Finding the index value corresponding to a value closest to 0 in an array

Hi,

I have an array with x amount of values. How can I find the index value of the element that is closest or equal to a certain value?

I tried it in the following manner, but it doesn't work when the value of the element in Temp is equal to the RefTemp value.

Temp = [-15.3, 0.2, 15.2, 30, 45.3];
RefTemp = 30; %Value to compare the Temp array values to 

for ii = 1:length(Temp)
         TempCalc(ii) = abs(Temp(ii) - RefTemp);
end

find(min(TempCalc));

Thank you very much for your help!

Answer 1

[Shashank Prasanna]

Do you have the Stats toolbox? if you do then do a nearest neighbor search as follows:

location = knnsearch(Temp',30);

If you don't have stats toolbox then use delauny to do nn search:

>> tri = delaunayn(Temp');
>> dsearchn(Temp',tri,30)

ans =

       4

[Mihai]

Interesting! I don't have the stats toolbox, and I've never seen either of those 2 functions before. It seems simple enough. I briefly tried playing around with the delaunayn function, and it seems it wouldn't work if 2 elements in the array were equal. This isn't a problem for me since the elements in my Temp array will be unique.

[Shashank Prasanna]

Nearest neighbor computation comes up in engineering and science more often then you can imagine, from pdf estimation to clustering. There are advanced data structures such as kdtrees which speed up neighbor search for higher dimension data. Also knnsearch has options on what to do during a tie.

Answer 2

[Cedric]

You have several options. The first question is: do you really need the index, or could you use a vector of logicals, e.g. for indexing something else. Look at the following; we want to extract all volumes associated with temperatures that are closest to a ref value:

 >> temp = [-15.3, 0.2, 15.2, 30, 45.3];
 >> volume = [4, 7, 28, 35, 20] ;
 >> ref = 27.2 ;
 >> dif = abs(temp-ref)
 dif =
   42.5000   27.0000   12.0000    2.8000   18.1000
 >> min(dif)
 ans =
    2.8000
 >> match = dif == min(dif)
 match =
     0     0     0     1     0                  % Vector of logicals indicate
                                                % where dif equals its min.
 >> idx = find(dif == min(dif))
 idx =
     4                                          % Index of element that 
                                                % the min matches

Now you can extract the corresponding volume with either the vector of logicals (which avoids using find()) or the index.

 >> volume(match)
 ans =
    35
 >> volume(idx)
 ans =
    35

This is one "vector" way to achieve what you want; without all the extra steps, this reduces to:

 >> dif = abs(temp-ref) ;
 >> volume(dif == min(dif))
 ans =
    35

[Mihai]

Thanks! I think the first part of your answer is along the same route I was going for and does what I need. I have to find the index value to use it with other arrays that correspond to the Temp array.

[Satish Tadepalli]

awesome explanation @ Cedric Wannaz.

Answer 3

[Kasper Ornstein-Mecklenburg]

The min function returns index as second output.

>> a = [-15.3, 0.2, 15.2, 30, 45.3]; %Define vector to analyze
>> aRef = 30; %Set reference
>> aDiff = abs(a - aRef); %Calculate diff
>> [minVal, minInd] = min(aDiff); %Find value closest to aRef
>> a(minInd)

ans =

      30

Answer 4

[rawand]

simply:

Temp = [-15.3, 0.2, 15.2, 30, 45.3];
RefTemp = 30; %Value to compare the Temp array values to 
for ii = 1:length(Temp)
         TempCalc(ii) = abs(Temp(ii) - RefTemp);
end
find(TempCalc == min(TempCalc));

[Stephen23]

@rawand: there is absolutely no point in wasting time writing slow and ugly loops as if this were some low-level language like C. MATLAB is a high-level language, and vectorized code is faster and neater:

>> TempCalc = abs(Temp-RefTemp);
>> find(TempCalc == min(TempCalc))
ans =
     4

Answer 5

[Robert Brandalik]

Also an Option:

unique([find(Temp-RefTemp == min(complex(Temp-RefTemp))),find(RefTemp-Temp == min(complex(RefTemp-Temp)))])