How to sum specific values from a loop and create a new vector?
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If I have this matrix M:
a=1:1:20;
b=[0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 2 0 0];
M=[a,b];
How could I find the i values where M(:,b)>0 (where the second column "b">0), and then take the values of the same row but the first column "a" M(a,:) and make a summation from that row i and i-2, i,e, [i + (i-1)+(i-2)].. like this:
X=[(5+6+7) (12+13+14) (18+17+16)] = [18 39 51]
Besides a vector Y which contains the summation from i-5 until i-3 like this:
Y=[(2+3+4) (9+10+11) (13+14+15)] = [9 30 42]
Both of them for all i values in a vector b of a matrix.
采纳的回答
Star Strider
2018-1-23
Use bsxfun and sum:
a=1:1:20;
b=[0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 2 0 0];
M=[a(:),b(:)];
idx = find(M(:,2) > 0);
X = sum(bsxfun(@plus, idx, -2:0),2)
Y = sum(bsxfun(@plus, idx, -5:-3),2)
X =
18
39
51
Y =
9
30
42
8 个评论
Philippe Corner
2018-1-23
Star Strider
2018-1-23
My pleasure!
The 2 is an argument to sum, and tells it to sum across columns. (The default behaviour is to sum across rows, or dimension 1.)
Philippe Corner
2018-1-23
编辑：Philippe Corner
2018-1-23
Star Strider
2018-1-23
Originally, ‘idx’ and ‘M(:,1)’ were the same. The code using ‘M(:,1)’ is:
X = sum(bsxfun(@plus, M(idx,1), -2:0),2)
Y = sum(bsxfun(@plus, M(idx,1), -5:-3),2)
giving the same result.
So this should work in your code:
f = sum(bsxfun(@plus, M(idx,1), -9:0),2)
My apologies for the confusion.
Philippe Corner
2018-1-23
Star Strider
2018-1-23
My pleasure.
I still do not understand the reason that the prototype code did not work with your actual data.
This works:
f = sum(arrayfun(@(x)sum(M(x,1)), bsxfun(@plus, idx', (-9:0)')));
I checked to be certain it produces the correct result with your data.
It works by calculating one column vector of indices into ‘M’ for each value of ‘idx’ and the offset of ‘(-9:0)’, calculated by the bsxfun call (that turned out to be correct after all), creating a matrix that here is (10x1881), the column size being the length of ‘idx’. It then uses the custom anonymous function ‘@(x)sum(M(x,1))’ to sum the first column of ‘M’ corresponding to the indices calculated by the bsxfun call. The arrayfun function contains an implicit loop (that I hope is more efficient than an explicit for loop). It then returns the result of the row sums for each column as a row vector.
Philippe Corner
2018-1-23
Star Strider
2018-1-23
Thank you!
If you want to have a moving sum for all the rows taken 10 at a time, use the movsum (link) function (introduced in R2016a).
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