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

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Mihai 2013 年 1 月 30 日
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!

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### 採用された回答

Shashank Prasanna 2013 年 1 月 30 日
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
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Shashank Prasanna 2013 年 1 月 30 日
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.

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### その他の回答 (4 件)

Cedric Wannaz 2013 年 1 月 30 日

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
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Satish Tadepalli 2016 年 11 月 11 日
awesome explanation @ Cedric Wannaz.

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Kasper Ornstein-Mecklenburg 2017 年 10 月 24 日

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
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rawand 2016 年 7 月 21 日

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));
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Stephen Cobeldick 2016 年 7 月 21 日
@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

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Robert Brandalik 2017 年 3 月 17 日
Also an Option:
unique([find(Temp-RefTemp == min(complex(Temp-RefTemp))),find(RefTemp-Temp == min(complex(RefTemp-Temp)))])

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