how to break large data into groups and run same steps in a loop for each group?

x= 1:100000000

data= [ sin(x), cos(x), tan(x), x]

how do I find mean of sin(x), cos(x), tan(x) and x in a loop (like M=mean(data1) and run this until I get mean of x) and plot each of these in a same plot ?

If I try to access data(1) it gives me a number and not an entire sin(x) value.

Answer 1

[Image Analyst]

Try this:

tic;
x = 1:100000000;
data= [ sin(x), cos(x), tan(x), x];
numElements = length(x);
theMeans = zeros(1, 4);
for k = 1 : 4
    index1 = (k - 1) * numElements + 1;
    index2 = k * numElements;
    theData = data(index1 : index2);
    theMeans(k) = mean(theData);
end
toc

Elapsed time is 36.811941 seconds.

theMeans % Show in command window.

theMeans = 1×4

1.0e+07 * 0.0000 0.0000 -0.0000 5.0000

% Obviously the means of the sin, cos, and tan will all be around 0
% while the mean of x will be gigantic.  So you'll only see the last bar.
bar(theMeans);
grid on;
fontSize = 14;
title('Means', 'FontSize', fontSize);
xlabel('Segment', 'FontSize', fontSize);
ylabel('Mean Value', 'FontSize', fontSize);

[Pragya Dhungel]

Thank you for the answer but I wanted plot of four different means and not their cumulative sum.

something like,

plot (x,meanofsin,x,meanofcos,meanoftan,x,meanofx)

with four different lines/bar for each mean.

[Image Analyst]

@Pragya Dhungel theMeans has all 4 values. Like I said, the mean of the first 4 (sin, cos, and tan) is of course very close to zero, while the mean of the x values is 50 million. Why do you think I'm getting a cumulative sum???? Did you overlook the part in my answer where it says

theMeans = 1×4

1.0e+07 * 0.0000 0.0000 -0.0000 5.0000

Note, that means there are 4 sums there, not only one. And so 3 of them have means of zero, and one of them has a mean of 50 million because your x goes from 1 to 100 million so the mean of x is 50 million.

[Pragya Dhungel]

My bad! Thank you for the elaboration.

[Pragya Dhungel]

I tried the code, it works for the bar plot but not for the line plot, i.e. plot(x,y1,x,y2,x,y3,x,y4) kind of command , where I want four different lines for each of the four function. Can we do that in a single line as we did for bar(theMeans) or code similar to @William Rose where the function are already seperated is only our option?

a=sin(2*pi*x); b=cos(2*pi*x); %compute arrays a, b

meana=mean(a); meanb=mean(b); meanx=mean(x); %compute means of the columns

plot(1:N,meana,'-rx',1:N,meanb,'-g+',1:N,meanx,'-b.'); %plot results

**apologies if you've already made it clear and I am still missing it.

[William Rose]

@Pragya Dhungel, Here is my understanding of what you want to do:

Array data has 4 long columns. Compute and plot the average of each column, using non-overlapping segments of length M:

N=1000; M=25;           %N=column length, M=segment length
L=fix(N/M);             %number of segments
%M does not need to be an integer divisor of N
x=(1:N)'*2*pi*4/N;
data=[sin(x),cos(x),tan(x),x];
means=zeros(L,4);  %allocate array for the means
for i=1:L
    for j=1:4
        means(i,j)=mean(data(1+(i-1)*M:i*M,j));
    end
end
figure;
subplot(211)
%disp([size(1:fix(N/M)),size(means(:,1))]);
plot(1:fix(N/M),means(:,1),'-r.',1:fix(N/M),means(:,2),'-g.',...
    1:fix(N/M),means(:,3),'-b.');
legend('sin','cos','tan'); ylabel('Mean of Segment'); 
title('Columns 1,2,3 of data()');
subplot(212), plot(1:fix(N/M),means(:,4),'-k.');
xlabel('Segment Number'); ylabel('Mean of Segment')
title('Column 4 of data()');

I plot the means for column 4 separately, due to the very different scale.

Answer 2

[William Rose]

@Pragya Dhungel,

The commands you gave create an array which is 1 row by 400000000 columns.

To illustrate, let us reduce the size of x:

x=1:10

x = 1×10

1 2 3 4 5 6 7 8 9 10

data=[sin(x), cos(x), tan(x), x];
fprintf('Size of data: %d by %d\n',size(data));

Size of data: 1 by 40

I expect this was not what you intended. Try this instead:

data=[sin(x); cos(x); tan(x); x];
fprintf('Size of data: %d by %d\n',size(data));

Size of data: 4 by 10

Each function is on a different row. Find the mean of each row:

disp(mean(data,2));  

    0.1411
   -0.1417
   -0.9016
    5.5000

mean(data,2) computes the mean of each row. mean(data) computes the mean of each column, which is probably not what you want, for this array.

To compute and display the mean of row 1:

disp(mean(data(1,:)));

    0.1411

To compute and display the mean of column 3:

disp(mean(data(:,3)));

    0.5021

[Pragya Dhungel]

Thank you. This worked well to display the mean.

BUT, what if the sin(x), cos(x) terms are not vectors but n*n matrices ?

how do I plot (x,meanofsin,x,meanofcos,meanoftan,x,meanofx) in a single plot but four different lines for each terms?

[William Rose]

@Pragya Dhungel,

N=20;
x=zeros(N);  %create N by N array
for i=1:N, for j=1:N, x(i,j)=2*(i+N*(j-1)-N^2/2)/N^2; end,end  %fill array
a=sin(2*pi*x); b=cos(2*pi*x); %compute arrays a, b
meana=mean(a); meanb=mean(b); meanx=mean(x); %compute means of the columns
plot(1:N,meana,'-rx',1:N,meanb,'-g+',1:N,meanx,'-b.'); %plot results
grid on; legend('mean(a)','mean(b)','mean(x)');

This seems to be what you requested.

[William Rose]

@Pragya Dhungel,

The following code illustrates the arrays a, b, x used in the example above:

N=20;
x=zeros(N);  %create N by N array
for i=1:N, for j=1:N, x(i,j)=2*(i+N*(j-1)-N^2/2)/N^2; end,end  %fill array
a=sin(2*pi*x); b=cos(2*pi*x); %compute arrays a, b
figure;
subplot(3,1,1), surf(1:N,1:N,a); zlabel('a')
subplot(3,1,2), surf(1:N,1:N,b); zlabel('b')
subplot(3,1,3), surf(1:N,1:N,x); zlabel('x')

Answer 3

[William Rose]

[I have moved this answer from the Comments to the Answers.]

@Pragya Dhungel,

N=20;
x=zeros(N);  %create N by N array
for i=1:N, for j=1:N, x(i,j)=2*(i+N*(j-1)-N^2/2)/N^2; end,end  %fill array
a=sin(2*pi*x); b=cos(2*pi*x); %compute arrays a, b
meana=mean(a); meanb=mean(b); meanx=mean(x); %compute means of the columns
plot(1:N,meana,'-rx',1:N,meanb,'-g+',1:N,meanx,'-b.'); %plot results
grid on; legend('mean(a)','mean(b)','mean(x)');

[Image Analyst]

Your initial question said nothing about having x be a 2-D array. You defined it as a 1-D array going from 1 to 100 million.

[William Rose]

@Image Analyst is exactly right, as usual.

@Pragya Dhungel, I wonder if you have a very long 1D array, and you want to compute the mean of it, one segment at a time. Perhaps you are thinking that by reshaping it into a rectangular array, you could compute the mean of each column. That could explain why you started with a 1D array, and then changed to 2D.

If you are trying to compute the means of successive segments of a long vector, then you are right: you could do it by reshaping, then computing the column means.

[Pragya Dhungel]

Yes. The data is too large and I am trying to break it down into groups and compute their mean. This discussion helped me a lot to understand it. Thanks!