computing montly averages when time series observations are irregural

8 vues (au cours des 30 derniers jours)
antonet
antonet le 28 Juil 2012
Dear all,
I have the following problem
Let
A ={
'02/11/08' [0.2150] [429.5674] [ 100] [2.2678e+04]
'30/11/08' [0.2117] [463.1360] [ 100] [2.4383e+04]
'28/12/08' [0.2209] [436.6316] [ 100] [2.3148e+04]
'25/01/09' [0.2212] [441.5430] [ 100] [2.2987e+04]
'22/02/09' [0.2201] [453.7015] [ 100] [2.3675e+04]
'22/03/09' [0.2185] [461.3925] [ 100] [2.3680e+04]
'19/04/09' [0.2104] [486.4095] [ 100] [2.5367e+04]
'17/05/09' [0.2162] [451.4833] [ 100] [2.3986e+04]
'14/06/09' [0.2158] [475.8620] [ 100] [2.4245e+04]
'12/07/09' [0.2211] [449.4574] [ 100] [2.2766e+04]
'09/08/09' [0.2221] [456.2507] [ 100] [2.2523e+04]
'06/09/09' [0.2175] [472.2659] [ 100] [2.3593e+04]
'04/10/09' [0.2182] [479.3408] [ 100] [2.4359e+04]
'01/11/09' [0.2286] [442.6719] [ 100] [2.1490e+04]
'29/11/09' [0.2211] [481.4548] [ 100] [2.4054e+04]
'27/12/09' [0.2259] [468.2757] [ 100] [2.3037e+04]
'31/01/10' [0.2300] [461.5581] [ 100] [2.2050e+04]
'28/02/10' [0.2259] [487.6257] [ 100] [2.3293e+04]
'28/03/10' [0.2200] [529.8777] [ 100] [2.5493e+04]
'25/04/10' [0.2315] [433.3039] [ 100] [2.0387e+04]
'23/05/10' [0.2274] [500.8603] [ 100] [2.4019e+04]
}
As you can see from the first column I have irregular time series observations. Each observation can represent a 4-week or a 5-week or a 6-week average. For example the observation [0.2117] which is the second value of the second column is the average of the values that have been recorded between ‘03/11/08' and '30/11/08' In order to be able to analyze econometrically these data and run my regressions I need to obtain regular time series observations. So I need to convert these values to monthly averages and have something like
Let A ={
'11/08' [mon. average] [mon. average] [mon. average]
'12/08' [mon. average] [mon. average] [mon. average]
'01/09' [mon. average] [mon. average] [mon. average]
'02/09' [mon. average] [mon. average] [mon. average]
'03/09' [mon. average] [mon. average] [mon. average]
'04/09' [mon. average] [mon. average] [mon. average]
'05/09' [mon. average] [mon. average] [mon. average]
'06/09' [mon. average] [mon. average] [mon. average]
'07/09' [mon. average] [mon. average] [mon. average]
'08/09' [mon. average] [mon. average] [mon. average]
'09/09' [mon. average] [mon. average] [mon. average]
'10/09' [mon. average] [mon. average] [mon. average]
'11/09' [mon. average] [mon. average] [mon. average]
'12/09' [mon. average] [mon. average] [mon. average]
}
Could interpolation be a solution to this problem? I would be grateful if you could provide some code.
cheers
  3 commentaires
antonet
antonet le 29 Juil 2012
Modifié(e) : antonet le 29 Juil 2012
Hi imager! yes as you can see there is overlapping across the dates as each value is the average of the values that have been observed 4 weeks (or 5 weeks or 6 weeks) back. Advice me what is the best approach to calculate the monthly averages. PLease ask me if you require further information
antonet
antonet le 29 Juil 2012
Imager HELP!
thanks

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per isakson
per isakson le 28 Juil 2012
Modifié(e) : per isakson le 28 Juil 2012
Hint:
  • vec = datevec( A{:,1}, 'dd/mm/yy' )
  • out = accumarray( vec(:,2), A(something), [], @mean )
Better
vec = datevec( char(A{:,1}), 'dd/mm/yy' );
out = accumarray( vec(:,2:3), transpose([A{:,2}]), [], @mean );
This returns the means of the values of each month respectively. There is a value for each month. Is that what you want?
  1 commentaire
antonet
antonet le 28 Juil 2012
I will check the outcome and I will tell you. Thank you so much so the code!

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