sort 2 matrices for minimum numbers sum and divide them
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Akzeptierte Antwort
Stephen23
am 4 Jun. 2019
Bearbeitet: Stephen23
am 4 Jun. 2019
This is MATLAB, so don't waste your time writing inefficient loops.
>> 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 Kommentare
Stephen23
am 5 Jun. 2019
Bearbeitet: Stephen23
am 5 Jun. 2019
>> 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
>> 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
Weitere Antworten (3)
Raj
am 4 Jun. 2019
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!!
0 Kommentare
Pullak Barik
am 4 Jun. 2019
Bearbeitet: Pullak Barik
am 4 Jun. 2019
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 Kommentare
Stephen23
am 4 Jun. 2019
Bearbeitet: Stephen23
am 4 Jun. 2019
@Pullak Barik: note that concatenation onto the result array like that is not considered good practice, and detrimentally affects efficiency:
The MATLAB documentation recommends preallocating arrays before the loop:
Andrei Bobrov
am 4 Jun. 2019
Bearbeitet: Andrei Bobrov
am 4 Jun. 2019
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);
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!