How can I add components of a vector given conditions from another vector of different dimensions?

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Mariana Kneppers
Mariana Kneppers 2016-7-4
评论： Mariana Kneppers 2016-7-5
I have the following columns of data:
buzz length start cue end cue
803 0.7 7 392.6
810.9 0.9 502.4 1002.4
1514.8 0.4 1231.6 1851.4
1516.2 0.9 2226 2782.2
1753 0.8 4024.2 4520
2470.6 2.9 5050.2 5536.2
4278.1 1.3 6059.2 6699.8
4415.9 2.9
5361.9 1.1
6244.1 5.7
6446.6 1.2
6459 2.7
I am trying to create a code that, like the SUMIFS function in Excel, will take the number of buzzes who's time (column 1) falls within the range of a start/end cue pair (columns 3 and 4) and sum their respective lengths (column 2). I'm sure the solution has something to do with indexing to refer to the index positions of some of the values, however I am not sure how to even approach this problem. Any help is appreciated! Thank you for your time!

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Stephen23
Stephen23 2016-7-4
编辑：Stephen23 2016-7-4
This is fast and easy to solve by using a simple mtimes.
Your data (as column vectors):
>> cueBeg = [7;502.4;1231.6;2226;4024.2;5050.2;6059.2];
>> cueEnd = [392.6;1002.4;1851.4;2782.2;4520;5536.2;6699.8];
>> datLen = [0.7;0.9;0.4;0.9;0.8;2.9;1.3;2.9;1.1;5.7;1.2;2.7];
>> datBuz = [803;810.9;1514.8;1516.2;1753;2470.6;4278.1;4415.9;5361.9;6244.1;6446.6;6459];
and the calculation:
>> datLen.' * (bsxfun(@gt,datBuz,cueBeg.')&bsxfun(@lt,datBuz,cueEnd.'))
ans =
0.00000 1.60000 2.10000 2.90000 4.20000 1.10000 9.60000
The output vector gives the sum of all "lengths" between each pair of cues. You might prefer to use ge and le instead of gt and lt.
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Stephen23
Stephen23 2016-7-5
编辑：Stephen23 2016-7-5
@Mariana Kneppers: Of course:
>> cueIdx = bsxfun(@gt,datBuz,cueBeg.') & bsxfun(@lt,datBuz,cueEnd.');
>> datLen.' * cueIdx % the total time length in each interval
ans =
0 1.6 2.1 2.9 4.2 1.1 9.6
>> sum(cueIdx,1) % the number of buzzes in each interval
ans =
0 2 3 1 2 1 3

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José-Luis
José-Luis 2016-7-4
编辑：José-Luis 2016-7-5
Shamelessly plagiarizing Stephen's data reconstruction, but providing a one liner instead:
cueBeg = [7;502.4;1231.6;2226;4024.2;5050.2;6059.2];
cueEnd = [392.6;1002.4;1851.4;2782.2;4520;5536.2;6699.8];
datLen = [0.7;0.9;0.4;0.9;0.8;2.9;1.3;2.9;1.1;5.7;1.2;2.7];
datBuz = [803;810.9;1514.8;1516.2;1753;2470.6;4278.1;4415.9;5361.9;6244.1;6446.6;6459];
result = arrayfun(@(x,y) sum(datLen(datBuz>x & datBuz < y)),cueBeg,cueEnd)
EDIT
and to get the count:
counts = arrayfun(@(x,y) sum(datBuz>x & datBuz < y),cueBeg,cueEnd)

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