how to find the first element that satisfy a condition in a matrix?

5 次查看（过去 30 天）
Rima Habib
Rima Habib 2018-11-12
编辑： Rima Habib 2018-11-13
Hallo,
I have a matrix describes the difference between secuencial elements Dwf (127*2398).
now I want to find out the first element (column-wise) that satisfy the condition < 0.01 then starting from this element I want to find the first element that is < 0 and make a new matrix limited with the elements bewteen this two conditions as in the following : ( first_element : last_element +1)
I have tried some coding but it didn't work out so far, I am still a beginner with MATLAB.
hope someone could help and thatnks in advance
for i=6:127 % the rows, I must start from the 6th row
for j = 1:2700 % the columns
first_element(:,j) = find(Dwf(:,j)> 0.01,1,'first');
last_element(:,j) = find(Dwf(:,j) < 0 & j > first_element,1,'first') ;
end
end
  4 个评论
显示 2更早的评论
Rima Habib
Rima Habib 2018-11-12
编辑：Rima Habib 2018-11-12
@Stephan, the idea was to let the programm runs for all the columns to find the requested conditions
i explained wrong above. the columns represent the time in other matrix ( that means I have 2398 values regarding specific time data stored in other matrix) but in the Dwf matrix they are actually the differences not the rows! sorry!
so i need to find the first element that satisfy the condition < 0.01 in each column then starting from this element I want to find the first element that is < 0
i used the nested loop thinking it might be easier
Rima Habib
Rima Habib 2018-11-12
编辑：Rima Habib 2018-11-12
@Matt J, you are right, i forgot about this essencial condition in matlab
in general for each column i want to get a matrix that meets the explained conditions so at the end i get n matrx each one satisfies : ( first_element : last_element +1).
so it doesnt have to be in one matrix.
i dont know if that possible?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Bruno Luong
Bruno Luong 2018-11-13
Try this
for j = 1:ny
idx_start(:,j) = find(Dwf(:,j)> 0.01,1,'first') ;
idx_end(:,j) = find (Dwf(:,j) < 0 & (1:nx)' > idx_start(:,j) , 1, 'first');
end

更多回答（1 个）

KSSV
KSSV 2018-11-13
编辑：KSSV 2018-11-13
YOu may follow something like below:
load('dwf.mat')
[nx,ny] = size(Dwf) ;
iwant = NaN(size(Dwf)) ;
for i = 1:ny
idx1 = find(Dwf(:,i)<0.01) ; idx1 = idx1(1) ;
idx2 = find(Dwf(:,i)<0.) ; idx2 = idx2(1) ;
iwant(idx1:idx2,i) = Dwf(idx1:idx2,i) ;
end
pcolor(iwant) ;
shading interp ;
colorbar
YOu have to check the conditions properly.
  3 个评论
显示 1更早的评论
KSSV
KSSV 2018-11-13
编辑：KSSV 2018-11-13
It works for the entire columns....see the loop.....it is across all the columns. I have edited the code for the second condition .
Rima Habib
Rima Habib 2018-11-13
编辑：Rima Habib 2018-11-13
I know i saw the loop, please bare with me, I am a beginner here
why it gives me one value idx=35 instead of a value for each column, which was i want
load('dwf.mat')
[nx,ny] = size(Dwf) ;
iwant = NaN(size(Dwf)) ;
for j = 1:ny
idx_start(:,j) = find(Dwf(:,j)> 0.01,1,'first') ;
idx_end(:,j) = find (Dwf(:,j) < 0 & j > idx_start(:,j) , 1, 'first');
end
I tried this code, for the first condition it works but for the second, i couldnt find a solution yet.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Labels and Annotations 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