Why sprint doesn't show a zero value from an array?
显示 更早的评论
I have the following code, and when I run everthing works perfectly except in the third column of the figure, the title of the histrogram should show zero instead of -14.
u_0=randi(4000,4000);
p=[0 3 7 30 33 36 56 60 64];
c=[-14 -12 -8;-8 -6 -4;0 0 0];
z=[60 40 20;60 40 20;60 40 20];
fig1=figure('Name','Fig 1');
fig1.WindowState = 'Maximized';
a=[1:450:2251];
for s=1:1:9
subplot(3,3,s)
for r=a(:)+p(s)
f1=u_0(:,r);
f1(f1==0)=NaN;
if s==1 || s==4 || s==7
edge=(0:1:25);
elseif s==2 || s==5 || s==8
edge=(0:4:90);
else s=3 | s==6 | s==9;
edge=(0:10:250);
end
hold on
h1=histogram(f1(:,1),"NumBins",20,'Normalization','pdf',"BinEdges",edge);
h2=histogram(f1(:,4),"NumBins",20,'Normalization','pdf',"BinEdges",edge);
h3=histogram(f1(:,6),"NumBins",20,'Normalization','pdf',"BinEdges",edge);
hold off
title(sprintf('D, (25^{o}C,%d,%d)',c(s),z(s)));
end
end
采纳的回答
Reduce the clutter:
a=[1:450:2251];
% Easier:
a = 1:450:2251;
[] is Matlab operator for a concatenation. [1:450:2251] concates the vector 1:450:2251 with nothing.
for r=a(:)+p(s)
a(:) is a column vector. Remember, that Matlab's for processes arrays columnwise. So r is set to a(:)+p(s) once and the loop does not run over the elements. Maybe you mean:
for r = a(:).' + p(s)
This is a real loop. Maybe you want this instead:
r = a(:) + p(s); % Without FOR
Whjat is the purpose of:
else s==7 | s==8 | s==9;
? It compares s with 7, 8 and 9, but does not produce an output. The else does not use a condition. So the "s==7 | s==8 | s==9" is dead code only without any effect.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Logical 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
