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)?

7 次查看（过去 30 天）
J2A2B2
J2A2B2 2017-12-13
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

请先登录，再回答此问题。

采纳的回答

Joel Miller
Joel Miller 2017-12-13
Similar to the previous answer, but without the meshgrid:
largelogical = largematrix <= 1;
[value, index] = max(largelogical);

更多回答（3 个）

Walter Roberson
Walter Roberson 2017-12-13
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.
Birdman
Birdman 2017-12-13
编辑：Birdman 2017-12-13
One approach:
outputvector=largematrix(largematrix<=1);
To find specific rows and columns for the values:
[r,c]=find(largematrix<=1);
Rik
Rik 2017-12-13
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

类别

在 Help Center 和 File Exchange 中查找有关 Partial Differential Equation Toolbox 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by