Calculating combinations of dates

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antonet
antonet on 30 Jun 2012
Dear all,
I have the following cell vector of dates
jji= '23/11/08'
'28/12/08'
'25/01/09'
that corresponds to the cell vector of values
values
[0.5637]
[0.6034]
[0.5943]
[0.4567 ]
and I want to create the following new vector of values
(23*0.5637+7*0.6034)/30
(28*0.6034+3*0.5943)/31
(25*5943+6*0.4567)/31
where in the first row of this vector the numerator (23*0.5637+7*0.6034) is the sum of two products.
In the first product which is 23*0.5637 the number 23 is actually the first two digits from '23/11/08'
In the second product which is 7*0.6034 the number 7 is actually the distance in days between '23/11/08' and the end date of that month, that is the difference
'23/11/08-30/11/08
The denominator in the first row (30) is the total number of days for month November 2008
Similar analysis holds for the rest of the rows.
In my case I have a huge vector of dates and values. is there any way to construct the new vector of values?
My proposal is this:
jji={ '23/11/08'
'28/12/08'
'25/01/09'
};
values ={
0.5637
0.6034
0.5943
0.4567 };
for i = 1:length(jji)
jjii{i,1} = jji{i,1}(1:2);
end
that produces
jjii =
'23'
'28'
'25'
TO find the end dates of each month i use
[Y, M] = datevec(jji,'dd/mm/yy'); enddates = datestr(datenum(Y,M+1,1)-1,'dd/mm/yy')
and I obtain
30/11/08 31/12/08 31/01/09
but then I need to calculate the differences between the jji and "endndates"
Any suggestions?
thank you
  1 Comment
antonet
antonet on 30 Jun 2012
I think I found how to calculate this difference
datenum(enddates,'dd/mm/yy') - datenum(jji,'dd/mm/yy')
But is there a better approach?
THanks

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Accepted Answer

Andrei Bobrov
Andrei Bobrov on 30 Jun 2012
Edited: Andrei Bobrov on 30 Jun 2012
EDIT
[Y M D] = datevec(jji,'dd/mm/yyyy');
ED = eomday(Y, M);
v = cell2mat(values);
n = numel(v);
out = sum([D, ED-D].*v(hankel(1:n-1,n+[-1 0])),2)./ED;
  5 Comments
antonet
antonet on 30 Jun 2012
Thanks Andrei.
I would like to ask what numel does in : n= numel(values);
Could we alternative have
kk=size(values);
n=kk(1,1);
thank you

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