Question

我想绘制正态分布图，但在同一图中通过循环绘制多个图。对于第一次迭代，它的精细，即一个图中的3个图，但在其他3个图中，它只绘制一个。代码

load iris.dat;
iris;
number_of_features = 4;
classes = iris(:,5);
number_of_classes = length(unique(classes));

for class_number = 1:number_of_classes
likelihood{class_number} = struct;

for feature_number = 1:number_of_features 
    likelihood{class_number}.pd{feature_number} = fitdist(iris(find(iris(:, 5) == class_number),feature_number),'Normal');
end
end
hold all
for feature_number = 1:number_of_features
for class_number=1:number_of_classes
  ppd=iris(find(iris(:, 5) == class_number),feature_number); 
     y=normpdf(ppd,likelihood{class_number}.pd{feature_number}.mu,likelihood{class_number}.pd{feature_number}.sigma);
  figure(feature_number)
  stem(ppd,y)
end
end

如果有人可以检查它并告诉我们代码中的问题

Answer 1

我不确定，但下面的代码可能有效。我刚改变了你的命令。

load iris.dat;
iris;
number_of_features = 4;
classes = iris(:,5);
number_of_classes = length(unique(classes));

for class_number = 1:number_of_classes
likelihood{class_number} = struct;

for feature_number = 1:number_of_features 
    likelihood{class_number}.pd{feature_number} = fitdist(iris(find(iris(:, 5) == class_number),feature_number),'Normal');
end
end

for feature_number = 1:number_of_features
figure(feature_number)
hold on
for class_number=1:number_of_classes
  ppd=iris(find(iris(:, 5) == class_number),feature_number); 
     y=normpdf(ppd,likelihood{class_number}.pd{feature_number}.mu,likelihood{class_number}.pd{feature_number}.sigma);
  stem(ppd,y)
end
hold off
end

通过matlab中的循环在一个图中绘制多个绘图

1 个答案: