data frequency conversion -cell matrix

Dear all,
I have the following cell matrix
A = {
1 ' ' [ NaN] [ NaN]
1 ' ' [ NaN] [ NaN]
1 ' ' [ NaN] [ NaN]
1 'MA 2009' [ 0] [ 0]
1 'MJ 2009' [ 0.2680] [ 3.0394]
1 'JA 2009' [ 0.0504] [ 0.6475]
1 'SO 2009' [ 14.0985] [ 148.2583]
1 'ND 2009' [ 0.1128] [ 1.1506]
1 'JF 2010' [ NaN] [ 148.2583]
1 'MA 2010' [ 2.5852] [ 34.0146]
1 'MJ 2010' [ 0.3220] [ 3.2846]
1 'JA 2010' [ 14.0985] [ 148.2583]
1 'SO 2010' [ 2.5852] [ NaN]
1 'ND 2010' [ 0.2938] [ 2.8540]
1 'JF 2011' [ 0.1128] [ 1.1506]
1 'MA 2011' [ 14.0985] [ 148.2583]
1 'MJ 2011' [ 2.1091] [ 15.0233]
1 'JA 2011' [ 0] [ 0]
2 ' ' [ NaN] [ NaN]
2 ' ' [ NaN] [ NaN]
2 ' ' [ NaN] [ NaN]
2 'MA 2009' [ 14.0985] [ 148.2583]
2 'MJ 2009' [ 2.7827] [ 18.9879]
2 'JA 2009' [ 11.8755] [ 126.4359]
2 'SO 2009' [ 0.0589] [ 0.6685]
2 'ND 2009' [ 11.8755] [ 126.4359]
2 'JF 2010' [ 0.0504] [ 0.6475]
2 'MA 2010' [ 11.8755] [ 126.4359]
2 'MJ 2010' [ 0.0504] [ 0.6475]
2 'JA 2010' [ 0] [ 0]
2 'SO 2010' [ 0.0248] [ 0.2823]
2 'ND 2010' [ 0] [ 0]
2 'JF 2011' [ 2.5852] [ 34.0146]
2 'MA 2011' [ 0.0207] [ 0.2282]
2 'MJ 2011' [ 11.8755] [ 126.4359]
2 'JA 2011' [ 14.0985] [ 148.2583]
3 ' ' [ NaN] [ NaN]
3 ' ' [ NaN] [ NaN]
3 ' ' [ NaN] [ NaN]
3 'MA 2009' [ 2.1091] [ 15.0233]
3 'MJ 2009' [ 0] [ 0]
3 'JA 2009' [ 0.1128] [ 1.1506]
3 'SO 2009' [ 0.0207] [ 0.2282]
3 'ND 2009' [ 0] [ 0]
3 'JF 2010' [ NaN] [ 1.1506]
3 'MA 2010' [ 0] [ 0]
3 'MJ 2010' [ 2.1091] [ 15.0233]
3 'JA 2010' [ 0] [ 0]
3 'SO 2010' [ 2.7827] [ NaN]
3 'ND 2010' [ 0] [ 0]
3 'JF 2011' [ 0.0207] [ 0.2282]
3 'MA 2011' [ 2.5852] [ 34.0146]
3 'MJ 2011' [ 0] [ 0]
3 'JA 2011' [ 11.8755] [ 126.4359]
}
I want to convert these data from Bimontly to monthly for each inividual i (first column).
I searched for relevant functions and I found for example: tomonthly() but I do not know how to exactly apply it to the above setting. Note that I have 30000 invividuals and 20 numerical columns instead of the last 2 that I display above
Any help is greately appreicated thanks in advance

9 comentarios

salva
salva el 26 de Jul. de 2012
If something is unclear please inform me
thanks
You write:
"I want to convert these data from Bimontly to monthly
for each inividual i (first column)."
but you don't write by what rule that should be done. Divide the bimonthly value by two?
salva
salva el 27 de Jul. de 2012
Editada: salva el 27 de Jul. de 2012
thank you per.
I was thinking somethink like
A = {
1 ' '
1 ' '
1 ' '
1 '3/2009'
1 '4/2009'
1 '5/2009'
1 '6/2009'
}
and so forth where as you can see the MA 2009 date is split up in 3/2009 and 4/2009. NOw for each month I have to change accordingly the last two columns. I think that one way to do that is to divide the bimontly values by two. The other way would be to interpolate each bimontly value to monthly values. Could I also use the tomonthly function?
thanks you
per isakson
per isakson el 27 de Jul. de 2012
Editada: per isakson el 27 de Jul. de 2012
Documentation of tomonthly says:
newfts = tomonthly(oldfts)
where
oldfts is Financial time series object
I don't have the Financial toolbox. However, the documentation indicates that tomonthly calculates monthly data from data with higher resolution, e.g. daily. I don't think it can take bimonthly data.
I cannot make sense of your example. Am I right that
1, 'MJ 2009', 0.2680, 3.0394
is data for May and June 2009, i.e. 5/2009 and 6/2009? In your example you assign these numbers to 4/2009.
salva
salva el 27 de Jul. de 2012
Editada: salva el 27 de Jul. de 2012
Yes, you are right. the last two colums are wrong. I just wanted to show the split up of the dates and I left the last two columns as they are.
So you propose to divide each bimontly value with 2?
But then for 3/2009 and 4/2009 I will have the same values? Am I right?Does it make sense to use interpolation?
thanks
salva
salva el 27 de Jul. de 2012
Do you need more information? Please let me know
thanks
What would you like to have instead of ?
A = {...
1 'MA 2009' [ 0] [ 0]
1 'MJ 2009' [ 0.2680] [ 3.0394]
1 'JA 2009' [ 0.0504] [ 0.6475]}
in
Aout = {...
1 '3/2009' [ ?] [ ?]
1 '4/2009' [ ?] [ ?]
1 '5/2009' [ ?] [ ?]
1 '6/2009' [ ?] [ ?]
1 '7/2009' [ ?] [ ?]
1 '8/2009' [ ?] [ ?]}
salva
salva el 27 de Jul. de 2012
yes exactly I want to start with A and obtain Aout. Please helppppp!
thanks
Oleg Komarov
Oleg Komarov el 27 de Jul. de 2012
@Andrei: he wants to divide it by 2.

Iniciar sesión para comentar.

 Respuesta aceptada

Andrei Bobrov
Andrei Bobrov el 28 de Jul. de 2012
n = cellfun(@(x)x/2,A(:,3:end),'un',0);
Out = [];
for ii = 1:size(A,1)
if strcmp(A(ii,2),' ')
Out = [Out; A(ii,:)];
else
[~,k] = ismember(A{ii,2}(1:2),{'JF','MA','MJ','JA','SO','ND'});
y = A{ii,2}(4:end);
Out = [Out;
[A(ii,1),{[sprintf('%d',k*2-1),'/',y]},n(ii,:);
A(ii,1),{[sprintf('%d',k*2),'/',y]},n(ii,:)]];
end
end

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