NaN when calculating average

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Pankaj
Pankaj el 22 de En. de 2012
I read data from a .xls file using xlsread. There are some 3000 rows with price data, all in number format in excel. I further try to calculate the average of previous 10 days for every day using mean2(prices(i-10,1):prices(i,1)). I do this for last 2990 rows. surprisingly a lot of these throw NaN. I have also run a check "isnan" on each element in prices and none of them tests positive. also stumbled on something:
isnumeric(NaN)
ans =
1
how is this working??

Respuesta aceptada

David Young
David Young el 22 de En. de 2012
Although a NaN is not a number, it is of class double, and so it is numeric. If a NaN was not numeric, it could not be held in a matrix with other numbers.
Your problem is that prices(i-10,1):prices(i,1) may be empty, so mean2 returns NaN. In fact, that expression doesn't look likely to be correct. I suspect you intended to use mean2(prices(i-10:i, 1)).

Más respuestas (2)

Andrew Newell
Andrew Newell el 22 de En. de 2012
You could use nanmean to ignore the NaN's while calculating means:
prices = rand(20,20);
n = size(prices,1); m = size(prices,2); window=10;
runningMean = zeros(n-window,m);
for ii=1:n-window
runningMean(ii,:) = nanmean(runningMean(ii:ii+window-1,:));
end
Note that I am interpreting your statement of "previous 10 days" as the ten days before the current day. Your version is actually averaging 11 days.
  6 comentarios
Atakan
Atakan el 22 de En. de 2012
This is my code:
function [randnormal]=atakan(a,b,m)
randnormal=[];
count=1;
while (count<=m)
R = normrnd(a,b);
u1=unifrnd(0,1);
u2=unifrnd(0,1);
y=tan(pi*(u1-1/2));
if (u2<=((sqrt(exp(1))/2)*(1+y^2)*(exp(1)^(-y^2/2))))
randnormal=[R;y];
end
count=count+1;
end
Pankaj
Pankaj el 23 de En. de 2012
Thanks Andrew - good solution - it helps!

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Atakan
Atakan el 22 de En. de 2012
Hi; I want to ask a question to the main page. But it gives an error such that "we are sorry but something went wrong". What can I do?
  3 comentarios
Atakan
Atakan el 22 de En. de 2012
I have tried many many times. But there is still a problem. If I send the question to you, can you ask the question for me? Because this is very important for me. thanks...
Jan
Jan el 22 de En. de 2012
Dear Atakan: The servers have severe problems currently.

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