Finding average data from a large data matrix
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Matlab_G
am 18 Nov. 2022
Kommentiert: Cris LaPierre
am 21 Nov. 2022
Hello,
I have a data matrix in the following format:
Row 1 is the years from 1990 to 2000
Row 2 is the month
Row 3 is the day
Row 4 is the data
Ex:
1990 1990 1990 1990…..until 2000
1 1 1 1 1 1 1 1…2 2 2 2 2 2….3 3 3 3
1 2 3 4 5 6
How can I calculate the average data of each month (ie. all the jan, feb, together)? The problem that I am encountering is finding a systematic way of dividing the data, given its format.
Thanks in advance
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Benjamin Thompson
am 18 Nov. 2022
An index vector over the available columns would work well here. Cannot be sure without a complete example of the data. Then pass the index vector as the second subscript argument for the submatrix of A that you want to pass to the "mean" function to get the average.
>> A = [1990 1990 1991 1991 1992 2000; 1 2 3 4 1 2; 1 2 3 4 5 6; 1 2 3 4 5 6]
A =
1990 1990 1991 1991 1992 2000
1 2 3 4 1 2
1 2 3 4 5 6
1 2 3 4 5 6
>> I_feb = A(2,:) == 2
I_feb =
1×6 logical array
0 1 0 0 0 1
>> mean(A(4,I_feb))
ans =
4
2 Kommentare
Benjamin Thompson
am 21 Nov. 2022
The index vector can use a more complex logic test:
I_two_conditions = (A(1,:) == 2000) & (A(2,:) == 1)
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Cris LaPierre
am 18 Nov. 2022
data = [1990 1990 1990 1990 1995 1995 1995 1995;
1 1 2 2 1 1 2 2;
1 2 1 2 1 2 1 2;
3 5 7 4 6 8 1 9]'
avgData = groupsummary(data(:,4),[data(:,1),data(:,2)],'mean')
This is probably easier to understand if you turn your data into a table.
dataT = array2table(data,'VariableNames',["Year","Month","Day","Data"])
avgDataT = groupsummary(dataT,["Year","Month"],"mean","Data")
One more may to do this is if you turn your date info into a datetime variable
dataT2 = table(datetime(data(:,1:3)),data(:,4),'VariableNames',["Date","Data"])
avgDataT2 = groupsummary(dataT2,"Date","month","mean","Data")
0 Kommentare
dpb
am 18 Nov. 2022
tData=array2table(X.','VariableNames',{'Year','Month','Day','Data'});
tMonthAvg=rowfun(@mean,tData,'GroupingVariables',{'Month'}, ...
'InputVariables','Data','OutputVariableNames','MonthAverage');
1 Kommentar
Cris LaPierre
am 21 Nov. 2022
A= [2000 2000 2000 2001 2001;
1 1 2 1 1;
5 3 8 9 7]'
groupsummary(A(:,3),[A(:,1) A(:,2)],'mean')
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