我觉得我在生成部分变化的块对角网格的解决方案中缺少一些直观的东西。无论如何,我想摆脱我的函数中的循环(为了挑战......)
给定参数元组、区间数和百分比变化:
params = [100 0.5 1
24 1 0.9];
nint = 1;
perc = 0.1;
所需的输出应该是:
pspacegrid(params,perc,nint)
ans =
90.0000 0.5000 1.0000
100.0000 0.5000 1.0000
110.0000 0.5000 1.0000
100.0000 0.4500 1.0000
100.0000 0.5000 1.0000
100.0000 0.5500 1.0000
100.0000 0.5000 0.9000
100.0000 0.5000 1.0000
100.0000 0.5000 1.1000
21.6000 1.0000 0.9000
24.0000 1.0000 0.9000
26.4000 1.0000 0.9000
24.0000 0.9000 0.9000
24.0000 1.0000 0.9000
24.0000 1.1000 0.9000
24.0000 1.0000 0.8100
24.0000 1.0000 0.9000
24.0000 1.0000 0.9900
您可以看到变化发生在此掩码表示的值处:
mask =
1 0 0
1 0 0
1 0 0
0 1 0
0 1 0
0 1 0
0 0 1
0 0 1
0 0 1
1 0 0
1 0 0
1 0 0
0 1 0
0 1 0
0 1 0
0 0 1
0 0 1
0 0 1
功能pspacegrid()
是:
function out = pspacegrid(params, perc, nint)
% PSPACEGRID Generates a parameter space grid for sensitivity analysis
% Size and number of variation steps
sz = size(params);
nsteps = nint*2+1;
% Preallocate output
out = reshape(permute(repmat(params,[1,1,nsteps*sz(2)]),[3,1,2]),[],sz(2));
% Mask to index positions where to place interpolated
[tmp{1:sz(2)}] = deal(true(nsteps,1));
mask = repmat(logical(blkdiag(tmp{:})),sz(1),1);
zi = cell(sz(1),1);
% LOOP per each parameter tuple
for r = 1:sz(1)
% Columns, rows, rows to interpolate and lower/upper parameter values
x = 1:sz(2);
y = [1; nint*2+1];
yi = (1:nint*2+1)';
z = [params(r,:)*(1-perc); params(r,:)*(1+perc)];
% Interpolated parameters
zi{r} = interp2(x,y,z, x, yi);
end
out(mask) = cat(1,zi{:});