find first time 0 again after higher 0

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Frederik Reese
Frederik Reese 2022-6-17
评论： Star Strider 2022-6-19
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

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Star Strider
Star Strider 2022-6-17
Ran in:
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 个评论
Frederik Reese
Frederik Reese 2022-6-19
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 2022-6-19
Ran in:
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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