MATLAB Answers

Union and Interseciton of Intervals

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Joris Lambrecht
Joris Lambrecht on 9 Jun 2011
Commented: Dan Bindman on 2 Oct 2020
I want to be able to quickly determine the interval resulting from unions and intersections of various intervals. The number of intervals may vary, so the method needs to be somewhat flexible.
In regular Matlab I have only found a way to intersect sets (e.g. {1,2,3} u {2,4}) not intervals (e.g. [0.1 3.3) n (1.1 2.9])
where 'u' indicates union and 'n' indicates intersection.
Example: ([1.1, 2.3] u [2.8 3.2]) n ([1.9, 3.1] u [4.2 4.3]) n [1.7, 3.2] = [1.9,2.3] u [2.8, 3.1]
In MuPad, you can do the following: (Dom::Interval([1.1, 2.3]) union Dom::Interval(2.8, 3.2)) intersect (Dom::Interval([1.9, 3.1]) union Dom::Interval(4.2, 4.3)) intersect (Dom::Interval([1.7, 3.2]))
Any ideas on how to do this without MuPad? I've seen the 'interval merging' file exchange entry but it only handles unions of intervals.
Thanks!

  3 Comments

Dan Bindman
Dan Bindman on 11 Apr 2020
Hi, I needed the union of intervals myself, so I wrote up a little code. It assumes that the intervals are closed to keep it simple.
function unionOfIntervalsMat = unionOfIntervals(intervalsMat)
%unionOfIntervalsMat = unionOfIntervals(intervalsMat) takes a set of
%(closed) intervals and returns the unions of those intervals.
%Each row of the input variable intervalsMat is an interval, with the first column
%giving the lower bounds for the intervals and the second column giving the upper bounds.
%Each row of the output variable unionOfIntervalsMat is also an interval, defined similarly.
intervalsMat=sortrows(intervalsMat,1);
nIntervals=size(intervalsMat,1);
unionOfIntervalsMat=nan(size(intervalsMat));
unionOfIntervalsMat(1,:)=intervalsMat(1,:);
unionCount=1;
for i=2:nIntervals
if intervalsMat(i,1) <= intervalsMat(i-1,2)
unionOfIntervalsMat(unionCount,2)=max(intervalsMat([i-1 i],2));
else
unionCount=unionCount+1;
unionOfIntervalsMat(unionCount,:)=intervalsMat(i,:);
end
end
unionOfIntervalsMat(isnan(unionOfIntervalsMat(:,1)),:)=[];
end
Here is an example:
>> intervalsMat=[4 10; 28 31;11 12; 1 7; 13 17; 16 23; 8 11; 25 29]
intervalsMat =
4 10
28 31
11 12
1 7
13 17
16 23
8 11
25 29
>>
>> unionOfIntervalsMat = unionOfIntervals(intervalsMat)
unionOfIntervalsMat =
1 12
13 23
25 31
Alex Bream
Alex Bream on 30 Jul 2020
Dan Bindman answer is wrong, e.g. unionOfIntervalsMat([1 10; 2 3; 5 6]) returns [1 10; 5 6] instead of [1 10]
Dan Bindman
Dan Bindman on 2 Oct 2020
Thanks Alex! Indeed I messed up a bit. In the for loop, I used i in some cases when I should have been using unionCount... anyhoo here is the new code:
function unionOfIntervalsMat = unionOfIntervals(intervalsMat)
%unionOfIntervalsMat = unionOfIntervals(intervalsMat) takes a set of
%(closed) intervals and returns the unions of those intervals.
%Each row of the input variable intervalsMat is an interval, with the first column
%giving the lower bounds for the intervals and the second column giving the upper bounds.
%Each row of the output variable unionOfIntervalsMat is also an interval, defined similarly.
intervalsMat=sortrows(intervalsMat,1);
nIntervals=size(intervalsMat,1);
unionOfIntervalsMat=zeros(size(intervalsMat));
unionOfIntervalsMat(1,:)=intervalsMat(1,:);
unionCount=1;
for i=2:nIntervals
if intervalsMat(i,1) <= unionOfIntervalsMat(unionCount,2)
unionOfIntervalsMat(unionCount,2)=max([intervalsMat(i,2) unionOfIntervalsMat(unionCount,2)]);
else
unionCount=unionCount+1;
unionOfIntervalsMat(unionCount,:)=intervalsMat(i,:);
end
end
unionOfIntervalsMat=unionOfIntervalsMat(1:unionCount,:);
end
I *think* it should now work in all situations. Here it is for your example:
>> unionOfIntervalsMat = unionOfIntervals([1 10; 2 3; 5 6])
unionOfIntervalsMat =
1 10
And here it is for the original example (still the same correct output):
intervalsMat =
4 10
28 31
11 12
1 7
13 17
16 23
8 11
25 29
>> unionOfIntervalsMat = unionOfIntervals(intervalsMat)
unionOfIntervalsMat =
1 12
13 23
25 31

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Answers (2)

Sean de Wolski
Sean de Wolski on 9 Jun 2011
doc intersect
doc union
and friends. Beware of floating points, however.
fpno = @(x)uint32(x*1000); %get rid of floating points
fpyes = @(x)double(x)/1000; %bring 'em back
fpyes(intersect(fpno([1.1 2.2]), fpno(1.2:.1:3.4)))
But I still don't understand how you're defining the interval as I can't get the same results as your example.

  2 Comments

Joris Lambrecht
Joris Lambrecht on 9 Jun 2011
intersect and union functions only operate on sets (not intervals). Like you say, beware floating points:
intersect([1.1 2.2], [1.2 3.4])
ans =
Empty matrix: 1-by-0
I want the result [1.2 2.2]
I can multiply my floating point numbers by a factor (determining precision, e.g. 1000 will result in 3 decimal place precsion), round them to make integers, construct sets from the integers, operate on these sets (using intersect and union), and then divide by my factor to get back to floating point. However this seems unnecessary and will generate large arrays to compare if I want high precision.
Sean de Wolski
Sean de Wolski on 9 Jun 2011
So what increment are you using to define an "interval"?

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Walter Roberson
Walter Roberson on 9 Jun 2011
Union and intersection of intervals can be done with min() and max() and appropriate logic; I remember implementing the logic about 6 years ago (probably in perl.) I did not, however, have to worry about open vs closed intervals: the representation and logic gets messier for those.

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