R help - Aug 2011 - [?]apply functions or for loop

Hello,

First time posting to this mail list.

I'd like to use R in the most efficient way. I'm accomplishing what I
want,
but feel there is a more R'ish way to do it. Just learning R.

*My goal: get ranks of value across rows with row names and column names
intact.*

I'm guessing one of the [?]apply functions will do what I need, but I
couldn't sort out which one (after a lot of searching).

Here is a simplified example of what I am doing. Again, I get the correct
result, but I assume there is a better way to do it.
> x = data.frame(1:4,4)
> x
  X1.4 X4
1    1  4
2    2  4
3    3  4
4    4  4
> ranks = matrix(0,nrow(x),ncol(x))
> ranks
     [,1] [,2]
[1,]    0    0
[2,]    0    0
[3,]    0    0
[4,]    0    0
> for(i in 1:nrow(x)){
+ ranks[i,] = rank(x[i,])
+ }
> ranks[i,]
[1] 1.5 1.5
> ranks
     [,1] [,2]
[1,]  1.0  2.0
[2,]  1.0  2.0
[3,]  1.0  2.0
[4,]  1.5  1.5
> rownames(ranks) = rownames(x)
> ranks
  [,1] [,2]
1  1.0  2.0
2  1.0  2.0
3  1.0  2.0
4  1.5  1.5
> rownames(ranks) = rownames(x)
> ranks
  [,1] [,2]
1  1.0  2.0
2  1.0  2.0
3  1.0  2.0
4  1.5  1.5
> colnames(ranks) = colnames(x)
> ranks
  X1.4  X4
1  1.0 2.0
2  1.0 2.0
3  1.0 2.0
4  1.5 1.5

Thanks,
Ben

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You should have been able to discern from the help pages that the generic
"apply" will do it.

E.g.,

apply(x,1,rank)

Now, you'll probably want to transpose the output of apply: it's a R
quirk/feature/bug/idiosyncrasy that apply(x,1,FUN) transposes the output and
most of the time I wind up switching it.

so r = t(apply(x,1,rank))

The syntax of apply is as follows: apply(THING TO APPLY, DIMENSION*,
FUNCTION, OTHERS)

THING TO APPLY -- Obvious
DIMENSION -- 1 for row-wise, 2 for column-wise
FUNCTION -- the function you are applying, here rank, without any arguments
or parentheses
OTHERS -- other arguments for more advanced functions (or to do things like
na.last=F or changing the ties.method with rank)

Hope this helps,

Michael Weylandt



On Fri, Aug 5, 2011 at 11:35 AM, Ben qant <ccquant@gmail.com> wrote:
> Hello,
>
> First time posting to this mail list.
>
> I'd like to use R in the most efficient way. I'm accomplishing what
I want,
> but feel there is a more R'ish way to do it. Just learning R.
>
> *My goal: get ranks of value across rows with row names and column names
> intact.*
>
> I'm guessing one of the [?]apply functions will do what I need, but I
> couldn't sort out which one (after a lot of searching).
>
> Here is a simplified example of what I am doing. Again, I get the correct
> result, but I assume there is a better way to do it.
>
> > x = data.frame(1:4,4)
> > x
>  X1.4 X4
> 1    1  4
> 2    2  4
> 3    3  4
> 4    4  4
> > ranks = matrix(0,nrow(x),ncol(x))
> > ranks
>     [,1] [,2]
> [1,]    0    0
> [2,]    0    0
> [3,]    0    0
> [4,]    0    0
> > for(i in 1:nrow(x)){
> + ranks[i,] = rank(x[i,])
> + }
> > ranks[i,]
> [1] 1.5 1.5
> > ranks
>     [,1] [,2]
> [1,]  1.0  2.0
> [2,]  1.0  2.0
> [3,]  1.0  2.0
> [4,]  1.5  1.5
> > rownames(ranks) = rownames(x)
> > ranks
>  [,1] [,2]
> 1  1.0  2.0
> 2  1.0  2.0
> 3  1.0  2.0
> 4  1.5  1.5
> > rownames(ranks) = rownames(x)
> > ranks
>  [,1] [,2]
> 1  1.0  2.0
> 2  1.0  2.0
> 3  1.0  2.0
> 4  1.5  1.5
> > colnames(ranks) = colnames(x)
> > ranks
>  X1.4  X4
> 1  1.0 2.0
> 2  1.0 2.0
> 3  1.0 2.0
> 4  1.5 1.5
>
> Thanks,
> Ben
>
>        [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>
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Hello,

I am using poLCA to run standard latent class analyses. Is there any way that I
can get parameters for each predictors, in terms of their association with the
latent variable ?

Thank you,

David Joubert
 		 	   		  
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