The FIND command returns an empty matrix for a number I know exists

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Dharmesh Nadhe
Dharmesh Nadhe 2011 年 1 月 28 日
コメント済み: Image Analyst 2018 年 4 月 16 日
I have sets of data in column vector form (around 70000 elements).
To find the maximum element I used "max(x)" and the value comes to around 0.0305.
I want to know the indices of this element (number of the max value element). I am using find command as
But I get answer as
"Empty matrix: 0-by-1".
It works when I use
and displays all indices with value less than 0.0305. Can someone explain what I am doing wrong?
  1 件のコメント
Sanjay 2013 年 10 月 22 日
I have a similar related problem. Above given question speaks about a unique problem where the indices of the maximum element has to be find. I have a data series contained in 'am1' with sample rate of 100 sps. The time 't' is of 359.79 secs so in total i have 35979 data points.
In my code I am using
trScl = find(t ==183.5); trScu = find(t ==213.5); where these index point have to be used to create a new time series
there is no error but output comes with an empty matrix likewise; trScl [] trScu []
Now if instead of above mentioned values I use trScl = find(t ==80); trScu = find(t ==98);
the code works providing me with the correct answer. i.e t <= 100 it works,!!!
Kindly help me on this subject


回答 (5 件)

Doug Eastman
Doug Eastman 2011 年 1 月 28 日
The issue here is that the value is not exactly 0.0305, it gets truncated when displayed in the command window. You need to save the maximum value in a variable and then use that variable to find the indices:
myMax = max(x);
i = find(x==myMax);
  1 件のコメント
Jiro Doke
Jiro Doke 2011 年 1 月 28 日
I recommend this answer. By saving the actual max value, you will get the full precision. Using the
[C,I] = max(...)
syntax for a vector will only give you the index of the first occurrence, not all occurrences.


Oleg Komarov
Oleg Komarov 2011 年 1 月 28 日
Look at second output:
[C,I] = max(...)
  1 件のコメント
Michelle Hirsch
Michelle Hirsch 2011 年 1 月 28 日
I recommend this answer - it gets right to the desired solution, with no worries about numeric precision.


Todd Flanagan
Todd Flanagan 2011 年 1 月 28 日
You are bumping into issues with floating point accuracy. Loren has a nice blog post about accuracy with floating point numbers.
You can get the result directly from the MAX command using the form that returns both the numerical and index results:
[C, I] = max(x);
Note that with [C,I] = max(...) if there are several identical maximum values, the index of the first one found is returned.
If you want to use find or need to use find because you expect more than 1 occurrence of the maximum, you'll need to take some approach that either brackets the number you are looking for or programatically gives you the exact floating point representation you are looking for.
For example:
a = [1 2 3 .3501/10 4];
myMin = min(a)
find(a == myMin)
or for other sorts of problems you may want to bracket like:
find(a<.0351 & a>.0350)

Image Analyst
Image Analyst 2013 年 10 月 22 日
  2 件のコメント
Image Analyst
Image Analyst 2013 年 12 月 9 日
I see. Dharmesh has not accepted any of the suggested answers, so perhaps he does not consider it solved. He should accept one or ask a follow up question. If Sanjay has a question, he should start his own discussion in a separate thread.


omar kammouh
omar kammouh 2018 年 4 月 15 日
I had the same problem. I solved it by simply copying and then pasting the column (or the matrix) from which I am trying to find the index of a value.
  1 件のコメント
Image Analyst
Image Analyst 2018 年 4 月 16 日
The function you should use now to compare floating point numbers is called ismembertol().


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