How to exclude certain values from an array?
显示 更早的评论
I am trying to plot a log equation, before I can I need to calculate x.
x can sometimes be returned as 0 or 1 so I need to exclude these points before I plot the log graph.
How do I achieve this?
So far I have tried this fairly inefficient way, and it doesnt seem to be excluding them from the array just not plotting them on the graph so that when i get to the last function to calculate p it doesnt calculate it for the whole array just one value:
xi1 = hits1./tot1
yi1=log(xi1./(1-xi1));
xplot1 = xi1
yplot1 = yi1;
xplot1(xplot1==0) = NaN;
yplot1(yplot1==0) = NaN;
xplot1(xplot1==1) = NaN;
yplot1(yplot1==1) = NaN;
xplot1(xplot1==-Inf) = NaN;
yplot1(yplot1==-Inf) = NaN;
xplot1(xplot1==Inf) = NaN;
yplot1(yplot1==Inf) = NaN;
figure(1)
plot(xplot1, yplot1,'linestyle','none','marker','o')
hold on
p1 = 4.8422;
p2 = -2.4067;
y = (p1.*xi1) + p2
plot(xi1, y)
hold off
p = 1/(1+exp(-((p1.*xi1) + p2)))
figure(3)
plot(xi1, p)
采纳的回答
Ameer Hamza
2020-11-16
编辑：Ameer Hamza
2020-11-16
If you want to remove them then use [] instead of NaN
xi1 = hits1./tot1
yi1=log(xi1./(1-xi1));
xplot1 = xi1
yplot1 = yi1;
mask = (xplot1==0) | (xplot1==1) | (isinf(xplot1));
xplot1(mask) = [];
yplot1(mask) = [];
figure(1)
plot(xplot1, yplot1,'linestyle','none','marker','o')
hold on
p1 = 4.8422;
p2 = -2.4067;
y = (p1.*xi1) + p2
plot(xi1, y)
hold off
p = 1/(1+exp(-((p1.*xi1) + p2)))
figure(3)
plot(xi1, p)
I haven't written a separate mask for yplot1 because I assume both have invalid elements at matching locations. If not, then you can similarly write a mask for yplot1
更多回答（1 个）
Rik
2020-11-16
1 个投票
Your current code is removing values from either the x or the y, instead of removing pairs. What you can do is define a logical array and mark all invalid positions. Then you can can remove the invalid point pairs (or set them to NaN, as you did).
For the last calculation you forgot to put in the dot for an element by element division.
类别
在 帮助中心 和 File Exchange 中查找有关 Annotations 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 
