MATLAB: How to find a value X of given Y close to 0, where the value X comes at the first place

findminimum

Hi,
I used to solved my previous question to find a value X of given Y close to 0, but there is a value X comes from the very late place, which is not what I need. How can I extract the earlier value of X?
For example,
338.00 339.00 340.00 341.00 342.00 343.00 … 964.00
1.00 -100.00 -100.00 -100.00 -100.00 -100.00 -100.00 … -100.00
2.00 100.00 100.00 100.00 100.00 100.00 100.00 … 100.00
3.00 0.28 0.12 -0.05 -0.21 -0.38 -0.55 … -0.02
4.00 0.28 0.12 -0.05 -0.21 -0.38 -0.55 … -0.02
5.00 8.21 8.24 8.26 8.28 8.30 8.32 … 8.35
6.00 8.21 8.24 8.26 8.28 8.30 8.32 … 8.35
Where the 340.00 is what I need, not the 964.00. How do I write the code to get 342.00?
The current code I am using is:
dr = damp_ratio(1:6 ,:);
[~,ix]=min(abs(dr),[],'all','linear');
[~,speed]=ind2sub(size(dr),ix);
Thanks in advance.
Terence

Best Answer

damp_ratio=[ -100.00  -100.00  -100.00  -100.00  -100.00  -100.00   -100.00
              100.00   100.00   100.00   100.00   100.00   100.00   +100
	      0.28      0.12 	  -0.05    -0.21    -0.38    -0.55   -0.02
              0.28 	0.12 	  -0.05    -0.21    -0.38    -0.55   -0.02
	      8.21    8.24 	   8.26     8.28     8.30     8.32   8.35 
	      8.21    8.24 	   8.26     8.28     8.30     8.32   8.35]      
damp_ratio = 6×7
 -100.0000 -100.0000 -100.0000 -100.0000 -100.0000 -100.0000 -100.0000
  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000  100.0000
    0.2800    0.1200   -0.0500   -0.2100   -0.3800   -0.5500   -0.0200
    0.2800    0.1200   -0.0500   -0.2100   -0.3800   -0.5500   -0.0200
    8.2100    8.2400    8.2600    8.2800    8.3000    8.3200    8.3500
    8.2100    8.2400    8.2600    8.2800    8.3000    8.3200    8.3500

dr = diff(  sign(damp_ratio(1:6 ,:)),1,2)~=0 ;
[val,ix]=max(dr,[],2);
speed=ix+1;
speed(val==0)=nan
speed = 6×1
   NaN
   NaN
     3
     3
   NaN
   NaN
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