finding the mean based on a specific value in other column
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Guys I have following data as an example.The data contain 4 coloumns.want to average 4th coloumn when 1st couloumn is equal to 527.1235 and third coloumn is 927.5![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/376146/image.jpeg)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/376146/image.jpeg)
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Asad (Mehrzad) Khoddam
2020-10-9
编辑:Asad (Mehrzad) Khoddam
2020-10-9
You can find the rows with the first condition and the other rows for the second condition. The intersection of the two rows, are the row numbers that satisfy both conditions:
rows = intersect(find(a(:,1)==527.1235), find(a(:,3)==927.5));
% average of the above rows
avg = mean(a(rows,4));
disp(avg)
or simpler :
rows = find(a(:,1)==527.1235 & a(:,3)==927.5);
% average of the above rows
avg = mean(a(rows,4));
disp(avg)
3 个评论
Asad (Mehrzad) Khoddam
2020-10-9
When data is missing in the data file, it shows as NaN. You can remove the lines that are NaN
use this option:
avg = mean(a(rows,4),'omitnan');
更多回答(2 个)
madhan ravi
2020-10-9
编辑:madhan ravi
2020-10-9
ix = (abs(column_1 - 527.1235) < 1e-4) &...
(abs(column_1 - 927.5) < 1e-1);
M = mean(column_4(ix))
0 个评论
Jon
2020-10-9
编辑:Jon
2020-10-9
Lets say you have put your data into an array X.
Find a logical index where the rows match your criteria using:
criteria = [527.1235 927.5]
idl = ismember(X(:,[1,3]),criteria,'rows')
then do the averaging on the 4th colulmn for the rows where the criteria matches
xMean = mean(X(idl,4))
5 个评论
Asad (Mehrzad) Khoddam
2020-10-9
When data is missing in the data file, it shows as NaN. You can remove the lines that are NaN
Jon
2020-10-9
I think you are trying to show that floating point comparisons could be a problem if they are not exact. I assumed they were exact, in any case given your example I get idl = 1 not 0
>> X = [527.1235 1.0000 927.5000],criteria =[527.1235 927.5000]
X =
527.1235 1.0000 927.5000
criteria =
527.1235 927.5000
>> idl = ismember(X(:,[1,3]),criteria,'rows')
idl =
logical
1
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