finding the mean based on a specific value in other column

50 visualizzazioni (ultimi 30 giorni)
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

Risposta accettata

Asad (Mehrzad) Khoddam
Asad (Mehrzad) Khoddam il 9 Ott 2020
Modificato: Asad (Mehrzad) Khoddam il 9 Ott 2020
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 Commenti
Najam us Saqib Fraz
Najam us Saqib Fraz il 9 Ott 2020
Thank You !! But I get NaN while use both the approaches !! The data I showed is a little part of my huge data...While importing it in Matlab and using your suggested solution it gives me NaN.....
Asad (Mehrzad) Khoddam
Asad (Mehrzad) Khoddam il 9 Ott 2020
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');

Accedi per commentare.

Più risposte (2)

madhan ravi
madhan ravi il 9 Ott 2020
Modificato: madhan ravi il 9 Ott 2020
ix = (abs(column_1 - 527.1235) < 1e-4) &...
(abs(column_1 - 927.5) < 1e-1);
M = mean(column_4(ix))

Jon
Jon il 9 Ott 2020
Modificato: Jon il 9 Ott 2020
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 Commenti
Asad (Mehrzad) Khoddam
Asad (Mehrzad) Khoddam il 9 Ott 2020
When data is missing in the data file, it shows as NaN. You can remove the lines that are NaN
Jon
Jon il 9 Ott 2020
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

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by