下面显示的用于绘制Mandelbrot set的代码,我认为我的代码有点冗余来构建 Matrix M
。在 Python 中,我知道有一种干净的方法,
M = [[mandel(complex(r, i)) for r in np.arange(-2, 0.5,0.005) ] for i in np.range(-1,1,0.005)]
在Matlab中是否有类似的方法?
function M=mandelPerf()
rr=-2:0.005:0.5;
ii=-1:0.005:1;
M = zeros(length(ii), length(rr));
id1 = 1;
for i =ii
id2 = 1;
for r = rr
M(id1, id2) = mandel(complex(r,i));
id2 = id2 + 1;
end
id1 = id1 + 1;
end
end
function n = mandel(z)
n = 0;
c = z;
for n=0:100
if abs(z)>2
break
end
z = z^2+c;
end
end
答案 0 :(得分:8)
你可以完全避免循环。您可以以矢量化方式执行迭代z = z.^2 + c
。为了避免不必要的操作,在每次迭代时都要跟踪哪些点c
已超过您的阈值,并继续仅使用其余点进行迭代(这是指数ind
和ind2
的目的。下面的代码):
rr =-2:0.005:0.5;
ii =-1:0.005:1;
max_n = 100;
threshold = 2;
c = bsxfun(@plus, rr(:).', 1i*ii(:)); %'// generate complex grid
M = max_n*ones(size(c)); %// preallocation.
ind = 1:numel(c); %// keeps track of which points need to be iterated on
z = zeros(size(c)); %// initialization
for n = 0:max_n;
z(ind) = z(ind).^2 + c(ind);
ind2 = abs(z(ind)) > threshold;
M(ind(ind2)) = n; %// store result for these points...
ind = ind(~ind2); %// ...and remove them from further consideration
end
imagesc(rr,ii,M)
axis equal
答案 1 :(得分:1)
你至少可以避免使用for循环:
function M=mandelPerf()
rr = -2:0.005:0.5;
ii = -1:0.005:1;
[R,I] = meshgrid(rr,ii);
M = arrayfun(@(x) mandel(x), R+1i*I);
end
function n = mandel(z)
n = 0;
c = z;
for n=0:100
if abs(z)>2
break
end
z = z^2+c;
end
end