sum of consecutive months value in an array

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Poulomi Ganguli
Poulomi Ganguli am 11 Sep. 2018
Kommentiert: Poulomi Ganguli am 11 Sep. 2018
Hello: I have an array where the first, second and third columns show year, month and values. I need to sum the values of consecutive months. The matrix X is as below
1988 6 223.12
1994 7 1033.0
1994 8 464.67
1995 9 251.33
1997 6 121.41
1997 8 624.58
1999 9 221.60
2001 7 782.04
The desired output:
1988 223.12
1994 1497.67
1995 251.33
1997 121.41
1997 624.58
1999 221.60
2001 782.04
I have tried like this:
dt = datetime([X(:,1:2), repmat([0 0 0 0],length(X),1)]);
Monthgroup = cumsum([1; months(diff(datetime([X(:,1:2), repmat([0 0 0 0],length(X),1)]))) == 1]);
where same year with a different month, example, 1994 in this case, will suppose to show a common index but I am getting some error. Based on this common index, could I use accumarray function? Any help will be appreciated. Thanks,
  2 Kommentare
jonas
jonas am 11 Sep. 2018
I guess it needs to work for any number of consecutive months?
Poulomi Ganguli
Poulomi Ganguli am 11 Sep. 2018
yes but for the same year not for different years.

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Akzeptierte Antwort

Andrei Bobrov
Andrei Bobrov am 11 Sep. 2018
Bearbeitet: Andrei Bobrov am 11 Sep. 2018
a = [1988 6 223.12
1994 7 1033.0
1994 8 464.67
1995 9 251.33
1997 6 121.41
1997 8 624.58
1999 9 221.60
2001 7 782.04 ];
b = datetime(a(:,1),a(:,2),1);
lo = [true; b(1:end-1) + calmonths(1) ~= b(2:end)];
out = [a(lo,1), accumarray(cumsum(lo),a(:,3))];
or
b2 = a(:,1:2)*[12;1];
lo2 = [true;diff(b2) ~= 1];
out2 = [a(lo2,1), accumarray(cumsum(lo2),a(:,3))];
  6 Kommentare
Rik
Rik am 11 Sep. 2018
If you want to merge them, use either of the methods Andrei posted, if not, use the adapted code I posted in my comment. Which is the correct version for you depends on your application.
Poulomi Ganguli
Poulomi Ganguli am 11 Sep. 2018
Thanks! Now it's clear.

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