Or maybe simpler I could just pick the first and the last day of the month and do a geometric mean of all values in between. Any ideas on how I can do that? Thank you very much
geometric mean month by month
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Dear All,
I need some help please.
I have two columns in a matrix B.
B=[A, X];
X is a column vector that contains different daily observations.
A is a column vector that contains the number of the month (for example for January it has "1" for february "2" etc till December where it has "12").
Column X has daily data for January, Feb till december. Thus Column A has numbers that change according to the month number. For instance, for january, Column A will have thirty one "1" corresponding to 31 days in january and twenty nine "2" corresponding to 29 days in February until the end of the year then i will start again with january with "1" and february with "2" etc till december .
What I would like to do is to do the gemetric mean of the values of X for similar values of A. Basically, it is like if Iam doing a geometric mean month by month untill the end of my data series. How can I do that please? I am stuck in this. ANy help is much appreciated. Best
S
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Réponses (3)
Eli Duenisch
le 23 Mai 2013
Modifié(e) : Azzi Abdelmalek
le 23 Mai 2013
Use logical indexing to obtain indices belonging to the same month and calculate the geometric mean for each month:
if true
for i=1:12, res(i)=geomean(X(A==i)); end
end
If you have no statistics toolbox installed you also need to write a function geomean() that calculates the geometric mean.
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Andrei Bobrov
le 23 Mai 2013
Modifié(e) : Andrei Bobrov
le 25 Mai 2013
out = accumarray(B(:,2),B(:,1),[],@geomean);
ADD
if the second column of your data is the same as:
b = arrayfun(@(x)x*ones(randi(7),1),repmat((1:12)',3,1),'un',0);
b = cat(1,b{:});
and all your array (B):
B = [randi(7,numel(b),1), b];
solution:
out = accumarray(B(:,2:3),B(:,1),[],@geomean);
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