How can I use the "diff" function instead of a for loop?

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Ashfaq Ahmed
Ashfaq Ahmed 2022-1-4
评论： Matt J 2022-1-4
Dear altruists,
How can I get the similar result from this few lines of code using the MATLAB built in "diff" function instead of using the for loop?
g = 9.807;
for i=2:size(lon,1)-1
for j=2:size(lat,2)-1
dx=double(m_lldist([lon(i+1) lon(i-1)],[lat(j) lat(j)]))*1000;
dy=double(m_lldist([lon(i) lon(i)],[lat(j+1) lat(j-1)]))*1000;
f=2*(omega)*sin((lat(j)*(pi/180)));
u(i,j)=-g/f*(ssh(i,j+1)-ssh(i,j-1))/dy;
v(i,j)= g/f*(ssh(i+1,j)-ssh(i-1,j))/dx;
end
end
the size of the variables: lat, lon, and ssh is the same. (size:71*294)
Any feedback from you will be highly appreciated. Happy new year all!

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采纳的回答

Matt J
Matt J 2022-1-4
编辑：Matt J 2022-1-4
[m,n]=size(ssh);
zm=zeros(m,1);
zn=zeros(1,n);
gf=g./(2*omega*sind( lat ) );
Du=diff(ssh(:,2:end),1,2)+diff(ssh(:,1:end-1),1,2);
Du=-gf.*[zm,Du,zm];
Dv=diff(ssh(2:end,:),1,1)+diff(ssh(1:end-1,:),1,1);
Dv=gf.*[zn;Dv;zn];
[dx,dy]=deal(ones(size(Du)));
for i=2:size(lon,1)-1
for j=2:size(lat,2)-1
dx(i,j)=double(m_lldist([lon(i+1) lon(i-1)],[lat(j) lat(j)]))*1000;
dy(i,j)=double(m_lldist([lon(i) lon(i)],[lat(j+1) lat(j-1)]))*1000;
end
end
u=Du./dy;
v=Dv./dx;
  2 个评论
Ashfaq Ahmed
Ashfaq Ahmed 2022-1-4
Hi Matt,
Thank you so much for your effort on my question. Unfortunately, I fogrot to include that, the g in my code is a variable with a constant value.
g = 9.807;
I ran your code and it is not giving me the values that I am expecting. Do you think it was because of the g?

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