how to treat the last observation? Forecasting??

1 回表示 (過去 30 日間)
antonet
antonet 2012 年 8 月 18 日
Dear all,
I have the following cell matrix
A={
'24/09/2000' [4.1583]
'22/10/2000' [3.9389]
'19/11/2000' [4.3944]
'17/12/2000' [3.9313]
'14/01/2001' [3.9313]
'11/02/2001' [3.9313]
'11/03/2001' [4.0339]}
I can easily convert the above data points to monthly averages; through the calculation of the weighted average between successive observations.
For instance for the first observation I have
24*4.1583+ 6*3.9389
where 24 is the 24 days from month Octomber and 6=(last date of the month-24=30-24) The only problem is how to treat the last observation:
11*4.0339+?
One way is to extrapolate but I do not consider this method reliable. The above values are prices. So could I apply some other way? Maybe forecasting taking into account seasonality?
Any help/code provided will greatly appreciated
thanks
  4 件のコメント
Walter Roberson
Walter Roberson 2012 年 8 月 18 日
You do not have enough data to forecast with, not with any meaningful accuracy.
antonet
antonet 2012 年 8 月 18 日
Hi Walter. I have. the above matrix is an example

サインインしてコメントする。

採用された回答

Oleg Komarov
Oleg Komarov 2012 年 8 月 18 日
A ={'24/09/2000' 4.1583
'22/10/2000' 3.9389
'19/11/2000' 4.3944
'17/12/2000' 3.9313
'14/01/2001' 3.9313
'11/02/2001' 3.9313
'11/03/2001' 4.0339};
% Convert to tractable double matrix, i.e. use serial dates
A = [datenum(A(:,1),'dd/mm/yyyy') cat(1,A{:,2})];
A =
730753 4.1583
730781 3.9389
730809 4.3944
730837 3.9313
730865 3.9313
730893 3.9313
730921 4.0339
% Create a list of end-of-month dates (using two tricks):
  1. create first-day-of-next-month dates and subtract one
  2. datenum() automatically adjusts the date in case one of the units exceeds the max accepted value
xi = datenum(2000,10:16,1)-1;
% Interpolate
B = interp1(A(:,1),A(:,2), xi,'linear','extrap');
% Plot
plot(A(:,1),A(:,2),'-dr',xi,B,'-ob')
set(gca,'Xtick',sort([A(:,1); xi']),'xgrid','on')
datetick('x','dd/mm','keepticks')

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeDates and Time についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by