<=1);

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Answer 4

Rik 2017-12-13

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编辑：Rik 2017-12-13

It will be much faster to use find, but you don't even need to.

You can use meshgrid to generate indices (like repmat((1:1000)',1,1000), but faster and clearer). Then use logical indexing to set all positions in the grid to inf for values>1, then using min will give you a vector with indices.

PS this will only find you 1 index per column