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

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Dmytro
Dmytro 2012-10-1
编辑： Walter Roberson 2017-11-12
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
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Matt J
Matt J 2012-10-2
编辑：Matt J 2012-10-2
This routine greatly outperforms all other (in my calculations it was always almost independent of N, as well as faster, while all others linearly grow with N). So, I think, the problem is sorted. Thank you all for your answers, I am really glad I received help here and I greatly appreciate it!
Maybe, but just so you know, there are mex-driven versions of this available on the FEX
http://www.mathworks.com/matlabcentral/fileexchange/23049-findidx
See also my 3rd Answer below for how to incorporate it.

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Matt J
Matt J 2012-10-2
编辑：Matt J 2012-10-2
Here's another approach that uses the find_idx function from the FEX
http://www.mathworks.com/matlabcentral/fileexchange/23049-findidx
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

更多回答（7 个）

Matt Fig
Matt Fig 2012-10-1
ar=[102 243 453 768 897 243 653 23];
KEY = 1:length(ar);
KEY(ar==243)
  2 个评论
Walter Roberson
Walter Roberson 2012-10-1
Each of the == operations in your code will take N comparisons. Dmytro also wants to do this for each of the N elements, for a total of N*N = O(N^2) operations. Dmytro hopes to do it more efficiently.
Matt Fig
Matt Fig 2012-10-1
编辑：Matt Fig 2012-10-1
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 2012-10-1
编辑：Matt J 2012-10-1
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.
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Dmytro
Dmytro 2012-10-1
Thank you, that's a good idea! Unfortunately, values might be larger than 2^96 (square of maximum array size) for large time-series. So for now probably the best approach will be to write and use recursive O(log N)-cost search at each step, as I wrote in the comment under the question.
Walter Roberson
Walter Roberson 2012-10-1
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 2012-10-1
编辑：Matt J 2012-10-1
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);
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Dima
Dima 2012-10-2
Yes, I am sure, the code is
css=zeros(500,1); NN=1000*(1:500);
for cn=1:500
RM=unique(randi(NN,NN,1));
aaa=cputime;
for kn=1:10000, nn=randi(NN); [~,bb]=histc(nn,RM); end
css(cn)=cputime-aaa;
end
plot(NN,css,'o');
Matt J
Matt J 2012-10-2
编辑：Matt J 2012-10-2
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
http://www.mathworks.com/matlabcentral/newsreader/view_thread/323511

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Walter Roberson
Walter Roberson 2012-10-1
If each element is present only once, consider using the three-output form of unique()
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Matt Fig
Matt Fig 2012-10-1
I think it is not equivalent, because UNIQUE sorts first for ar of any real length (>1000). Time-wise this all adds up to more of it!
Matt J
Matt J 2012-10-2
编辑：Matt J 2012-10-2
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 2017-10-19
编辑：Walter Roberson 2017-10-19
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 个评论
Walter Roberson
Walter Roberson 2017-10-19
That would print out the result of each isequal() . Better would be to use
function posX = findPosition(x,y)
posX=[];
for i = 1:length(x)
p = x - y;
if isequal(p(i),0)
posX=[posX,i]
end
end
Which optimizes to
function posX = findPosition(x,y)
posX=[];
p = x - y;
for i = 1:length(x)
if isequal(p(i),0)
posX=[posX,i]
end
end
But then that could be simplified further as
function posX = findPosition(x,y)
posX=[];
for i = 1:length(x)
if x(i) == y
posX=[posX,i]
end
end
However, this requires continually expanding posX, which could be expensive. Less expensive would be:
function posX = findPosition(x,y)
posX = 1 : length(x);
posX = posX( x == y );
On the other hand, this is an order N calculation anyhow, and since the indices have to be found for each entry in the array, using this repeatedly would still end up being O(N^2) which the poster was trying to avoid.
dipanka tanu sarmah
dipanka tanu sarmah 2017-10-19
yeah. u are awesome. I didnt go thriugh the optimization part. thnk you

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Walter Roberson
Walter Roberson 2017-10-19
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 2017-11-12
编辑：Walter Roberson 2017-11-12
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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