MATLAB: Determination between Inf and -Inf

detecting -inf

Hello,

I have a matrix that contains both inf and -inf points. I'm currently using isinf() to find the locations of these points however I have no way of knowing whether it's -inf or +inf.

The reason for this is because I later replace each inf with a predetermine value (5000) and I would like to be able to replace -inf with -5000 (but so far I haven't been able to detect them. Sample code below:

data(isinf(data)) = max(model.ub); %the max here will return 5000;

Thanks Al

Best Answer

a=[1 2 inf 3 -inf -4]
idx=isinf(a)
sign(a(idx))

Related Solutions

MATLAB: Nanmean returns -Inf

Hi Balsip

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

while

A=[NaN NaN -12 NaN NaN NaN NaN -1 -5 -8 NaN NaN NaN NaN -20 NaN NaN -3];
mean(A)
median(A)
nanmean(A)
nanmedian(A)
   =
     NaN
   =
     NaN
   =
     -8.1667
   =
     -6.5000

as expected, mean and median ignore NaN

mean([-12 -1 -5 -8 -20 -3])
 =
   -8.1667
 median([-12 -1 -5 -8 -20 -3])
 =
   -6.5000

however when 1/0

A=[NaN NaN -12 NaN NaN NaN NaN -1 -5 -8 NaN NaN 1/0 NaN -20 NaN NaN -3];
mean(A)
median(A)
nanmean(A)
nanmedian(A)
ans =
   NaN
ans =
   NaN
ans =
   Inf
ans =
    -5

or Inf are part of vector A,

A=[NaN NaN -12 NaN NaN NaN NaN -1 -5 -8 NaN NaN Inf NaN -20 NaN NaN -3];
mean(A)
median(A)
nanmean(A)
nanmedian(A)
ans =
   NaN
ans =
   NaN
ans =
   Inf
ans =
    -5

nanmean takes into account Inf values but nanmedian doesn't. This is because

Y1

and/or

X2

have one or more null values, introducing Inf s elements in A.

A=(X1./Y1)./(X2./Y2);

to avoid this, either directly correct Infs to NaNs

A(find(A==Inf))=NaN

or on Y1 and X2, remove their nulls

tol=.000001;
Y1(find(Y1==0))=tol;
X2(find(X2==0))=tol;

Let tol be a really small, small enough so it can be ignored.

It could also be that Y2 and/or X1 elements take Inf values but I assumed from the question that such is not the case, so only one or more elements of Y1 and X2 are null.

Balsip, please note that although

A=(X1./Y1)./(X2./Y2);

and

for i=1:length(X1)
        A(i)=(X1(i)./Y1(i))./(X2(i)./Y2(i));
end

are mathematically the same, the time consumption of the compact expression is 1 order of magnitude better than the for loop

L=1e7;
X1=randi([1 1e4],1,L);Y1=randi([1 1e4],1,L);
X2=randi([1 1e4],1,L);Y2=randi([1 1e4],1,L);
tic
A=(X1./Y1)./(X2./Y2);
toc
Elapsed time is 0.038204 seconds.
>> tic 
for i=1:length(X1)
        A(i)=(X1(i)./Y1(i))./(X2(i)./Y2(i));
end
toc
Elapsed time is 0.236449 seconds.

the operator ./ is optimised against the for loop you attempted to use a possible solution.

Balsip

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

MATLAB: Why i receive Nan for integral -inf to -inf

The integral from -inf to -inf over exp(-(x.^2)/2) is the same as calculating the area of the rectangle bounded by -inf and -inf with a height e(-(x.^2)/2) evaluated at -inf(which is 0).The difference between -infs will return an NaN in MATLAB . The NaN times the height(0) results in NaN as well. The documentation link for properties of NaN : https://www.mathworks.com/help/matlab/ref/nan.html

Best Answer

Related Solutions

MATLAB: Nanmean returns -Inf

MATLAB: Why i receive Nan for integral -inf to -inf

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