## Mean averaging separate rows using a certain section of each row.

on 24 May 2012

### Honglei Chen (view profile)

I need to taken a mean average of a certain section of a row of a vector - and I need to do that for every row - giving each row's mean average.

What I have is a <300000x17> vector and when the 17 rows are plotted separately they all level off at there own approximate value - maybe by about 40% along the x-axis. Right at the very right hand edge the values seem to do a weird peak so I want to avoid the very end too.

At the end of this I'd hopefully be left with 17 values only.

Any help would be much appreciated as always!

Matt Fig

### Matt Fig (view profile)

on 2 Nov 2012

I need to taken a mean average of a certain section of a row of a vector - and I need to do that for every row - giving each row's mean average.

What I have is a 300000x17 vector and when the 17 rows are plotted separately they all level off at there own approximate value - maybe by about 40% along the x-axis. Right at the very right hand edge the values seem to do a weird peak so I want to avoid the very end too.

At the end of this I'd hopefully be left with 17 values only.

Any help would be much appreciated as always!

## Products

### Honglei Chen (view profile)

on 24 May 2012

Hi Tom, your example has 17 columns, not 17 rows. I'll assume you mean a 17x300000 matrix. It's not clear if the region for every row is the same, but I'll assume you want to take first 40% of each row, i.e., 120000 points. You can do the following:

```mu = mean(x(:,1:120000),2)
```

This should be a good starting point.

Tom

### Tom (view profile)

on 24 May 2012

Hi Honglei, thanks again for the answers. It's actually 17 columns. I tried using the x=x(:) to switch it around but it didn't work. Do you think you could let me know how to do it's 300000x17? Thanks.

Honglei Chen

### Honglei Chen (view profile)

on 25 May 2012

Then you just need to switch the dimensions

mu = mean(x(1:120000,:),1)

You actually don't need that 1 in the end as the first dimension is the default dimension. I included it anyway to show the parallelism among the syntax.

Tom

### Tom (view profile)

on 26 May 2012

That's great thanks Honlgei - I got it working well in the end. :)

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