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

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Najam us Saqib Fraz 2020 年 10 月 9 日
コメント済み: Jon 2020 年 10 月 9 日
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 サインインしてコメントする。

### 採用された回答

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 件のコメント

madhan ravi 2020 年 10 月 9 日
Note this answer doesn't include the floating point comparison. Use of find() for the second approach is not necessary ,a well use of logical indexing is more than enough.
Najam us Saqib Fraz 2020 年 10 月 9 日
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.....
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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### その他の回答 (2 件)

madhan ravi 2020 年 10 月 9 日

ix = (abs(column_1 - 527.1235) < 1e-4) &...
(abs(column_1 - 927.5) < 1e-1);
M = mean(column_4(ix))

#### 0 件のコメント

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Jon 2020 年 10 月 9 日

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 件のコメント

Najam us Saqib Fraz 2020 年 10 月 9 日
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...
When data is missing in the data file, it shows as NaN. You can remove the lines that are NaN
Jon 2020 年 10 月 9 日
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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