Finding the index value corresponding to a value closest to 0 in an array
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Mihai
il 30 Gen 2013
Modificato: Kasper Ornstein-Mecklenburg
il 24 Ott 2017
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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Shashank Prasanna
il 30 Gen 2013
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
2 Commenti
Shashank Prasanna
il 30 Gen 2013
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.
Più risposte (4)
Cedric
il 30 Gen 2013
Modificato: Cedric
il 30 Gen 2013
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
2 Commenti
Kasper Ornstein-Mecklenburg
il 24 Ott 2017
Modificato: Kasper Ornstein-Mecklenburg
il 24 Ott 2017
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
il 21 Lug 2016
Modificato: Stephen23
il 21 Lug 2016
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));
1 Commento
Stephen23
il 21 Lug 2016
Modificato: Stephen23
il 12 Nov 2016
@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
Robert Brandalik
il 17 Mar 2017
Also an Option:
unique([find(Temp-RefTemp == min(complex(Temp-RefTemp))),find(RefTemp-Temp == min(complex(RefTemp-Temp)))])
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