count number of values greater than threshold.

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Jay Hanuman
Jay Hanuman 2016-10-28
编辑： dpb 2016-10-29
I have time as x axis and values as y axis. I want to count no. of values greater than 30 for x axis between 1 and 60 and then between 61 & 100. count between 1 & 60 ,between 61 & 100 should be stored separately.
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James Tursa
James Tursa 2016-10-28
Do you mean for x indexes between 1 and 60 (i.e. x(1:60)), or do you mean for x values between 1 and 60 (i.e. x>=1 & x <=60)?

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回答（2 个）

James Tursa
James Tursa 2016-10-28
编辑：James Tursa 2016-10-28
From your description:
x = your x-axis values
y = your y-axis values (same size as x)
threshold = 30;
g = x>=1 & x <=60;
count1 = sum(y(g) > threshold);
g = x>=61 & x <=100;
count2 = sum(y(g) > threshold);
or to mash it all together:
count1 = sum(x>= 1 & x <= 60 & y > threshold);
count2 = sum(x>=61 & x <=100 & y > threshold);
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Image Analyst
Image Analyst 2016-10-29
What do you mean by "count the values of y"? James showed you how to count the NUMBER of y values. Do you want to SUM the y values themselves (instead of count them)? If so, just get the indexes and use sum():
indexRange1 = x>= 1 & x <= 60 & y > threshold;
sum1 = sum(y(indexRange1));
indexRange2 = x>=61 & x <=100 & y > threshold;
sum2 = sum(y(indexRange2));
dpb
dpb 2016-10-29
Yeah, but those rely on logical tests and are difficult to generalize; either of the solutions I provided automagically handle whatever number of element groups there and the length of the y vector...

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dpb
dpb 2016-10-28
编辑：dpb 2016-10-29
n=sum(reshape(y,60,[])>30).';
presuming mod(length(y),60)==0
If there are an indivisible number of elements, then
n=sum(reshape([y zeros(1,20)],60,[])>30).'; % augment y to multiple of 60
Or, if you don't like the reliance on memory storage order...
>> y=randi(50,100,1); % some sample data
>> N=60; % a number over which to group
>> subs=fix([1:length(y)].'/N-eps)+1; % indexing subscript for groups of N
>> subs(N-1:N+1) % show we got the breakpoint desired...
ans =
1 1 2
>> accumarray(subs,y,[],@(x) sum(x>30)) % the compute engine
ans =
19
14
>> sum(y(1:N)>30) % check got right answer for first grouping...
ans =
19
>>
NB: The subs vector must be column vector, note the .' transpose operator. Also note the -eps compensation on the calculation of the index; this is needed to ensure the roundoff is down for the evenly-divisible element(s) in the array, else't it'd count 1:N-1 into the first group instead 1:N.
ADDENDUM
Just to show the first solution works...
>> n=sum(reshape([y.' zeros(1,20)],60,[])>30).'
n =
19
14
>>
And, it can be generalized, too...
>> n=sum(reshape([y.' zeros(1, N-mod(length(y),N))],N,[])>30).'
n =
19
14
>>
to automate the augmentation process based on y,N

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