我试图将我的CFD数据(它以标量N数组的形式,N对应于Y,M到x,以及P到z)在时间步骤的子集上进行平均。下面我尝试简化我想要的平均过程的描述。
我有一个代码,它似乎能正确地完成我想做的事情,但要完成计算需要太长时间。我想让它运行得更快,如果有必要的话,我愿意修改代码。下面是我的代码版本,它可以处理较小版本的数据。
x = -5:5; % x - position data
y = -2:.5:5; % y - position data
z = -5:5; % z - position data
% my grid is much bigger actually
[X,Y,Z] = meshgrid(x,y,z); % mesh for plotting data
dX = diff(x)'; dX(end+1) = dX(end); % x grid intervals
dY = diff(y)'; dY(end+1) = dY(end); % y grid intervals
dZ = diff(z)'; dZ(end+1) = dZ(end); % z grid intervals
TestPoints = combvec(x,y,z)'; % you need the Matlab Neural Network Toolbox to run this
dXYZ = combvec(dX',dY',dZ')';
% TestPoints is the unrotated grid
M = length(x); % size of grid x - direction
N = length(y); % size of grid y - direction
P = length(z); % size of grid z - direction
D = randi([-10,10],N,M,P,3); % placeholder for data for 3 time steps (I have more than 3, this is a subset)
D2{3,M*N*P} = [];
PosAll{3,2} = [];
[xSph,ySph,zSph] = sphere(50);
c = 0.01; % 1 cm
nu = 8e-6; % 8 cSt
s = 3*c; % span for Aspect Ratio 3
r_g = s/sqrt(3);
U_g = 110*nu/c; % velocity for Reynolds number 110
Omega = U_g/r_g; % angular velocity
T = (2*pi)/Omega; % period
dt = 40*T/1920; % time interval
DeltaRotAngle = ((2*pi)/T)*dt; % angle interval
timesteps = 121:123; % time steps 121, 122, and 123
for ti=timesteps
tj = find(ti==timesteps);
Theta = ti*DeltaRotAngle;
Rotate = [cos(Theta),0,sin(Theta);...
0,1,0;...
-sin(Theta),0,cos(Theta)];
PosAll{tj,1} = (Rotate*TestPoints')';
end
for i=1:M*N*P
aa = TestPoints(i,1);
bb = TestPoints(i,2);
cc = TestPoints(i,3);
rs = 0.8*sqrt(dXYZ(i,1)^2 + dXYZ(i,2)^2 + dXYZ(i,3)^2);
handles.H = figure;
hs = surf(xSph*rs+aa,ySph*rs+bb,zSph*rs+cc);
[Fs,Vs,~] = surf2patch(hs,'triangle');
close(handles.H)
for ti=timesteps
tj = find(timesteps==ti);
f = inpolyhedron(Fs,Vs,PosAll{tj,1},'FlipNormals',false);
TestPointsR_ti = PosAll{tj,1};
PointsInSphere = TestPointsR_ti(f,:);
p1 = [aa,bb,cc];
p2 = [PointsInSphere(:,1),...
PointsInSphere(:,2),...
PointsInSphere(:,3)];
w = 1./sqrt(sum(...
(p2-repmat(p1,size(PointsInSphere,1),1))...
.^2,2));
D_ti = reshape(D(:,:,:,tj),M*N*P,1);
D2{tj,i} = [D_ti(f),w];
end
end
D3{M*N*P,1} = [];
for i=1:M*N*P
D3{i} = vertcat(D2{:,i});
end
D4 = zeros(M*N*P,1);
for i=1:M*N*P
D4(i) = sum(D3{i}(:,1).*D3{i}(:,2))/...
sum(D3{i}(:,2));
end
D_ta = reshape(D4,N,M,P);我期望得到一个N,x,M,x,P数组,其中每个索引是所有点的加权平均值,包括在未旋转网格中特定位置的所有时间步长。正如你所看到的,这正是我所得到的。然而,主要的问题是,当我使用更大的“真实”数据集时,需要多长时间才能做到这一点。上面的代码只需几分钟就能运行,但是当M= 120、N= 24和P= 120时,时间步骤数为24,这可能会花费更长的时间。根据我的估计,完成整个计算大约需要25+天。
发布于 2019-01-04 01:48:40
没关系,我可以帮你学数学。你在这里要做的是在一个球体里找到东西。你有一个很好定义的球体,所以这很容易。只要找出所有点与中心点之间的距离。不需要绘制或使用多面体。注意,第66行,我根据球的中心点修改点,计算这些点的距离,并与球体的半径进行比较。
% x = -5:2:5; % x - position data
x = linspace(-5,5,120);
% y = -2:5; % y - position data
y = linspace(-2,5,24);
% z = -5:2:5; % z - position data
z = linspace(-5,5,120);
% my grid is much bigger actually
[X,Y,Z] = meshgrid(x,y,z); % mesh for plotting data
dX = diff(x)'; dX(end+1) = dX(end); % x grid intervals
dY = diff(y)'; dY(end+1) = dY(end); % y grid intervals
dZ = diff(z)'; dZ(end+1) = dZ(end); % z grid intervals
TestPoints = combvec(x,y,z)'; % you need the Matlab Neural Network Toolbox to run this
dXYZ = combvec(dX',dY',dZ')';
% TestPoints is the unrotated grid
M = length(x); % size of grid x - direction
N = length(y); % size of grid y - direction
P = length(z); % size of grid z - direction
D = randi([-10,10],N,M,P,3); % placeholder for data for 3 time steps (I have more than 3, this is a subset)
D2{3,M*N*P} = [];
PosAll{3,2} = [];
[xSph,ySph,zSph] = sphere(50);
c = 0.01; % 1 cm
nu = 8e-6; % 8 cSt
s = 3*c; % span for Aspect Ratio 3
r_g = s/sqrt(3);
U_g = 110*nu/c; % velocity for Reynolds number 110
Omega = U_g/r_g; % angular velocity
T = (2*pi)/Omega; % period
dt = 40*T/1920; % time interval
DeltaRotAngle = ((2*pi)/T)*dt; % angle interval
timesteps = 121:123; % time steps 121, 122, and 123
for ti=timesteps
tj = find(ti==timesteps);
Theta = ti*DeltaRotAngle;
Rotate = [cos(Theta),0,sin(Theta);...
0,1,0;...
-sin(Theta),0,cos(Theta)];
PosAll{tj,1} = (Rotate*TestPoints')';
end
tic
for i=1:M*N*P
aa = TestPoints(i,1);
bb = TestPoints(i,2);
cc = TestPoints(i,3);
rs = 0.8*sqrt(dXYZ(i,1)^2 + dXYZ(i,2)^2 + dXYZ(i,3)^2);
% handles.H = figure;
% hs = surf(xSph*rs+aa,ySph*rs+bb,zSph*rs+cc);
% [Fs,Vs,~] = surf2patch(hs,'triangle');
% close(handles.H)
for ti=timesteps
tj = find(timesteps==ti);
% f = inpolyhedron(Fs,Vs,PosAll{tj,1},'FlipNormals',false);
f = sqrt(sum((PosAll{tj,1}-[aa,bb,cc]).^2,2))<rs;
TestPointsR_ti = PosAll{tj,1};
PointsInSphere = TestPointsR_ti(f,:);
p1 = [aa,bb,cc];
p2 = [PointsInSphere(:,1),...
PointsInSphere(:,2),...
PointsInSphere(:,3)];
w = 1./sqrt(sum(...
(p2-repmat(p1,size(PointsInSphere,1),1))...
.^2,2));
D_ti = reshape(D(:,:,:,tj),M*N*P,1);
D2{tj,i} = [D_ti(f),w];
end
if ~mod(i,10)
toc
end
end
D3{M*N*P,1} = [];
for i=1:M*N*P
D3{i} = vertcat(D2{:,i});
end
D4 = zeros(M*N*P,1);
for i=1:M*N*P
D4(i) = sum(D3{i}(:,1).*D3{i}(:,2))/...
sum(D3{i}(:,2));
end
D_ta = reshape(D4,N,M,P);就运行时而言,在我的计算机上,运行旧代码需要57个小时。新代码需要2个小时。在这一点上,主要的计算是距离,所以我怀疑你会变得更好。
https://stackoverflow.com/questions/54028084
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