find first time 0 again after higher 0

Hi,

probably a simple question, but How can I find the Value (index) in each row of a matrix, which is after some values higher than 0 is 0 again?

for example this row in a matrix:

0 0 0 0.0619999999999550 0.108000000000004 0.129999999999995 0.156999999999982 0.176999999999964 0.185000000000002 0.189999999999998 0.187999999999988 0.187999999999988 0.185999999999979 0.182999999999993 0.180999999999983 0.176999999999964 0.170999999999992 0.164999999999964 0 0

I want to know the index of this row at the big black 0.

Thanks

Answer 1

[Star Strider]

One approach —

M = [0	0	0	0.0619999999999550	0.108000000000004	0.129999999999995	0.156999999999982	0.176999999999964	0.185000000000002	0.189999999999998	0.187999999999988	0.187999999999988	0.185999999999979	0.182999999999993	0.180999999999983	0.176999999999964	0.170999999999992	0.164999999999964		0	0 ]

M = 1×20

0 0 0 0.0620 0.1080 0.1300 0.1570 0.1770 0.1850 0.1900 0.1880 0.1880 0.1860 0.1830 0.1810 0.1770 0.1710 0.1650 0 0

idx = M~=0;                                                             % Logical Index
Out = strfind(idx, [1 0])+1                                             % Desired Index

Out = 19

Check = M((-1:1)+Out)                                                   % Check Result

Check = 1×3

0.1650 0 0

Thius should be reasonably robust to similar problems.

.

[Frederik Reese]

Thanks forr your help.

If i put a for loop on it to find all the values in different rows the for loop stops when it hits a row with only 0 values. Do you know how to solve this problem?

for i=1:29
idx = D_HQ5000_WH(i,:)~=0;                                              % Logical Index
Out = strfind(idx, [1 0])+1                                             % Desired Index
% Check = D_HQ5000_WH((-1:1)+Out)                                         % Check Result
Zeit_Flutende (i,1) = Out 
end

[Star Strider]

As always, my pleasure!

The only approach that comes quickly to mind is to add an if, elseif, else block —

D_HQ5000_WH = [randi([0 1], 1, 10); zeros(1,10); randi([0 1], 1, 10); ones(1,10)]

D_HQ5000_WH = 4×10

1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1

for i = 1:size(D_HQ5000_WH,1)
idx = D_HQ5000_WH(i,:)~=0;                                              % Logical Index
Out = NaN;
    if any(idx) & ~all(idx)
        Out = strfind(idx, [1 0])+1;                                    % Desired Index
    end
    Check = Out                                                         % Delete Later
end

Check = 1×2

4 6

Check = NaN

Check = 1×2

3 9

Check = NaN

I assigned ‘Out’ to NaN if either ‘idx’ has all the elements true or all the elements false. This traps conditions where all the elements are zero or none are zero, and sets those results to NaN. (It does not trap situations where there is more than one transition, so I assume that never occurs in practise.)

.