假设nodeA是nodeB的子节点。更改nodeB的zRotation值会有效地旋转nodeA，但nodeA的位置（因为它相对于nodeB）保持不变。
现在假设nodeA已从nodeB中删除，但其位置在屏幕中保持不变。 nodeC被添加到nodeB并旋转。
如何检测nodeC和nodeA之间的重叠？ SKNode as described here中的containsPoint无法正常工作，因为nodeC和nodeA位置都不会因轮换而发生变化。
答案 0 :(得分：2)
您可以测试与function [features,weights] = MI(features,labels,Q)
if nargin <3
Q = 12;
end
edges = zeros(size(features,2),Q+1);
for k = 1:size(features,2)
minval = min(features(:,k));
maxval = max(features(:,k));
if minval==maxval
continue;
end
quantlevels = minval:(maxval-minval)/500:maxval;
N = histc(features(:,k),quantlevels);
totsamples = size(features,1);
N_cum = cumsum(N);
edges(k,1) = -Inf;
stepsize = totsamples/Q;
for j = 1:Q-1
a = find(N_cum > j.*stepsize,1);
edges(k,j) = quantlevels(a);
end
edges(k,j+2) = Inf;
end
S = zeros(size(features));
for k = 1:size(S,2)
S(:,k) = quantize(features(:,k),edges(k,:))+1;
end
I = zeros(size(features,2),1);
for k = 1:size(features,2)
I(k) = computeMI(S(:,k),labels,0);
end
[weights,features] = sort(I,'descend');
%% EOF
function [I,M,SP] = computeMI(seq1,seq2,lag)
if nargin <3
lag = 0;
end
if(length(seq1) ~= length(seq2))
error('Input sequences are of different length');
end
lambda1 = max(seq1);
symbol_count1 = zeros(lambda1,1);
for k = 1:lambda1
symbol_count1(k) = sum(seq1 == k);
end
symbol_prob1 = symbol_count1./sum(symbol_count1)+0.000001;
lambda2 = max(seq2);
symbol_count2 = zeros(lambda2,1);
for k = 1:lambda2
symbol_count2(k) = sum(seq2 == k);
end
symbol_prob2 = symbol_count2./sum(symbol_count2)+0.000001;
M = zeros(lambda1,lambda2);
if(lag > 0)
for k = 1:length(seq1)-lag
loc1 = seq1(k);
loc2 = seq2(k+lag);
M(loc1,loc2) = M(loc1,loc2)+1;
end
else
for k = abs(lag)+1:length(seq1)
loc1 = seq1(k);
loc2 = seq2(k+lag);
M(loc1,loc2) = M(loc1,loc2)+1;
end
end
SP = symbol_prob1*symbol_prob2';
M = M./sum(M(:))+0.000001;
I = sum(sum(M.*log2(M./SP)));
function y = quantize(x, q)
x = x(:);
nx = length(x);
nq = length(q);
y = sum(repmat(x,1,nq)>repmat(q,nx,1),2);
交叉的节点。但在这种情况下，节点需要位于同一节点树中。也许你不应该在测试之前从nodeB中删除nodeA？