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Is there a way to obtain desired index without using 'find'?

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Dmytro
Dmytro on 1 Oct 2012
Edited: Walter Roberson on 12 Nov 2017
Dear all,
Suppose there is some array
ar=[102 243 453 768 897 ...]
Is there any way to obtain indices of an element having value e.g. 243 without using find?
For each element I need to find its index, so the computational cost will be O(N^2), which is too much. One possibility is using sparse matrices, e.g.
smatr=sparse(ar,1,1:length(ar))
and then index can be retrieved simply as ind=smatr(243) and so on, reducing computational cost to O(N). However, the problem is that in my computations values of 'ar' might exceed maximum size of matrix. So is there any additional way to relate values of 'ar' to indices so the latter can be obtained in one operation?
Thanks,
Dima
  9 Comments

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Accepted Answer

Matt J
Matt J on 2 Oct 2012
Edited: Matt J on 2 Oct 2012
Here's another approach that uses the find_idx function from the FEX
It seems to overcome the O(N) overhead of HISTC and is virtually independent of N in speed,
ar=[102 243 453 768 897 102];
[sar,i,j]=unique(ar);
ind=j(floor(find_idx(768,sar))), %search for 768 for example

More Answers (7)

Matt Fig
Matt Fig on 1 Oct 2012
ar=[102 243 453 768 897 243 653 23];
KEY = 1:length(ar);
KEY(ar==243)
  2 Comments
Matt Fig
Matt Fig on 1 Oct 2012
This is the quickest way to do it in MATLAB, that I know of anyway! Not, of course, counting a customized MEX function.

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Matt J
Matt J on 1 Oct 2012
Edited: Matt J on 1 Oct 2012
Here's a modification of the sparse matrix approach that might ease the difficulty with maximum matrix sizes limits,
armax=max(ar);
armin=min(ar);
arc=ar-armin+1;
arcmax=armax-armin+1;
N=ceil(sqrt(arcmax));
[I,J]=ind2sub([N,N], arc);
smatr=sparse(I,J,1:length(arc),N,N);
Now, whenever you need to look up an index, e.g. 243, you would do
[i,j]=ind2sub([N,N],243-armin+1);
ind=smatr(i,j);
The difference is that now the matrix dimensions of smatr are roughly the square root of what they were in your approach. Even less, actually, since we offset the ar values by armin in its construction.
  2 Comments
Walter Roberson
Walter Roberson on 1 Oct 2012
2^96 cannot be handled as array indexes for MATLAB. The limit is 2^48 if I recall correctly. 2^96 exceeds 64 bit representation.

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Matt J
Matt J on 1 Oct 2012
Edited: Matt J on 1 Oct 2012
Here is an implementation of your O(log_2(N)) idea using HISTC
ar=[102 243 453 768 897 102];
[sar,i,j]=unique(ar);
[~,bin]=histc(768,sar); %search for 768 for example
ind=j(bin);
  5 Comments
Matt J
Matt J on 2 Oct 2012
OK, through the Newsgroup, I think it's been identified why HISTC is O(N). It's because it does a pre-check the monotonicity of the RM values in
[~,bb]=histc(nn,RM);
The search itself is O(log(N)). See also

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Walter Roberson
Walter Roberson on 1 Oct 2012
If each element is present only once, consider using the three-output form of unique()
  5 Comments
Matt J
Matt J on 2 Oct 2012
Sure. I only meant that the output is the same and that unique() doesn't really accomplish anything here, seeing as how the elements are already unique.

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dipanka tanu sarmah
dipanka tanu sarmah on 19 Oct 2017
Edited: Walter Roberson on 19 Oct 2017
you can use the following function :
function posX = findPosition(x,y)
posX=[];
for i =1:length(x)
p=x-y;
isequal(p(i),0)
if ans==1
posX=[posX,i]
end
end
  2 Comments
dipanka tanu sarmah
dipanka tanu sarmah on 19 Oct 2017
yeah. u are awesome. I didnt go thriugh the optimization part. thnk you

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Walter Roberson
Walter Roberson on 19 Oct 2017
myMap = containers.Map( ar, 1:length(ar) )
This uses a hash structure. I am not sure which one it uses, so I do not know the creation time costs; certainly no less than O(N) as the 1:length(ar) is O(N). O(N*log(N)) I suspect.
Once hashed, each lookup is nominally constant time (I do not know if the hash structure it uses can have collisions that might need to be resolved.)

Christian Troiani
Christian Troiani on 12 Nov 2017
Edited: Walter Roberson on 12 Nov 2017
for k = 1:length(ar)
if ar(k)==243 % Using your example of the 243 element
posX=k;
break
else
continue
end
end

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