# finding the mean based on a specific value in other column

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Najam us Saqib Fraz 2020년 10월 9일
댓글: Jon 2020년 10월 9일
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 댓글을 달려면 로그인하십시오.

### 채택된 답변

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)
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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');

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### 추가 답변(2개)

편집: madhan ravi 2020년 10월 9일
ix = (abs(column_1 - 527.1235) < 1e-4) &...
(abs(column_1 - 927.5) < 1e-1);
M = mean(column_4(ix))
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Jon 2020년 10월 9일
편집: Jon 2020년 10월 9일
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))
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Jon 2020년 10월 9일
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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