count number of values greater than threshold.

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.

[James Tursa]

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)?

Answer 1

[James Tursa]

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);

[Jay Hanuman]

how to count values of y greater than 30 for x indexes 1 to 60, and then for 61 to 100.

[Image Analyst]

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]

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...

Answer 2

[dpb]

 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