多维线性插值的预计算权重
多维线性插值的预计算权重
提问于 2017-09-09 08:14:39
我有一个沿D维度的非均匀矩形网格，网格上的逻辑值矩阵V，以及查询数据点X的矩阵。网格点的数量因维度而异。
我对相同的网格G和查询X多次运行插值，但对不同的值V。
目标是预先计算插值的索引和权重，并重用它们，因为它们总是相同的。
这是一个二维的例子，每次在循环中我都必须计算索引和值，但我想在循环之前只计算一次。我保留应用程序中的数据类型(主要是单一的和逻辑的gpuArrays)。
% Define grid
G{1} = single([0; 1; 3; 5; 10]);
G{2} = single([15; 17; 18; 20]);
% Steps and edges are reduntant but help make interpolation a bit faster
S{1} = G{1}(2:end)-G{1}(1:end-1);
S{2} = G{2}(2:end)-G{2}(1:end-1);
gpuInf = 1e10;
% It's my workaround for a bug in GPU version of discretize in Matlab R2017a.
% It throws an error if edges contain Inf, realmin, or realmax. Seems fixed in R2017b prerelease.
E{1} = [-gpuInf; G{1}(2:end-1); gpuInf];
E{2} = [-gpuInf; G{2}(2:end-1); gpuInf];
% Generate query points
n = 50; X = gpuArray(single([rand(n,1)*14-2, 14+rand(n,1)*7]));
[G1, G2] = ndgrid(G{1},G{2});
for i = 1 : 4
% Generate values on grid
foo = @(x1,x2) (sin(x1+rand) + cos(x2*rand))>0;
V = gpuArray(foo(G1,G2));
% Interpolate
V_interp = interpV(X, V, G, E, S);
% Plot results
subplot(2,2,i);
contourf(G1, G2, V); hold on;
scatter(X(:,1), X(:,2),50,[ones(n,1), 1-V_interp, 1-V_interp],'filled', 'MarkerEdgeColor','black'); hold off;
end
function y = interpV(X, V, G, E, S)
y = min(1, max(0, interpV_helper(X, 1, 1, 0, [], V, G, E, S) ));
end
function y = interpV_helper(X, dim, weight, curr_y, index, V, G, E, S)
if dim == ndims(V)+1
M = [1,cumprod(size(V),2)];
idx = 1 + (index-1)*M(1:end-1)';
y = curr_y + weight .* single(V(idx));
else
x = X(:,dim); grid = G{dim}; edges = E{dim}; steps = S{dim};
iL = single(discretize(x, edges));
weightL = weight .* (grid(iL+1) - x) ./ steps(iL);
weightH = weight .* (x - grid(iL)) ./ steps(iL);
y = interpV_helper(X, dim+1, weightL, curr_y, [index, iL ], V, G, E, S) +...
interpV_helper(X, dim+1, weightH, curr_y, [index, iL+1], V, G, E, S);
end
end回答 2
Stack Overflow用户
发布于 2017-09-10 08:49:50
我找到了一种方法来做到这一点，并将其发布在这里，因为(到目前为止)还有两个人感兴趣。它只需要对我的原始代码稍作修改(见下文)。
% Define grid
G{1} = single([0; 1; 3; 5; 10]);
G{2} = single([15; 17; 18; 20]);
% Steps and edges are reduntant but help make interpolation a bit faster
S{1} = G{1}(2:end)-G{1}(1:end-1);
S{2} = G{2}(2:end)-G{2}(1:end-1);
gpuInf = 1e10;
% It's my workaround for a bug in GPU version of discretize in Matlab R2017a.
% It throws an error if edges contain Inf, realmin, or realmax. Seems fixed in R2017b prerelease.
E{1} = [-gpuInf; G{1}(2:end-1); gpuInf];
E{2} = [-gpuInf; G{2}(2:end-1); gpuInf];
% Generate query points
n = 50; X = gpuArray(single([rand(n,1)*14-2, 14+rand(n,1)*7]));
[G1, G2] = ndgrid(G{1},G{2});
[W, I] = interpIW(X, G, E, S); % Precompute weights W and indexes I
for i = 1 : 4
% Generate values on grid
foo = @(x1,x2) (sin(x1+rand) + cos(x2*rand))>0;
V = gpuArray(foo(G1,G2));
% Interpolate
V_interp = sum(W .* single(V(I)), 2);
% Plot results
subplot(2,2,i);
contourf(G1, G2, V); hold on;
scatter(X(:,1), X(:,2), 50,[ones(n,1), 1-V_interp, 1-V_interp],'filled', 'MarkerEdgeColor','black'); hold off;
end
function [W, I] = interpIW(X, G, E, S)
global Weights Indexes
Weights=[]; Indexes=[];
interpIW_helper(X, 1, 1, [], G, E, S, []);
W = Weights; I = Indexes;
end
function [] = interpIW_helper(X, dim, weight, index, G, E, S, sizeV)
global Weights Indexes
if dim == size(X,2)+1
M = [1,cumprod(sizeV,2)];
Weights = [Weights, weight];
Indexes = [Indexes, 1 + (index-1)*M(1:end-1)'];
else
x = X(:,dim); grid = G{dim}; edges = E{dim}; steps = S{dim};
iL = single(discretize(x, edges));
weightL = weight .* (grid(iL+1) - x) ./ steps(iL);
weightH = weight .* (x - grid(iL)) ./ steps(iL);
interpIW_helper(X, dim+1, weightL, [index, iL ], G, E, S, [sizeV, size(grid,1)]);
interpIW_helper(X, dim+1, weightH, [index, iL+1], G, E, S, [sizeV, size(grid,1)]);
end
endStack Overflow用户
发布于 2017-09-10 20:26:57
为了完成这项任务，除了计算内插值之外，应该完成整个插值过程。这是一个从Octave c++ source翻译过来的解决方案。除了不需要v数组之外，输入的格式与interpn函数的frst签名相同。此外，X应该是矢量，并且不应该是ndgrid格式。输出W (权重)和I (位置)的大小都是(a ,b)，a是网格上的点的邻居数量，b是要插值的请求点的数量。
function [W , I] = lininterpnw(varargin)
% [W I] = lininterpnw(X1,X2,...,Xn,Xq1,Xq2,...,Xqn)
n = numel(varargin)/2;
x = varargin(1:n);
y = varargin(n+1:end);
sz = cellfun(@numel,x);
scale = [1 cumprod(sz(1:end-1))];
Ni = numel(y{1});
index = zeros(n,Ni);
x_before = zeros(n,Ni);
x_after = zeros(n,Ni);
for ii = 1:n
jj = interp1(x{ii},1:sz(ii),y{ii},'previous');
index(ii,:) = jj-1;
x_before(ii,:) = x{ii}(jj);
x_after(ii,:) = x{ii}(jj+1);
end
coef(2:2:2*n,1:Ni) = (vertcat(y{:}) - x_before) ./ (x_after - x_before);
coef(1:2:end,:) = 1 - coef(2:2:2*n,:);
bit = permute(dec2bin(0:2^n-1)=='1', [2,3,1]);
%I = reshape(1+scale*bsxfun(@plus,index,bit), Ni, []).'; %Octave
I = reshape(1+sum(bsxfun(@times,scale(:),bsxfun(@plus,index,bit))), Ni, []).';
W = squeeze(prod(reshape(coef(bsxfun(@plus,(1:2:2*n).',bit),:).',Ni,n,[]),2)).';
end测试：
x={[1 3 8 9],[2 12 13 17 25]};
v = rand(4,5);
y={[1.5 1.6 1.3 3.5,8.1,8.3],[8.4,13.5,14.4,23,23.9,24.2]};
[W I]=lininterpnw(x{:},y{:});
sum(W.*v(I))
interpn(x{:},v,y{:})感谢@SardarUsama的测试和他的有用的评论。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/46126019
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