sort 2 matrices for minimum numbers sum and divide them

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Tino
Tino 2019-6-4
编辑： Stephen23 2019-6-5
I have r = [ 1 3 4 5 ......n]
and x = [ 5 8 9 4 ......n]
I will like to do this computation
when m = 1
sort the lowest number of r and divide it with the lowest number of x till the end n
for example =[ 1/4, 3/5, 4/8, 5/9.......n]
when m = 2
= [(1 + 3) / (5 + 4) , (4 +5)/(8+9) , ...........n]
when m =3
=[ (1 + 3 + 4)/( 4 + 5 + 8),................n]
Thanks for your help in advance
Tino
  2 个评论
Stephen23
Stephen23 2019-6-4
编辑：Stephen23 2019-6-4
And what should be the last term if n is not exactly divisble by m ?
Tino
Tino 2019-6-4
Hi Stephen
if the last term is not divisible it is ignored. That is to say the computation stops when it is not divisible by mHope to hear from you soonest
Tino

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采纳的回答

Stephen23
Stephen23 2019-6-4
编辑：Stephen23 2019-6-4
This is MATLAB, so don't waste your time writing inefficient loops.
Learn to write simpler vectorized code, just like experienced MATLAB users do:
>> r = sort([1,3,4,5,10]); % sorted!
>> x = sort([5,8,9,4,10]); % sorted!
>> m = 1;
>> n = m*fix(numel(r)/m);
>> sum(reshape(r(1:n),m,[]),1) ./ sum(reshape(x(1:n),m,[]),1)
ans =
0.25000 0.60000 0.50000 0.55556 1.00000
>> m = 2;
>> n = m*fix(numel(r)/m);
>> sum(reshape(r(1:n),m,[]),1) ./ sum(reshape(x(1:n),m,[]),1)
ans =
0.44444 0.52941
>> m = 3;
>> n = m*fix(numel(r)/m);
>> sum(reshape(r(1:n),m,[]),1) ./ sum(reshape(x(1:n),m,[]),1)
ans =
0.47059
  7 个评论
显示 5更早的评论
Tino
Tino 2019-6-4
Thanks for your help Stephen
How do I make it
0.25 4 1
0.4 1 1.5
0.5 0.5 0.66667
You ve been a big help
Hoping to hear from you soon
Stephen23
Stephen23 2019-6-5
编辑：Stephen23 2019-6-5
>> S = size(A);
>> m = 1;
>> n = m*fix(S(1)/2/m);
>> r = A(1:n,:)
>> x = A(1+n:2*n,:)
>> Z = sum(reshape(r,m,[]),1) ./ sum(reshape(x,m,[]),1);
>> Z = reshape(Z,[],S(2)).'
Z =
0.25000 0.40000 0.50000
4.00000 1.00000 0.50000
1.00000 1.50000 0.66667
Or in one fell swoop:
>> Z = sum(permute(reshape(r,m,[],S(2)),[3,2,1]),3) ./ ...
sum(permute(reshape(x,m,[],S(2)),[3,2,1]),3)
Z =
0.25000 0.40000 0.50000
4.00000 1.00000 0.50000
1.00000 1.50000 0.66667

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更多回答（3 个）

Raj
Raj 2019-6-4
I am assuming n is a multiple of m. In that case this works:
r=sort(r);
x=sort(x);
m=input('enter value of m:');
if m==1
for ii=1:numel(r)
Answer(1,ii)=r(1,ii)/x(1,ii);
end
Answer
elseif m==2
Answer(1,1)=(r(1,1)+r(1,2))/(x(1,1)+x(1,2));
for ii=3:2:numel(r)
Answer(1,(ii+1)/2)=(r(1,ii)+r(1,ii+1))/(x(1,ii)+x(1,ii+1));
end
Answer
elseif m==3
Answer(1,1)=(r(1,1)+r(1,2)+r(1,3))/(x(1,1)+x(1,2)+x(1,3));
n=0;
for ii=4:3:numel(r)
Answer(1,(ii-n)/2)=(r(1,ii)+r(1,ii+1)+r(1,ii+2))/(x(1,ii)+x(1,ii+1)+x(1,ii+2));
n=n+1;
end
Answer
else
disp('Invalid Value of m')
end
There may be better and optimized way of doing this also.
P.S: How about adding a 'Homework' tag next time and showing what you attempted in addition to 'Hope to hear from you soonest'? Everybody wil not be as free as i am today! Cheers!!
Pullak Barik
Pullak Barik 2019-6-4
编辑：Pullak Barik 2019-6-4
Hi!
I assume that r and x are just vectors, and not matrices with more than one dimension.
I guess the following function will work for you-
function result = sort_sum_and_divide(r, x, m)
r = sort(r);
x = sort(x);
n = length(r);
if(length(r) == length(x)) %To check if the input is wrong
i = 1:m:n;
if(i(end) + m - 1 > n)
i(end) = []; %To drop the elements at the end if they can not be grouped
end
result = zeros(1, length(i)); %preallocation of result array
for idx = 1:length(i)
result(idx) = sum(r(i(idx):i(idx)+m-1))./sum(x(i(idx):i(idx)+m-1));
end
else
disp('Length of r and x are not the same');
end
end
  2 个评论
Stephen23
Stephen23 2019-6-4
编辑：Stephen23 2019-6-4
@Pullak Barik: note that concatenation onto the result array like that is not considered good practice, and detrimentally affects efficiency:
https://www.mathworks.com/help/matlab/matlab_prog/techniques-for-improving-performance.html
The MATLAB documentation recommends preallocating arrays before the loop:
https://www.mathworks.com/help/matlab/matlab_prog/preallocating-arrays.html

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Andrei Bobrov
Andrei Bobrov 2019-6-4
编辑：Andrei Bobrov 2019-6-4
m = 3;
r = [ 1 3 4 5 30];
x = [ 5 8 9 4 78];
out = funt(r,x,3);
function out = funt(r,x,m)
ad = nan(mod(-numel(r),m),1);
a = sort(cat(3,[r(:);ad],[x(:);ad]));
b = sum(reshape(a,m,[],2),'omitnan');
out = b(:,:,1)./b(:,:,2);
end
add
rx = [1
3
4
5
6
7
8
9
2
6];
m = 2;
rrxx = sum(sort(reshape([reshape(rx,[],2);nan(mod(-numel(rx)/2,m),2)],...
m,[],2)),'omitnan');
out = rrxx(:,:,1)./rrxx(:,:,2);
  2 个评论
Tino
Tino 2019-6-4
Thanks Andrei Just one more help
if I have a data with various n columns and wishes to partition those column into 2 to perform this computation how do I go about it
for instance
1
3
4
5
6
7
8
9
2
6
and wish to divide this one column into halfs (r and x) to carry out the computation accross several columns
How can I do this intitial code
Thanks again
Tino
Tino
Tino 2019-6-4
Hi Andrei
Your code is givng me the right answer
I have r = [ 1 3 4 5 ......n]
and x = [ 5 8 9 4 ......n]
I will like to do this computation
when m = 1
sort the lowest number of r and divide it with the lowest number of x till the end n
for example =[ 1/4, 3/5, 4/8, 5/9.......n]
when m = 2
= [(1 + 3) / (5 + 4) , (4 +5)/(8+9) , ...........n]
when m =3
=[ (1 + 3 + 4)/( 4 + 5 + 8),................n]
Thanks for your help in advance
Tino

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