# finding average of a percentile of rows in a matrix?

Latest activity Commented on by Ahmad on 31 Jul 2012

Hi All,

In the below given matrix, I wish to calculate the mean/average of first 20% rows and subtract it from the mean of the last 20% rows.

```x =
1   2   3   4   5
2   4   6   8  10
3   6   9  12  15
4   8  12  16  20
5  10  15  20  25
1   2   3   4   5
2   4   6   8  10
3   6   9  12  15
4   8  12  16  20
5  10  15  20  25
```

this is a sample matrix, so there can be 100 rows and 100 columns but I need to average the values in the first 20% and subtract it from the average of last 20%. No sorting is needed in my matrix. therefore, in the above matrix, I would need to average the first two rows and subtract the resultant from the average of the last two rows.

any help will be appreciated.

Regards,

AMD.

## Products

Answer by Azzi Abdelmalek on 26 Jul 2012
Edited by Azzi Abdelmalek on 26 Jul 2012
```   n=size(x,1);n1=round(n/5);row_mean=mean(x(n-n1+1:end,:))-mean(x(1:n1,:))
result_mean=mean(row_mean)```

% you can use the result: row_mean wich is a line % or the result: result_mean which is the mean of the line row_mean

Hi Azzi,

Sorry for being late. Couldn't reply due to illness.

My data has NaNs and the answer results in a NaN. How can I avoid NaNs in this calculation.

Regards,

AMD.

Azzi Abdelmalek on 30 Jul 2012

can we replace them with 0?

Answer by Wayne King on 26 Jul 2012
Edited by Wayne King on 26 Jul 2012

```numrows = round(0.2*size(A,1));
lastrowstart = size(A,1)-numrows+1;
lowermean = mean(A(1:numrows,:),2);
uppermean = mean(A(lastrowstart:end,:),2);
uppermean-lowermean
```

Or did you mean take the mean over all elements in the bottom 20% of rows and upper 20% of rows, so you end up with a single number?

If that is the case:

```    numrows = round(0.2*size(A,1));
lastrowstart = size(A,1)-numrows+1;
lowermean = mean(mean(A(1:numrows,:),2));
uppermean = mean(mean(A(lastrowstart:end,:),2));
uppermean-lowermean```

Hi Wayne,

Sorry for being late. Couldn't reply due to illness. Yes there are NaNs in my data. Please tell how to skip NaNs in this mean calculation.

Regards,

AMD.

Oleg Komarov on 30 Jul 2012

Thanks.

Answer by Azzi Abdelmalek on 30 Jul 2012
Edited by Azzi Abdelmalek on 30 Jul 2012

%replacing nan by zero

```x(find(isnan(x)))=0     % add this to replace nan by zero
```

% the previous code

```   n=size(x,1);n1=round(n/5);row_mean=mean(x(n-n1+1:end,:))-mean(x(1:n1,:))
result_mean=mean(row_mean)```

%in case you want calculate the mean, ignoring nan , that means: if i have [1 3 nan 4]; the mean will not (1+3+0+4)/4, but (1+3+4)/3. in this case add this code

``` ind=arrayfun(@(y) ~isnan(y),x)
x(find(isnan(x)))=0;
n=size(x,1);n1=round(n/5);
row_mean1=sum(x(n-n1+1:end,:))./sum(ind(n-n1+1:end,:))-sum(x(1:n1,:))./sum(ind(1:n1,:))```

## 1 Comment

Azzi Abdelmalek on 30 Jul 2012

if you dn't want replace nan by 0, what do you want to do?