adding counts of ordered pairs
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I have a sequence of
by 3 arrays, say
,
, …,
that are generated within a loop. That is, the number of rows of each array varies, but each array has three columns. The first two columns represent ordered pairs. The third column is the count of those ordered pairs. I want to "add" these together so that I get an array that accumulates the counts of these ordered pairs into a new array that is
with the same structure. For example, if
is
by 3 arrays, say
,
, …,
isA1=[1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
and
is
isA1=[1 5 1;
13 5 1;
13 7 3;
14 1 5];
then the cumulative matrix should be
A=[1 4 3;
1 5 1;
3 5 1;
12 4 7;
13 5 3;
13 7 3;
14 1 6];
The arrays being "summed" in this way have hundreds of entries and are themselves summaries of arrays with thousands of entries, so efficiency matters.
採用された回答
その他の回答 (1 件)
A1=[1 4 3;
3 5 1;
12 4 7;
13 5 2;
14 1 1];
A2=[1 5 1;
13 5 1;
13 7 3;
14 1 5];
A12=[A1; A2];
[A12u,~,J]=unique(A12(:,1:2),'rows','stable');
A=[A12u,accumarray(J,A12(:,3))]
7 件のコメント
Barbara Margolius
2022 年 9 月 6 日
編集済み: Barbara Margolius
2022 年 9 月 6 日
Bruno Luong
2022 年 9 月 6 日
編集済み: Bruno Luong
2022 年 9 月 6 日
I hope you don't have such case where Bjorn's method fails
A1=[1 1 1];
A2=[1 1 -1];
or pairs are not integer >= 1.
Barbara Margolius
2022 年 9 月 6 日
編集済み: Barbara Margolius
2022 年 9 月 6 日
Bruno Luong
2022 年 9 月 6 日
floor can returns 0 no?
Barbara Margolius
2022 年 9 月 6 日
Bjorn Gustavsson
2022 年 9 月 6 日
@Bruno Luong: The amount of subconsious/implicit assumptions I make when writing QD-solutions is a bit frightening.
Bruno Luong
2022 年 9 月 6 日
That's called "intuition".
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