Overlapping time-intervals

31 Ansichten (letzte 30 Tage)
Rostislav Teryaev
Rostislav Teryaev am 12 Nov. 2017
Bearbeitet: Ted Shultz am 24 Aug. 2020
I have two date-time arrays a and b.
a for start time and b for end time (above horizontal line on the picture).
What is the way to parse them and get what is beneath horizontal line on the picture?

Akzeptierte Antwort

David Goodmanson
David Goodmanson am 14 Nov. 2017
Bearbeitet: David Goodmanson am 14 Nov. 2017
Hi Rostislav
It looks like you want the union of closed intervals. Here is some code that I think does the job. When the ends of intervals have the same value, it works because the sort function is stable and preserves ordering for ties.
% Nx2 matrix of endpoints x1, x2 of intervals
x = [8 10; 2 4; 4 5; 5 7; 9 11; 15 16; 14 17; 10 12]
% -----plot it first
nline = repmat((1:size(x,1))',1,2);
plot(x',nline','o-')
ylim([-.5*nrow 1.5*nrow])
% ----- find union of intervals
x = sort(x,2);
nrow = size(x,1);
[x ind] = sort(x(:));
n = [(1:nrow) (1:nrow)]';
n = n(ind);
c = [ones(1,nrow) -ones(1,nrow)]';
c = c(ind);
csc = cumsum(c); % =0 at upper end of new interval(s)
irit = find(csc==0);
ilef = [1; irit+1];
ilef(end) = []; % no new interval starting at the very end
% y matrix is start and end points of the new intervals, y1,y2
%
% ny matrix is the corresponding indices of the start and end points
% in terms of what row of x they occurred in.
y = [x(ilef) x(irit)]
ny = [n(ilef) n(irit)]
  2 Kommentare
Rostislav Teryaev
Rostislav Teryaev am 14 Nov. 2017
Bearbeitet: Rostislav Teryaev am 14 Nov. 2017
Thank you for your reply. Works like a charm! I did my one with loops, but your many times faster.
function [ a,b ] = intervalsSort( a, b )
tf = 1;
f = 1;
[a, ind] = sort(a);
b = b(ind);
for i = 1:length(a)
if i == 1
c(1) = a(1);
d(1) = b(1);
else
f = 1;
for j = 1:length(c)
if c(j)<=a(i) & a(i)<=d(j)
if d(j) < b(i)
d(j) = b(i);
end
f = 0;
end
end
if f
c = [c a(i)];
d = [d b(i)];
end
end
end
a = c;
b = d;
end
Ted Shultz
Ted Shultz am 21 Aug. 2020
Bearbeitet: Ted Shultz am 24 Aug. 2020
This is really great code. I fixed a small type, and added comments so others can learn from it as well.
intervalsIn = [8 10; 2 4; 4 5; 5 7; 9 11; 15 16; 14 17; 10 12]
% -----plot it first
nline = repmat((1:size(intervalsIn,1))',1,2);
nrow = size(intervalsIn,1);
plot(intervalsIn',nline','o-')
ylim([-.5*nrow 1.5*nrow])
% ----- find union of intervals
intervalsIn = sort(intervalsIn,2); % make the pairs always increasing pairs
[intervalsIn, ind] = sort(intervalsIn(:)); % sorts all the input values, and keepts track of how they moved around
c = [ones(1,nrow) -ones(1,nrow)]'; % this is a matrix that keeps track of when intervals start (1) and stop (-1)
c = c(ind); % put in order of occurrence
csc = cumsum(c); %sum up starts (1) and stops (-1) , will be =0 at upper end of new interval(s)
irit = find(csc==0); % find index locations of 0 (ends of intervals)
ilef = [1; irit+1]; % start of intervals index is at the very start (1) and one after all the other ends
ilef(end) = []; % no new interval starting at the very end
% spansOut matrix is start and end points of the new intervals, y1,y2
spansOut = [intervalsIn(ilef) intervalsIn(irit)]

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Hung Doan
Hung Doan am 19 Jul. 2019
nrow is undefined in the initial plotting but other than that, works like a charm!
  2 Kommentare
Rostislav Teryaev
Rostislav Teryaev am 23 Jul. 2019
I want to mention that poposed solution is very nice, but I remember that it has some kind of a bug. I can not point in which case, but I faced it in my project which I was doing that times. My solutions is slower but did not gave an error.
You can use any solution you want, but consider what I said.
Ted Shultz
Ted Shultz am 21 Aug. 2020
Bearbeitet: Ted Shultz am 24 Aug. 2020
Small typo in original answer, code could be be:
intervalsIn = [8 10; 2 4; 4 5; 5 7; 9 11; 15 16; 14 17; 10 12]
% -----plot it first
nline = repmat((1:size(intervalsIn,1))',1,2);
nrow = size(intervalsIn,1);
plot(intervalsIn',nline','o-')
ylim([-.5*nrow 1.5*nrow])
% ----- find union of intervals
intervalsIn = sort(intervalsIn,2); % make the pairs always increasing pairs
[intervalsIn, ind] = sort(intervalsIn(:)); % sorts all the input values, and keepts track of how they moved around
c = [ones(1,nrow) -ones(1,nrow)]'; % this is a matrix that keeps track of when intervals start (1) and stop (-1)
c = c(ind); % put in order of occurrence
csc = cumsum(c); %sum up starts (1) and stops (-1) , will be =0 at upper end of new interval(s)
irit = find(csc==0); % find index locations of 0 (ends of intervals)
ilef = [1; irit+1]; % start of intervals index is at the very start (1) and one after all the other ends
ilef(end) = []; % no new interval starting at the very end
% spansOut matrix is start and end points of the new intervals, y1,y2
spansOut = [intervalsIn(ilef) intervalsIn(irit)]

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by