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How many times does each number appear?

Asked by Hamad on 1 Apr 2013

Let a

 a = 
     1 4
     3 5
     4 7
     6 8

be integer intervals

    [1 2 3 4]
    [3 4 5]
    [4 5 6 7]
    [6 7 8]

I'm looking for an output b to count the number of times each element appears. It is clear that 1 appears one time, 2 appears one time, 3 appears twice, and so on...

So the output b would list the element and how many times it appears:

 b = 
     1 1
     2 1
     3 2
     4 3
     5 2
     6 2
     7 2
     8 1

The challenge is, a is an extremely large nx2 array, so using find even with parallel looping is inefficient. Is there a fast way to yield b given a? If it helps, we can be sure that the elements down the left-hand column of a are sorted, but not necessarily so for the right-hand column. Thanks in advance!


Jan Simon on 2 Apr 2013

Some contributors spend a lot of time here to solve the problem. What about voting for this question to get the interst of others?

Hamad on 2 Apr 2013

First thing to say is that I'm a little overwhelmed by the breadth of effort I have received with this coding challenge. For that reason, I wish to thank each contributor individually.

Firstly, Image Analyst, thank you for a very speedy initial response. And referring to "why do you need this?"... there are ~1e8 milliseconds in a 24-hour period. Highly-traded assets such as index futures contracts yield price updates every few milliseconds during the day and every few minutes at night, averaging around 10,000 observations per day. The product of observations for a pair of these is therefore also ~1e8.

Brian B, you deserve thanks for offering creative contributions based on filters, and for following up with others' contributions.

Jan Simon, you deserve thanks for offering a wide range of efficient solutions along with performance metrics.

Last but by no means least, Cedric Wannaz deserves thanks for offering very detailed intuitive explanations to his contributions. You make impenetrable code seem easy.


I have tested all solutions to the same standard, measured by efficiency (minor coding modifications excepted). The most efficient algorithm which solves the problem on my system setup and to my particular needs is provided by Teja Muppirala, and this is followed closely by the FOR-loop provided by Jan Simon. Thank you!!

For this reason, I gratefully accept Teja Muppirala's contribution. At the same time, I offer +1 votes to every contribution here. You have helped me greatly. My sincerest thanks to you all.

Best wishes,




No products are associated with this question.

7 Answers

Answer by Teja Muppirala on 2 Apr 2013
Accepted answer

Assuming that the second column of a is always >= the first column of a, then this is an efficient solution.

 mA = max(a(:));
 b = (1:mA)';
 b = [b cumsum(accumarray(a(:,1),1,[mA 1]) + ...
               accumarray(min(a(:,2)+1,mA),-1,[mA 1]))];
 b(end) = b(end) +1;

1 Comment

Cedric Wannaz on 2 Apr 2013

+10! Just needs a small correction on the last element.

Teja Muppirala
Answer by Brian B on 2 Apr 2013
Edited by Brian B on 2 Apr 2013

Like Cedric, I keep thinking of iterating over interval length. Another way to look at the vector b is as the sum of the outputs of a bunch of FIR filters with rectangular impulse responses to appropriate inputs. If you cut the filter length in half at each iteration, you can avoid iterating over all n interval lengths, and instead loop about log2(n) times.

Here I generate intervals of up to 1000, so the loop runs about 10 times.

a=[randi(M,N,1) randi(1e3,N,1)]*[1 1; 0 1];
idx = 1:max(a(:));
d = diff(a, 1, 2) + 1 ; % interval length
s = a(:,1); % interval start point
L = ceil(max(d)/2);
b = 0*idx;
while L>0
    iL = (d>=L);
    u = hist(s(iL),idx);
    %b = b + conv(u, ones(1,L),'same');
    b = b + filter(ones(1,L),1,u);
    s(iL) = s(iL)+L;
    d(iL) = d(iL)-L;    
    L = ceil(max(d)/2);

This ran in 243 seconds on my machine; Cedric's solution exhausted my memory after 577 seconds....

EDIT: Replaced the call to conv() with a call to filter(), since conv() was returning the central part of the convolution, not the initial part, as we need.

1 Comment

Brian B
Answer by Jan Simon on 2 Apr 2013
Edited by Jan Simon on 2 Apr 2013

Are the values in the first and second column of a distinct? If so:

c = zeros(1, max(a(:, 2)) + 1);
c(a(:, 1)) = 1;
a2     = a(:, 2) + 1;  % [EDITED: +1 inserted]
c(a2)  = c(a2) - 1; 
c      = cumsum(c);
c(end) = [];

If the values are not unique, instead of 1 and -1 histc determines the values:

q = 1:max(a(:, 2));
c = histc(a(:, 1), q);
c = c - histc(a(:, 2) + 1, q);  % [EDITED: +1 inserted]
c = cumsum(c);

How efficient is simple FOR loop?

n = max(a(:, 2));
c = zeros(1, n + 1);
for ia = 1:length(a)
  a1    = a(ia, 1);
  c(a1) = c(a1) + 1;
  a2    = a(ia, 2) + 1;  % [EDITED: +1 inserted]
  c(a2) = c(a2) - 1;
c(end) = [];
c      = cumsum(c);

Converting this to C would be interesting also. Most of all the limitation of CUMSUM to the type double would not matter, such that UINT16 might reduce the data size, if sufficient.


Jan Simon on 2 Apr 2013

Interesting timings:

N = 1e7; M = 1e6;
a = [randi(M,N,1) randi(1e3,N,1)]*[1 1; 0 1];
FOR-loop:   1.90 seconds
HISTC:      9.64 seconds (see also Brian B's HISTC solution)
CUMSUM:     1.71 seconds (columns of [a] must be unique!)
CONV:      41.00 seconds (Brian B's loop with HIS and CONV)

The relation my change for the real data. But for the test data, the simple FOR loop can compete with the vectorized methods (R2009a/64, Win7).

Hamad on 2 Apr 2013

My rankings concur in general with yours. Teja's method is more efficient, but a for-loop is more flexible. This was a very tough decision. Please see my final comment to the question above. Apologies for the delay in getting back to you and everyone here. My sincerest thanks for your help. Best wishes, Hamad.

Jan Simon
Answer by Image Analyst on 1 Apr 2013
Edited by Image Analyst on 1 Apr 2013

Are they known to be integers? I'd use histc(). You could do this in about 4 lines of code -- 1 (or maybe a 3 line for loop) to construct a 1D list of all numbers you gave in the seconds array you showed, 1 to construct the hist bin edges, one to call histc() and one to construct "b" from the result of histc(). Give it a try and come back if you have trouble.

1 Comment

Hamad on 2 Apr 2013

Image Analyst, many thanks for your answer. Please see my comment to Cedric Wannaz's answer below, which also pertains to your answer. Many thanks! Hamad.

Image Analyst
Answer by Cedric Wannaz on 1 Apr 2013
Edited by Cedric Wannaz on 1 Apr 2013

Hi Hamad,

Well, how large is "extremely large"? Image Analyst might have given you the optimal solution already. As an alternative, you could go for a solution involving ACCUMARRAY or SPARSE, where you would use these integers as indices and a vector of ones as values. This way you get counts per index/integer. For example:

 >> buffer = arrayfun(@(r) a(r,1):a(r,2), 1:size(a,1), 'UniformOutput', false) ;
 >> buffer = [buffer{:}].' ;
 >> counts = accumarray(buffer, ones(numel(buffer),1))
 counts =

There is still room for improvement though (in particular, I guess that a PARFOR loop would be more efficient than ARRAYFUN for building the vector of all integers, but I kept the latter because my point was to illustrate ACCUMARRAY). If not efficient enough after you improve it, it is probably something that you could optimize writing a few lines of C/MEX, in particular for building the vector of all integers.


Hamad on 2 Apr 2013

Many thanks, Cedric. You are a wizard.

Cedric Wannaz on 3 Apr 2013

Well, not sure about that, but you're welcome! The thread started with a homework tag and ended up being one of the most animated that I've seen a while .. I'd bet that your next question won't be mistaken for a HW ;-)



Hamad on 3 Apr 2013

With careful descriptions including words like {let, suppose, given, solution}... I bet it will be mistaken again! Look out for my next soon! Take care.

Cedric Wannaz
Answer by Brian B on 2 Apr 2013
Edited by Brian B on 2 Apr 2013

Jan Simon's solution is interesting because cumsum can be though of as an IIR filter, and so this idea can further improve the FIR approach above, without looping:

a=[randi(M,N,1) randi(1e3,N,1)]*[1 1; 0 1];
idx = 1:max(a(:));
b = hist(a(:,1),idx);
b = b - hist(a(:,2)+1,idx);
b = cumsum(b);

This works even if the values are not distinct.


Jan Simon on 2 Apr 2013

I've posted a very similar solution, but without the "+1" in the 2nd histc() call - I've fixed my bug now.

Brian B on 2 Apr 2013

Yes, I posted this before you edited yours to handle the non-distinct case! I gave you a vote, since you obviously had the key idea.

Brian B
Answer by Brian B on 2 Apr 2013
Edited by Brian B on 2 Apr 2013

You can avoid saving all of the intermediate intervals (e.g., [1 2 3 4]) if you just want the count:

 a = [1 4;3 5;4 7;6 8];
 b = zeros(max(max(a)),1);
 parfor k=1:length(b)
    b(k) = sum( a(:,1)<=k & k<=a(:,2) );


Brian B on 2 Apr 2013

Or, if the intervals tend to be small compared to max(max(a)), you could iterate over interval length:

 a = [1 4;3 5;4 7;6 8];
 idx = 1:max(max(a));
 while ~isempty(a)
    b = b + hist(a(:,1),idx);
    % Then adjust the a matrix by increasing the lower bound
    % by one and removing any rows of a which are no longer valid intervals.

I'll let you work out the details, since I just noticed this is tagged as homework.

Hamad on 2 Apr 2013

Elegant solution but the data overhead costs more than the parallelism saves, due to array of logicals of size a being calculated over each k. Thanks very much for your answer!

Hamad on 2 Apr 2013

lol, why does every one of my posts get tagged as homework??

Many thanks for your help, Brian. Best wishes, Hamad.

Brian B

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