Monthly sum function with nansum not being used correctly

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Augusto Gabriel da Costa Pereira
Augusto Gabriel da Costa Pereira 2022 年 11 月 8 日
コメント済み: Mathieu NOE 2022 年 11 月 9 日
I use the function below to perform the monthly precipitation accumulator, however, when there are days without NaN measurements in a given month, the result for that month is zero. But this result is wrong, the correct series for this month is NaN and not zero.
Another problem seen in the function is the result of months. From January 2000 to December 2020 there are 252 months. when I use the function, the result is only 251 months, that is, one month to go
Below is:
1 - the function used for the monthly sum.
2 - rainfall data, in column 1 the days of the year and in columns 2, 3, 4 and 5 the rainfall (INMET_Diario.mat);
function Y = monthly_sum(time_X,X)
[l,c] = size(X);
grp = [];
for y = 2000:2020 % Change period here
for m = 1:12
grp = [grp datenum(y,m,1)];
end
end
X_hour = zeros(length(grp)-1,c);
for i = 1:length(grp)-1
if(mod(i,500)==0)
disp(i)
end
for j = 1:c
X_hour(i,j) = nansum(X(time_X>=grp(i) & time_X<grp(i+1),j));
end
end
Y = [grp(1:length(grp)-1)', X_hour];
Best Regards,
AP
  3 件のコメント
Augusto Gabriel da Costa Pereira
Augusto Gabriel da Costa Pereira 2022 年 11 月 8 日
If my goal is to AVERAGE, replace @sum with @mean on the last line?
Mathieu NOE
Mathieu NOE 2022 年 11 月 9 日
yes that works also
A = [19260702 0.026 0.000 NaN 1.175
19260705 0.036 0.002 NaN 2.175
19260715 0.044 0.003 NaN 3.175
19260816 0.042 0.007 NaN 25.928];
% Matrix to table conversion
T = array2table(A,'VariableNames',{'DATE','S1','S2','S3','S4'});
% We add a new column "month"
T.MONTH = floor(T.DATE/100);
% varfun can apply a custom function to your table and group the result according
% to one (or more) variable(s)
Result_sum = varfun(@sum,T,'InputVariables',{'S1','S2','S3','S4'},'GroupingVariables','MONTH')
Result_mean = varfun(@mean,T,'InputVariables',{'S1','S2','S3','S4'},'GroupingVariables','MONTH')

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採用された回答

Jan
Jan 2022 年 11 月 8 日
編集済み: Jan 2022 年 11 月 8 日
function Y = monthly_sum(T, X)
grp = [datenum(2000, 1:252, 1), Inf];
ngrp = numel(grp) - 1;
c = size(X, 2);
X_hour = nan(ngrp, c);
for i = 1:ngrp
for j = 1:c
TX = X(T >= grp(i) & T < grp(i+1), j);
if ~isempty(TX)
X_hour(i,j) = nansum(TX);
end
end
end
Y = [grp(1:ngrp).', X_hour];
end
  2 件のコメント
Augusto Gabriel da Costa Pereira
Augusto Gabriel da Costa Pereira 2022 年 11 月 8 日
thank you Jan
problem fully resolved
Jan
Jan 2022 年 11 月 8 日
You are welcome.
datenum(2000, 1:252, 1) is tricky, but working fine.

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