find first time 0 again after higher 0

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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

Accepted Answer

Star Strider
Star Strider on 17 Jun 2022
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.
.
  2 Comments
Frederik Reese
Frederik Reese on 19 Jun 2022
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
Star Strider on 19 Jun 2022
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.)
.

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