I have a code for forecasting the 11th data based on the first 10 data by "exponential smoothing".
Now I want to modify this code to estimate NaN values in matrix z_new with the same method based on historical data.
I do not know how to relate these two problems. Can u help me?
Here is the original code.
z=[1,5,7,4,3,9,2,6,8,0]
n=length(z);a=[0.9 ];
yi1=[];
yi1(1)=NaN;
yi1(2)=z(1);
for i=3:11
yi1(i)=a(1)*z(i-1)+(1-a(1))*yi1(i-1);
end
RMSE1=...
sqrt((1/(n-1)*sum((z(2:end)-yi1(2:10)).^2)));
RM=[NaN,NaN,RMSE1];
i=1:11;
fprintf('\ti\t\tYi\t\t\tY^i(0.9)\n')
t=[i;z,NaN;yi1]';
t=[t;RM];
disp(t)
hold on
plot(i(1:10),z,'o:')
plot(i(2:11),yi1(2:11),'*g-')
legend('actual values','forecast:0.9')
hold off
% Now, how to estimate NaN in z_new with the same concept?
% z_new=[1,5,7,4,NaN,9,2,6,8,NaN]
The NaNs in your problem at least in the exemple are just there to adjust array size for plotting and could all be removed.
do you realise that this code
clear
z=[1,5,7,4,3,9,2,6,8,0]
n=length(z);a=[0.9 ];
yi1=[];
yi1(1)=z(1);
for i=2:10
yi1(i)=a(1)*z(i)+(1-a(1))*yi1(i-1);
end
figure(1);
plot(1:10,z,'o:');hold on
plot(2:11,yi1,'*g-')
legend('actual values','forecast:0.9')
hold off
produces the same plot with no NaN tricks / mess ?
if your data in z contains NaN,
you need to remove them using command isnan that gives you back a boolean
I have no clue what you wanna do but here is an example
clear
z=[1,5,NaN,4,3,9,2,6,8,0]
n=length(z);a=[0.9 ];
index = [];
yi1(1)=z(1);
for i=2:10
if ~isnan(z(i))
yi1(i)=a(1)*z(i)+(1-a(1))*yi1(i-1);
else
yi1(i)=yi1(i-1); % if z value is NaN then best prediction is that it did not change
index = [index,i]; %store the index where I had no clue
end
end
yi1(index) = NaN; % attributes NaN to all the value where I had no clue so maybe better not plot them.
figure(1);
plot(1:10,z,'o:');hold on
plot(2:11,yi1,'*g-')
legend('actual values','forecast:0.9')
hold off
- Badges
- Science topic