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From: Walter Roberson on 25 Jul 2010 16:17
Roger Stafford wrote:

> Without the statistics toolbox you can use
>
> sum(m(m~=0),2)./sum(m~=0,2)

Or just

sum(m,2)./sum(m~=0,2)

as the 0's will not change the sum of m.

If all of the columns happen to be 0 then both of these will yield nan
for that row; the OP did not indicate what value they would like in that
situation.
From: us on 25 Jul 2010 16:19
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <i2i59n$mat$1(a)fred.mathworks.com>...
> "Evan " <Evan(a)fake.com> wrote in message <i2i3qr$lb8$1(a)fred.mathworks.com>...
> > I can't seem to figure out how to find the mean of the non-zero elements in the rows of a 2-D array without using a for loop..
> >
> > Suppose I have a 2D array
> >
> > Bob =
> > 0 50 54 0
> > 0 50 52 0
> > 0 52 54 0
> > 0 58 60 0
> >
> > Is there an easy way to find the mean of each row while keeping the result a 2D array like:
> >
> > Result: 52
> > 51
> > 53
> > 59
> >
> > I've tried the nonzeros command, but it keeps flattening the array before the mean command and I get the mean of the whole array rather than a row at a time. Is there some simple way to do this without using a for loop to parse through each row at a time?
> >
> > Thanks!
> - - - - - - - -
> Without the statistics toolbox you can use
>
> sum(m(m~=0),2)./sum(m~=0,2)
>
> Roger Stafford

ROGER(!)...
after three (3!) replies...

:-)
urs
From: Evan on 25 Jul 2010 16:50
Thanks for such quick responses,

Sadly I don't have access to the statistics toolbox and I can't seem to figure out the correct syntax yet. I guess I my example was flawed from the beginning, what I should have had was something like:

m = 10 50 54 00
00 50 52 40
00 52 54 00
00 12 00 00

Where I don't know which columns will have zeros. Then I can't limit myself to simply averaging the two middle columns. In my example there's always 4 columns, and there's a minimum of one non-zero number, but the column locations and number of non-zero numbers changes.

I can't seem to get either of the solutions to work... Thanks!
From: Roger Stafford on 25 Jul 2010 16:52
"us " <us(a)neurol.unizh.ch> wrote in message <i2i67n$kjm$1(a)fred.mathworks.com>...
> ROGER(!)...
> after three (3!) replies...
>
> :-)
> urs
- - - - - - - - -
To tell the awful truth Urs my decrepit 84-year-old brain is getting slower and more erratic all the time. When the time comes that my replies are giving more trouble than they're worth I'll have to quit contributing to CSSM. For example if I try to prove that 1 = 2, you'll know it's time.

That reminds me, here is a (in)famous proof of that very assertion. Suppose we have two numbers, a and b, that are equal. Then go through the following reasoning:

a = b
a*b = b*b
a*b-a*a = b*b-a*a
a*(b-a) = (b+a)*(b-a)
a = b+a
a = a+a
1*a = 2*a
1 = 2

Should I quit CSSM now?

Roger Stafford
From: Evan on 25 Jul 2010 16:53
As always, I replied too soon.

sum(m,2)./sum(m~=0,2)

Seems to do the trick for me, thanks a ton!
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