MatLab:从采样数据创建 3D 直方图 [英] MatLab: Create 3D Histogram from sampled data
问题描述
我在数组 transitions=zeros(101,101) 中的区间 [0,1] 中采样了数据，我想将其绘制为 3D 直方图.transitions 填充了类似于本线程末尾提供的示例数据的数据.
I have sampled data in the interval [0,1] in an Array transitions=zeros(101,101) which I want to plot as a 3D-histogram. transitions is filled with data similar to the example data provided at the end of this thread.
第一列是第一个观察变量X，第二列是第二个变量Y，第三列是归一化频率.IE.对于第一行:观察到的变量对 (0,0) 的归一化频率为 0.9459.(0,Y) 的归一化频率的总和是 1.
The first columns refers to the first observed variable X, the second column to the second variable Y and the third column is the normalized frequency. I.e. for the first row: the observed normalized frequency of the variable pair (0,0) is 0.9459. The sum of the normalized frequencies for (0,Y)thus is 1.
我尝试使用以下代码制作(某种)3D 直方图:
I tried to make (sort of) a 3D histogram with the following code:
x_c = (transitions(:,1) * 100)+1;
y = (transitions(:,2) * 100)+1;
z = transitions(:,4);
%A = zeros(10,10);
A = zeros(max(x_c),max(y));
for i = 1:length(x_c)
try
if(z(i)>0)
A(int32(x_c(i)), int32(y(i))) = abs(log(z(i)));
else
% deal with exceptions regarding log(0)
A(int32(x_c(i)), int32(y(i))) = 0;
end
catch
disp('');
end
end
bar3(A);
然而，由于它是离散空间 A 中的采样数据，因此输出如下图所示.这在某种程度上具有误导性，因为图中存在间隙"(对于没有采样数据的坐标，z 值 = 0).我宁愿将采样数据分配给相应的图，从而产生真实"的 3d 直方图.
However, since it is sampled data in a discrete space A the output looks like the plot below. This is somehow misleading as there are 'gaps' in the plot (z-value = 0 for coordinates where I have no sampled data). I rather would like to have the sampled data being assigned to their corresponding plots, thus resulting in a 'real' 3d histogram.
顺便说一下，由于我创建 A 的hack"，x-、y- 和 z-比例也不正确.3D 直方图的轴(所有三个)应该在 [0,1] 的区间内.
By the way, as a result of my 'hack' of creating A also the x-,y- and z-scale is not correct. The 3D histogram's axes (all three) should be in the interval of [0,1].
ans =
0 0 0.9459
0 0.0500 0.0256
0 0.1000 0.0098
0 0.1100 0.0004
0 0.1500 0.0055
0 0.1600 0.0002
0 0.2000 0.0034
0 0.2100 0.0001
0 0.2500 0.0024
0 0.2600 0.0001
0 0.3000 0.0018
0 0.3200 0.0000
0 0.3700 0.0000
0 0.4000 0.0010
0 0.4200 0.0000
0 0.4500 0.0007
0 0.5000 0.0007
0 0.5300 0.0000
0 0.5500 0.0005
0 0.6000 0.0005
0 0.6300 0.0000
0 0.7000 0.0002
0 0.7400 0
0 0.7500 0.0003
0 0.7900 0.0000
0 0.8000 0.0002
0 0.8400 0.0000
0 0.8500 0.0002
0 0.8900 0.0000
0 0.9000 0.0002
0 0.9500 0.0001
0 1.0000 0.0001
0.0500 0 0.0235
0.0500 0.0500 0.0086
0.0500 0.1000 0.0045
. . .
. . .
. . .
. . .
. . .
0.9500 0.9000 0.0035
0.9500 0.9500 0.0066
0.9500 1.0000 0.0180
1.0000 0 0.0001
1.0000 0.0500 0.0001
1.0000 0.1000 0.0001
1.0000 0.1100 0.0000
1.0000 0.1500 0.0001
1.0000 0.1600 0.0000
1.0000 0.2000 0.0001
1.0000 0.2100 0.0000
1.0000 0.2500 0.0001
1.0000 0.2600 0.0000
1.0000 0.3000 0.0001
1.0000 0.3200 0.0000
1.0000 0.3700 0.0000
1.0000 0.4000 0.0002
1.0000 0.4200 0
1.0000 0.4500 0.0002
1.0000 0.5000 0.0003
1.0000 0.5300 0.0000
1.0000 0.5500 0.0004
1.0000 0.6000 0.0004
1.0000 0.6300 0.0000
1.0000 0.7000 0.0007
1.0000 0.7400 0.0000
1.0000 0.7500 0.0010
1.0000 0.7900 0.0000
1.0000 0.8000 0.0015
1.0000 0.8400 0.0001
1.0000 0.8500 0.0024
1.0000 0.8900 0.0002
1.0000 0.9000 0.0042
1.0000 0.9500 0.0111
1.0000 1.0000 0.3998
推荐答案
我通过处理非聚合数据找到了解决方案.特别地，数据集transitions 的每一行都包含对X 和Y 的一个观察.我使用下面的代码生成规范化的 3D 直方图(和 2D 地图)，如下所示:
I found a solution by working on the non-aggregated data. In particular each row of the data set transitions contains one observation of Xand Y. I used the code below to produce a normalized 3D histogram (and a 2D map) as folllows:
function createHistogram(transitions)
uniqueValues = unique(transitions(:,1));
biases = cell(numel(uniqueValues),1);
for i = 1:numel(uniqueValues)
start = min(find(transitions(:,1) == uniqueValues(i)));
stop = max(find(transitions(:,1) == uniqueValues(i)));
biases(i) = mat2cell(transitions(start:stop,2));
end
combinedBiases = padcat(biases{1},biases{2},biases{3},biases{4},...
biases{5},biases{6},biases{7},biases{8},biases{9},biases{10},...
biases{11},biases{12},biases{13},biases{14},biases{15},biases{16},...
biases{17},biases{18},biases{19});
bins = 0:0.1:1;
[f, x] = hist(combinedBiases, bins);
%
% normalize
%
for i = 1:numel(f(1,:))
for j = 1:numel(f(:,i))
f(j,i) = f(j,i)/numel(biases{i});
end
end
bHandle = bar3(x, f);
ylim([-0.04,1.04])
for k = 1:length(bHandle)
zdata = get(bHandle(k),'ZData');
set(bHandle(k),'CData',zdata, 'FaceColor','interp');
end
colormap('autumn');
hcol = colorbar();
axis('square');
cpos=get(hcol,'Position');
cpos(4)=cpos(4)/3; % Halve the thickness
cpos(2)=0.4; % Move it down outside the plot#
cpos(1)=0.82;
set(hcol, 'Position',cpos);
xlabel('Enrollment biases');
ylabel('Aging biases');
zlabel('Bias transition probability');
title(strcat('Probability mass function of bias transitions (', device,')'));
set(gca,'XTick',0:2:20);
set(gca,'XTickLabel',0:0.1:1);
print('-dpng','-r600',strcat('tau_PMF3D_enrollment-ageing-', device));
view(2);
cpos(1)=0.84;
set(hcol, 'Position',cpos);
print('-dpng','-r600',strcat('tau_PMF2D_enrollment-ageing-', device));
end
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