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Sum the daily data based on month and year

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Ryn
Ryn on 29 Apr 2019
Answered: ABHILASH SINGH on 13 Mar 2020
I have a three dimensional matrix of gridded data (16x18x7669) where 16 is the latitudes, 18 the longitudes and 7669 is the daily data from 1/1/1998 to 31/12/2018. I would like to get the monthly and yearly totals for each grid. I am a newbie in Matlab and I will appreciate suggestions on how to go about it. Thank you in advance

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Accepted Answer

Stephane Dauvillier
Stephane Dauvillier on 29 Apr 2019
Edited: Stephane Dauvillier on 29 Apr 2019
Lets assume your 3D variable is called data
Frist get the group id : which width of data correspond to the same month of the same year:
time=datetime(1998,1,1):datetime(2018,12,31);
grp = findgroups(year(time),month(time));
Then cget the unique list of group
uniqueGr = unique(grp);
Initialize the result and loop on unique groupe element
result = zeros(size(data,1),size(data,2),numel(uniqueGr)) ;
for iGroup = 1:numel(uniqueGr)
Compute the sum on the 3rd dimension but only for the data which corresponding date to the current group
result(:,:,iGroup) = sum(data(:,:,grp==uniqueGr(iGroup)),3);
end
Note there is the function splitapply that makes this easier if you have 2D array

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Ryn
Ryn on 30 Apr 2019
Hi Stephane,
This works but Iam getting unrealistic values and I cannot figure out what I am doing wrong. This is what I have tried for the yearly totals:
clear;close all;clc;
load CHIRPS2
%%yearlytotals
yGroup = findgroups(year(dateCHIRPS));
uniqueGrYr=unique(yGroup);
%Initialize the result and loop on unique groupe element
pCHIRPS_Y=zeros(size(precipCHIRPS2,1),size(precipCHIRPS2,2),numel(uniqueGrYr));
for iGroup = 1:numel(uniqueGrYr)
pCHIRPS_Y(:,:, iGroup) = sum (precipCHIRPS2 (:,:, yGroup == uniqueGrYr (yGroup)), 3);
end
Stephane Dauvillier
Stephane Dauvillier on 30 Apr 2019
What do you mean by "unrealistic values" ?
You ask to sum the whatever data you have for each year.
Let's just take a tiny example
years = [1990,1990,1991];
data(:,:,1) = [1,2;3,4;5,6] ;
data(:,:,2) = [7,8;9,10;11,12];
data(:,:,3) = [13,14;15,16;17,18];
Then you will have as a result
for year 1990
result(:,:,1) = data(:,:,1) + data(:,:,2) = [8,10;12,14;16,18]
and for year 1991
result(:,:,2) = data(:,:,3) = [13,14;15,16;17,18];

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