Retrieving Indices of a vector whose elements fulfil a criteria

[mikeph]

I have a 100x1 vector A, and want to know the ten indices i for which A(i) are the greatest 10 values in A.

I can find the values by:

B = sort(A)
B(1:10);

But I don't know how to find the indices, apart from starting a new if loop which searches the entire matrix (which seems very time consuming).

As another example, I have a matrix B which is 10,000x2 and I want to know the value of B(i,1) for which B(i,2) is the maximum value in the column vector B(:,2). Eg.

>> B = [(5:0.1:5.9)',rand(10,1)]
B =
   5.000000000000000   0.696266337082995
   5.100000000000000   0.093820026774866
   5.200000000000000   0.525404403859336
   5.300000000000000   0.530344218392863
   5.400000000000000   0.861139811393332
   5.500000000000000   0.484853333552102
   5.600000000000001   0.393456361215266
   5.700000000000000   0.671431139674026
   5.800000000000001   0.741257943454206
   5.900000000000000   0.520052467390387

>> for i = 1:10
if B(i,2) == max(B(:,2))
B(i,1)
end
end
ans =
   5.400000000000000

However I think this is taking absolutely ages in my code (I need to find this 20bn times so it's definitely an area for optimisation). Is there a simpler way to find the indices?

 

[D H]

It looks like you are using Matlab. Matlab's sort function can return two values, the sorted array and an array of permutation indices.

[B,idx] = sort (A);

 

[mikeph]

I didn't realize sort had that second output! I've been using it for ages as well.

Thanks for the fast response!

 

[mikeph]

Following on from this, is there any way to quickly find a specific row in a vector?

For example, searching for the half-width half-maximum of a Gaussian

>> x = -20:0.0001:20;
>> f = 4*exp(-x.^2/3);
>> time = cputime; for i = 200000:400001
if abs(f(i) - max(f)/2) < 0.0001
x(i)
break
end
end; cputime-time
ans =
    1.4420
ans =
   45.2031

ie. I can find the value but it takes a long time. 45 seconds is ordinarily fine but I need to perform this many, many times.

Is there no "search" function?

BTW. In the example I realize that there is probably no i such that f(i) = 2 exactly. If there is no answer to this question, would there be an answer if there did exist an i where f(i) = 2?

 

[D H]

In the above example, you are calling max(f) 200,001 times, each invocation of which requires searching over a list of 400,001 elements. Of course that is slow. Cache the value of max(f) outside the loop and you will see a dramatic speed bump.

 

[matonski]

What about instead of the loop, doing something like this (using your above definitions of f and x):

[junk, halfwidthIndex] = min(abs(f - max(f)/2));
x(halfwidthIndex)