Question about expand.grid function in R

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I don't think you can do it within expand.grid() but something like
this might work:

rownames(x) <- apply(x, 1, paste, collapse = "")

Michael
Hi,

I am extremely new to R, and was wondering if someone would be able to help me with a question regarding the expand.grid function. When I input

expand.grid.rep <- function(x, n=1) do.call(expand.grid, rep(list(x),n))
expand.grid.rep(c("a", "b", "c"), 3)

my output is as follows,

Var1 Var2 Var3
1 ? ? a ? ?a ? ?a
2 ? ? b ? ?a ? ?a
3 ? ? c ? ?a ? ?a
4 ? ? a ? ?b ? ?a
5 ? ? b ? ?b ? ?a
6 ? ? c ? ?b ? ?a
7 ? ? a ? ?c ? ?a
8 ? ? b ? ?c ? ?a
9 ? ? c ? ?c ? ?a
10 ? ?a ? ?a ? ?b
11 ? ?b ? ?a ? ?b
12 ? ?c ? ?a ? ?b
13 ? ?a ? ?b ? ?b
14 ? ?b ? ?b ? ?b
15 ? ?c ? ?b ? ?b
16 ? ?a ? ?c ? ?b
17 ? ?b ? ?c ? ?b
18 ? ?c ? ?c ? ?b
19 ? ?a ? ?a ? ?c
20 ? ?b ? ?a ? ?c
21 ? ?c ? ?a ? ?c
22 ? ?a ? ?b ? ?c
23 ? ?b ? ?b ? ?c
24 ? ?c ? ?b ? ?c
25 ? ?a ? ?c ? ?c
26 ? ?b ? ?c ? ?c
27 ? ?c ? ?c ? ?c

I was wondering if there was anyway I can change the row numbers to labels that indicate what is in each row. Instead of a 1, I'd like to have a label saying "aaa", etc. I'm not sure if this is even possible within the expand.grid function but any help would be appreciated. Thanks.
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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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I don't think you can do it within expand.grid() but something like
this might work:

rownames(x) <- apply(x, 1, paste, collapse = "")

Also

rownames(x) <- do.call(paste, c(x, sep=""))

or, in recent versions,

rownames(x) <- do.call(paste0, x)
Michael

On Tue, May 1, 2012 at 5:05 AM, Kelly Cool <kellycool79 at yahoo.com> wrote:
Hi,

I am extremely new to R, and was wondering if someone would be able to help me with a question regarding the expand.grid function. When I input

expand.grid.rep <- function(x, n=1) do.call(expand.grid, rep(list(x),n))
expand.grid.rep(c("a", "b", "c"), 3)

my output is as follows,

Var1 Var2 Var3
1     a    a    a
2     b    a    a
3     c    a    a
4     a    b    a
5     b    b    a
6     c    b    a
7     a    c    a
8     b    c    a
9     c    c    a
10    a    a    b
11    b    a    b
12    c    a    b
13    a    b    b
14    b    b    b
15    c    b    b
16    a    c    b
17    b    c    b
18    c    c    b
19    a    a    c
20    b    a    c
21    c    a    c
22    a    b    c
23    b    b    c
24    c    b    c
25    a    c    c
26    b    c    c
27    c    c    c

I was wondering if there was anyway I can change the row numbers to labels that indicate what is in each row. Instead of a 1, I'd like to have a label saying "aaa", etc. I'm not sure if this is even possible within the expand.grid function but any help would be appreciated. Thanks.
       [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com
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I was wondering if there was a quick way to extract out certain rows from a data set in R?
I have a data.frame, LOG, ?where in one column, sample_data_tx, there is a list of 62 different types of treatment. I've sub-selected the rows that contain the names, "PLO" and "NOY" to make a new vector which I call, Test.

?Here's my code so far,

##In LOG data set, Test set is every treatment, PLO and NOY##

##Select rows in the LOG data set that contain Noy##

Noy <- which(LOG$sample_data_tx == "Noy")

##Select rows in the LOG data set that contain PLO##

PLO <- which(LOG$sample_data_tx == "PLO")

##Make Test Set##
?Test <- c(Noy, PLO)

Test
[1] ?8 24 50 23 29 46 55

Within the data.frame, LOG, I would like to now make another vector, "Training", that contains every row in the column, "sample_data_tx", except rows 8, 24, 50, 23, 29, 46, 55.
I think you're looking for negative indexing (which is, in my opinion,
pretty much the best thing ever)

E.g.,

x <- letters[1:10]

x[1:3] # First three letters

x[-(1:3)] # Without the first three letters

x[-4] # Leave out "d"

etc.

Of course, for this case, you might also want the subset function:

subset(LOG, sample_data_tx %in% c("Noy","PLO"))
"Test" is also an integer and I am hoping to make a hierarchical plot with both the "Test" and "Training" vectors so I am not sure if I first need to convert the data from integer to numeric form?
No, almost always these sorts of conversions will be taken care of you
automatically

Best,
Michael
 I am new to R so all help is appreciated. Thanks in advance.
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______________________________________________
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

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This is a well-known limitation. You have to group categorical attributes together to work around.

--------------------------------------------------------------------------------------
Weifeng (aaron) liu ?|??retail systems pricing? | ?sr research scientist

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Kelly Cool
Sent: Tuesday, May 29, 2012 7:47 AM
To: r-help at r-project.org
Subject: [R] Question about random Forest function in R

Hello,?

I am trying to run the random Forest function on a data.frame using the following code..

myrf <- randomForest (y=sample_data_metal, x=Train, importance=TRUE, proximity=TRUE)

However, an error occurs saying, "can not handle categorical predictors with more than 32 categories".?

My "x=Train" data.frame is quite large and my "y=sample_data_metal" is one column.?

I'm not sure how to go about fixing this error or if there is even a way to get around this error. Thanks in advance for any help.?
Hi Kelly,

The function has a limitation that it cannot handle any column in your "x" that is a categorical variable with more than 32 categories.  One possibility is to see if you can "bin" some of the categories into one to get below 32 categories.

Andy 

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Kelly Cool
Sent: Tuesday, May 29, 2012 10:47 AM
To: r-help at r-project.org
Subject: [R] Question about random Forest function in R

Hello,?

I am trying to run the random Forest function on a data.frame using the following code..

myrf <- randomForest (y=sample_data_metal, x=Train, importance=TRUE, proximity=TRUE)

However, an error occurs saying, "can not handle categorical predictors with more than 32 categories".?

My "x=Train" data.frame is quite large and my "y=sample_data_metal" is one column.?

I'm not sure how to go about fixing this error or if there is even a way to get around this error. Thanks in advance for any help.?

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