date:20130514


On 13-05-14 4:52 AM, Luca Nanetti wrote:

Dear experts,

I wanted to signal a peculiar, unexpected behaviour of 'apply'. It is not a
bug, it is per spec, but it is so counterintuitive that I thought it could
be interesting.

I have an array, let's say test, dim=c(7,5).


test - array(1:35, dim=c(7, 5))
test


  [,1] [,2] [,3] [,4] [,5]
[1,]18   15   22   29
[2,]29   16   23   30
[3,]3   10   17   24   31
[4,]4   11   18   25   32
[5,]5   12   19   26   33
[6,]6   13   20   27   34
[7,]7   14   21   28   35

I want a new array where the content of the rows (columns) are permuted,
differently per row (per column)

Let's start with the columns, i.e. the second MARGIN of the array:

test.m2 - apply(test, 2, sample)
test.m2


  [,1] [,2] [,3] [,4] [,5]
[1,]1   10   18   23   32
[2,]79   16   25   30
[3,]6   14   17   22   33
[4,]4   11   15   24   34
[5,]2   12   21   28   31
[6,]58   20   26   29
[7,]3   13   19   27   35

perfect. That was exactly what I wanted: the content of each column is
shuffled, and differently for each column.
However, if I use the same with the rows (MARGIIN = 1), the output is
transposed!


test.m1 - apply(test, 1, sample)
test.m1


  [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]12345   13   21
[2,]   22   30   17   18   19   20   35
[3,]   15   23   24   32   26   27   14
[4,]   29   16   31   25   33   34   28
[5,]89   10   11   1267

In other words, I wanted to permute the content of the rows of test, and
I expected to see in the output, well, the shuffled rows as rows, not as
column!

I would respectfully suggest to make this behavior more explicit in the
documentation.


It's is already very explicit:  If each call to FUN returns a vector of 
length n, then apply returns an array of dimension c(n, dim(X)[MARGIN]) 
if n  1.  In your first case, sample is applied to columns, and 
returns length 7 results, so the shape of the final result is c(7, 5). 
In the second case it is applied to rows, and returns length 5 results, 
so the shape is c(5, 7).


Duncan Murdoch

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Re: [R] Unexpected behavior of apply when FUN=sample

2013-05-14 Thread Enrico Schumann

On Tue, 14 May 2013, Luca Nanetti luca.nane...@gmail.com writes:

 Dear experts,

 I wanted to signal a peculiar, unexpected behaviour of 'apply'. It is not a
 bug, it is per spec, but it is so counterintuitive that I thought it could
 be interesting.

 I have an array, let's say test, dim=c(7,5).

 test - array(1:35, dim=c(7, 5))
 test

  [,1] [,2] [,3] [,4] [,5]
 [1,]18   15   22   29
 [2,]29   16   23   30
 [3,]3   10   17   24   31
 [4,]4   11   18   25   32
 [5,]5   12   19   26   33
 [6,]6   13   20   27   34
 [7,]7   14   21   28   35

 I want a new array where the content of the rows (columns) are permuted,
 differently per row (per column)

 Let's start with the columns, i.e. the second MARGIN of the array:
 test.m2 - apply(test, 2, sample)
 test.m2

  [,1] [,2] [,3] [,4] [,5]
 [1,]1   10   18   23   32
 [2,]79   16   25   30
 [3,]6   14   17   22   33
 [4,]4   11   15   24   34
 [5,]2   12   21   28   31
 [6,]58   20   26   29
 [7,]3   13   19   27   35

 perfect. That was exactly what I wanted: the content of each column is
 shuffled, and differently for each column.
 However, if I use the same with the rows (MARGIIN = 1), the output is
 transposed!

 test.m1 - apply(test, 1, sample)
 test.m1

  [,1] [,2] [,3] [,4] [,5] [,6] [,7]
 [1,]12345   13   21
 [2,]   22   30   17   18   19   20   35
 [3,]   15   23   24   32   26   27   14
 [4,]   29   16   31   25   33   34   28
 [5,]89   10   11   1267

 In other words, I wanted to permute the content of the rows of test, and
 I expected to see in the output, well, the shuffled rows as rows, not as
 column!

 I would respectfully suggest to make this behavior more explicit in the
 documentation.

As you said yourself, this behaviour is documented:

  If each call to ‘FUN’ returns a vector of length ‘n’, then ‘apply’
  returns an array of dimension ‘c(n, dim(X)[MARGIN])’ [...]

And it has nothing to do with 'sample'. Try:

  apply(test, 1, function(x) x)
  apply(test, 2, function(x) x)

The result is only counterintuitive (or inconvenient, perhaps) in the
special case in which apply is supposed to return an array that has the
same dimension as its input.  More generally, you will do something like 

  apply(test, 1, median)
  apply(test, 1, function(x) list(sum = sum(x), values = x))

and in such cases, apply does not return an array.


  
-- 
Enrico Schumann
Lucerne, Switzerland
http://enricoschumann.net

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Re: [R] Unexpected behavior of apply when FUN=sample

2013-05-14 Thread Rui Barradas


Hello,

The problem is that apply returns the results vector by vector and in R 
vectors are column vectors. This is not exclusive of apply with sample 
as the function to be called, but of apply in general. Try, for instance


apply(test, 1, identity)  # transposes the array

The rows are returned as column vectors. And you should expect this 
behavior from apply with MARGIN = 1.

And this is in fact documented, in the Value section of ?apply:

Value

If each call to FUN returns a vector of length n, then apply returns an 
array of dimension c(n, dim(X)[MARGIN]) if n  1.


The length of the returned vector is the number of rows and the number 
of columns is the dim corresponding to MARGIN...


Hope this helps,

Rui Barradas

Em 14-05-2013 09:52, Luca Nanetti escreveu:

Dear experts,

I wanted to signal a peculiar, unexpected behaviour of 'apply'. It is not a
bug, it is per spec, but it is so counterintuitive that I thought it could
be interesting.

I have an array, let's say test, dim=c(7,5).


test - array(1:35, dim=c(7, 5))
test


  [,1] [,2] [,3] [,4] [,5]
[1,]18   15   22   29
[2,]29   16   23   30
[3,]3   10   17   24   31
[4,]4   11   18   25   32
[5,]5   12   19   26   33
[6,]6   13   20   27   34
[7,]7   14   21   28   35

I want a new array where the content of the rows (columns) are permuted,
differently per row (per column)

Let's start with the columns, i.e. the second MARGIN of the array:

test.m2 - apply(test, 2, sample)
test.m2


  [,1] [,2] [,3] [,4] [,5]
[1,]1   10   18   23   32
[2,]79   16   25   30
[3,]6   14   17   22   33
[4,]4   11   15   24   34
[5,]2   12   21   28   31
[6,]58   20   26   29
[7,]3   13   19   27   35

perfect. That was exactly what I wanted: the content of each column is
shuffled, and differently for each column.
However, if I use the same with the rows (MARGIIN = 1), the output is
transposed!


test.m1 - apply(test, 1, sample)
test.m1


  [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]12345   13   21
[2,]   22   30   17   18   19   20   35
[3,]   15   23   24   32   26   27   14
[4,]   29   16   31   25   33   34   28
[5,]89   10   11   1267

In other words, I wanted to permute the content of the rows of test, and
I expected to see in the output, well, the shuffled rows as rows, not as
column!

I would respectfully suggest to make this behavior more explicit in the
documentation.

Kind regards,
Luca Nanetti



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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Unexpected behavior of apply when FUN=sample

2013-05-14 Thread Ted Harding

On 14-May-2013 09:46:32 Duncan Murdoch wrote:
 On 13-05-14 4:52 AM, Luca Nanetti wrote:
 Dear experts,

 I wanted to signal a peculiar, unexpected behaviour of 'apply'.
 It is not a bug, it is per spec, but it is so counterintuitive
 that I thought it could be interesting.

 I have an array, let's say test, dim=c(7,5).

 test - array(1:35, dim=c(7, 5))
 test

   [,1] [,2] [,3] [,4] [,5]
 [1,]18   15   22   29
 [2,]29   16   23   30
 [3,]3   10   17   24   31
 [4,]4   11   18   25   32
 [5,]5   12   19   26   33
 [6,]6   13   20   27   34
 [7,]7   14   21   28   35

 I want a new array where the content of the rows (columns) are
 permuted, differently per row (per column)

 Let's start with the columns, i.e. the second MARGIN of the array:
 test.m2 - apply(test, 2, sample)
 test.m2

   [,1] [,2] [,3] [,4] [,5]
 [1,]1   10   18   23   32
 [2,]79   16   25   30
 [3,]6   14   17   22   33
 [4,]4   11   15   24   34
 [5,]2   12   21   28   31
 [6,]58   20   26   29
 [7,]3   13   19   27   35

 perfect. That was exactly what I wanted: the content of each column is
 shuffled, and differently for each column.
 However, if I use the same with the rows (MARGIIN = 1), the output is
 transposed!

 test.m1 - apply(test, 1, sample)
 test.m1

   [,1] [,2] [,3] [,4] [,5] [,6] [,7]
 [1,]12345   13   21
 [2,]   22   30   17   18   19   20   35
 [3,]   15   23   24   32   26   27   14
 [4,]   29   16   31   25   33   34   28
 [5,]89   10   11   1267

 In other words, I wanted to permute the content of the rows of test, and
 I expected to see in the output, well, the shuffled rows as rows, not as
 column!

 I would respectfully suggest to make this behavior more explicit in the
 documentation.
 
 It's is already very explicit:  If each call to FUN returns a vector of 
 length n, then apply returns an array of dimension c(n, dim(X)[MARGIN]) 
 if n  1.  In your first case, sample is applied to columns, and 
 returns length 7 results, so the shape of the final result is c(7, 5). 
 In the second case it is applied to rows, and returns length 5 results, 
 so the shape is c(5, 7).
 
 Duncan Murdoch

And the (quite simple) practical implication of what Duncan points out is:

  test - array(1:35, dim=c(7, 5))
  test
  #  [,1] [,2] [,3] [,4] [,5]
  # [1,]18   15   22   29
  # [2,]29   16   23   30
  # [3,]3   10   17   24   31
  # [4,]4   11   18   25   32
  # [5,]5   12   19   26   33
  # [6,]6   13   20   27   34
  # [7,]7   14   21   28   35

# To permute the rows:
  t(apply(t(test), 2, sample))
  #  [,1] [,2] [,3] [,4] [,5]
  # [1,]   22   298   151
  # [2,]   30   16   2329
  # [3,]   10   31   243   17
  # [4,]   114   25   32   18
  # [5,]   265   12   33   19
  # [6,]   27   34   20   136
  # [7,]   35   28   147   21

which looks right!
Ted.

-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 14-May-2013  Time: 11:07:46
This message was sent by XFMail

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Re: [R] Unexpected behavior of apply when FUN=sample

2013-05-14 Thread Gabor Grothendieck

On Tue, May 14, 2013 at 4:52 AM, Luca Nanetti luca.nane...@gmail.com wrote:
 Dear experts,

 I wanted to signal a peculiar, unexpected behaviour of 'apply'. It is not a
 bug, it is per spec, but it is so counterintuitive that I thought it could
 be interesting.

 I have an array, let's say test, dim=c(7,5).

 test - array(1:35, dim=c(7, 5))
 test

  [,1] [,2] [,3] [,4] [,5]
 [1,]18   15   22   29
 [2,]29   16   23   30
 [3,]3   10   17   24   31
 [4,]4   11   18   25   32
 [5,]5   12   19   26   33
 [6,]6   13   20   27   34
 [7,]7   14   21   28   35

 I want a new array where the content of the rows (columns) are permuted,
 differently per row (per column)

 Let's start with the columns, i.e. the second MARGIN of the array:
 test.m2 - apply(test, 2, sample)
 test.m2

  [,1] [,2] [,3] [,4] [,5]
 [1,]1   10   18   23   32
 [2,]79   16   25   30
 [3,]6   14   17   22   33
 [4,]4   11   15   24   34
 [5,]2   12   21   28   31
 [6,]58   20   26   29
 [7,]3   13   19   27   35

 perfect. That was exactly what I wanted: the content of each column is
 shuffled, and differently for each column.
 However, if I use the same with the rows (MARGIIN = 1), the output is
 transposed!

 test.m1 - apply(test, 1, sample)
 test.m1

  [,1] [,2] [,3] [,4] [,5] [,6] [,7]
 [1,]12345   13   21
 [2,]   22   30   17   18   19   20   35
 [3,]   15   23   24   32   26   27   14
 [4,]   29   16   31   25   33   34   28
 [5,]89   10   11   1267

 In other words, I wanted to permute the content of the rows of test, and
 I expected to see in the output, well, the shuffled rows as rows, not as
 column!

 I would respectfully suggest to make this behavior more explicit in the
 documentation.

aaply in the plyr package works in the way you expected.

--
Statistics  Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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[R] Possible bug in 'data.table'

2013-05-14 Thread Manta

Dear R users,

I may have found a bug in the function 'data.table'. I have a similar
question as the one in this post:

http://stackoverflow.com/questions/3367190/aggregate-and-weighted-mean-in-r

I have a dataset with assets, quantity traded, date and time. I would like
to calculate the value weighted average time of trades, by date of trade.
The variables are as follows:

Classes ‘data.table’ and 'data.frame':  307787 obs. of  12 variables:
$ CODE: int 1 2 3 4 4 5 3 2 2 1
 $ DATE : int  20070102 20070102 20070102 20070102 20070102 20070102
20070102 20070102 20070102 20070102 ...
 $ TIME : chr  09:14:14 09:14:33 09:26:19 09:40:45 ...
 $ PRICE: num  105.2 105.2 96.8 96.9 96.8 ...
 $ QTY  : int  500 500 500 1000 500 500 1000
300 1000 500 ...
 $ DATE2: Date, format: 2007-01-02 2007-01-02 2007-01-02
2007-01-02 ...
 $ TIME2: chr  09:14:14 09:14:33 09:26:19 09:40:45 ...
 $ TIME3: POSIXlt, format: 2013-05-13 09:14:14 2013-05-13 09:14:33
2013-05-13 09:26:19 2013-05-13 09:40:45 ...
 - attr(*, .internal.selfref)=externalptr 

If I run the command 

temp[,list(weighted.mean(PRICE,QTY)),by=DATE2] 

I get

   DATE2V1
   1: 2007-01-02 100.67024
   2: 2007-01-03  99.89599
   3: 2007-01-04 100.54347
   4: 2007-01-05 100.82472
   5: 2007-01-08  99.39865
  --- 
1524: 2012-12-19 103.73392
1525: 2012-12-20 103.77344
1526: 2012-12-21 102.89063
1527: 2012-12-27 101.53089
1528: 2012-12-28 103.35999

While I run the command 

temp[,list(weighted.mean(TIME3,QTY)),by=DATE2]

R crashes. This behavior occurs in both R Studio and R GUI, regardless of
the version (2.14, 2.15, 3.0). Is it really a bug or am I doing something
wrong?



--
View this message in context: 
http://r.789695.n4.nabble.com/Possible-bug-in-data-table-tp4667025.html
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Re: [R] Unexpected behavior of apply when FUN=sample

2013-05-14 Thread Tsjerk Wassenaar

t(apply(test,1,sample)) will also do.
As the OP noted, the results are simply transposed. So if an operation is
to be applied to rows, yielding modified rows, simply transpose the results.

Cheers,

Tsjerk


On Tue, May 14, 2013 at 12:07 PM, Ted Harding ted.hard...@wlandres.netwrote:

 On 14-May-2013 09:46:32 Duncan Murdoch wrote:
  On 13-05-14 4:52 AM, Luca Nanetti wrote:
  Dear experts,
 
  I wanted to signal a peculiar, unexpected behaviour of 'apply'.
  It is not a bug, it is per spec, but it is so counterintuitive
  that I thought it could be interesting.
 
  I have an array, let's say test, dim=c(7,5).
 
  test - array(1:35, dim=c(7, 5))
  test
 
[,1] [,2] [,3] [,4] [,5]
  [1,]18   15   22   29
  [2,]29   16   23   30
  [3,]3   10   17   24   31
  [4,]4   11   18   25   32
  [5,]5   12   19   26   33
  [6,]6   13   20   27   34
  [7,]7   14   21   28   35
 
  I want a new array where the content of the rows (columns) are
  permuted, differently per row (per column)
 
  Let's start with the columns, i.e. the second MARGIN of the array:
  test.m2 - apply(test, 2, sample)
  test.m2
 
[,1] [,2] [,3] [,4] [,5]
  [1,]1   10   18   23   32
  [2,]79   16   25   30
  [3,]6   14   17   22   33
  [4,]4   11   15   24   34
  [5,]2   12   21   28   31
  [6,]58   20   26   29
  [7,]3   13   19   27   35
 
  perfect. That was exactly what I wanted: the content of each column is
  shuffled, and differently for each column.
  However, if I use the same with the rows (MARGIIN = 1), the output is
  transposed!
 
  test.m1 - apply(test, 1, sample)
  test.m1
 
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
  [1,]12345   13   21
  [2,]   22   30   17   18   19   20   35
  [3,]   15   23   24   32   26   27   14
  [4,]   29   16   31   25   33   34   28
  [5,]89   10   11   1267
 
  In other words, I wanted to permute the content of the rows of test,
 and
  I expected to see in the output, well, the shuffled rows as rows, not as
  column!
 
  I would respectfully suggest to make this behavior more explicit in the
  documentation.
 
  It's is already very explicit:  If each call to FUN returns a vector of
  length n, then apply returns an array of dimension c(n, dim(X)[MARGIN])
  if n  1.  In your first case, sample is applied to columns, and
  returns length 7 results, so the shape of the final result is c(7, 5).
  In the second case it is applied to rows, and returns length 5 results,
  so the shape is c(5, 7).
 
  Duncan Murdoch

 And the (quite simple) practical implication of what Duncan points out is:

   test - array(1:35, dim=c(7, 5))
   test
   #  [,1] [,2] [,3] [,4] [,5]
   # [1,]18   15   22   29
   # [2,]29   16   23   30
   # [3,]3   10   17   24   31
   # [4,]4   11   18   25   32
   # [5,]5   12   19   26   33
   # [6,]6   13   20   27   34
   # [7,]7   14   21   28   35

 # To permute the rows:
   t(apply(t(test), 2, sample))
   #  [,1] [,2] [,3] [,4] [,5]
   # [1,]   22   298   151
   # [2,]   30   16   2329
   # [3,]   10   31   243   17
   # [4,]   114   25   32   18
   # [5,]   265   12   33   19
   # [6,]   27   34   20   136
   # [7,]   35   28   147   21

 which looks right!
 Ted.

 -
 E-Mail: (Ted Harding) ted.hard...@wlandres.net
 Date: 14-May-2013  Time: 11:07:46
 This message was sent by XFMail

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 PLEASE do read the posting guide
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 and provide commented, minimal, self-contained, reproducible code.




-- 
Tsjerk A. Wassenaar, Ph.D.

[[alternative HTML version deleted]]

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[R] points overlay axis

2013-05-14 Thread Jonathan Phillips

Hi,
I'm trying to do quite a simple task, but I'm stuck.

I've set xaxs = 'i' as I want the origin to be (0,0), but unfortunately I
have points that are sat on the axis.  R draws the axis over the points,
which hides the points somewhat and looks unsightly.
Is there any way of getting a point to be drawn over the axis?

Thanks,
Jon Phillips

[[alternative HTML version deleted]]

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Re: [R] points overlay axis

Probably but since we don't know what you are doing, it is very hard to give 
any advice.  

Please read this for a start 
https://github.com/hadley/devtools/wiki/Reproducibility and give us a clear 
statement of the problem 

Thanks

John Kane
Kingston ON Canada


 -Original Message-
 From: 994p...@gmail.com
 Sent: Tue, 14 May 2013 13:34:35 +0100
 To: r-help@r-project.org
 Subject: [R] points overlay axis
 
 Hi,
 I'm trying to do quite a simple task, but I'm stuck.
 
 I've set xaxs = 'i' as I want the origin to be (0,0), but unfortunately I
 have points that are sat on the axis.  R draws the axis over the points,
 which hides the points somewhat and looks unsightly.
 Is there any way of getting a point to be drawn over the axis?
 
 Thanks,
 Jon Phillips
 
   [[alternative HTML version deleted]]
 
 __
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[R] Problem with R websocket package

2013-05-14 Thread malej.gregy

Hello to everybody,

I'm repeating my post as the previous one was posted in HTML - sorry guys! 

I seem to be in struggle with the websockets in R. I wanted to download the 
streaming data from the BitCoin exchange MtGox directly to R, but R cannot 
establish the connection.

The websocket specs are defined as:


Host: websocket.mtgox.com or socketio.mtgox.com
Port: 80 or 443 ( ssl )
Namespace:/mtgox (Including beginning slash)

url for more details: https://en.bitcoin.it/wiki/MtGox/API/Streaming

and my code is: 

 require(websockets)
 con = websocket(https://socketio.mtgox.com/mtgox,port=443)

and I always end up with an error:
  con = websocket(https://socketio.mtgox.com/mtgox,port=443)
 Error in websocket(https://socketio.mtgox.com/mtgox;, port = 443) : 
 Connection error

Does anyone have an idea what is wrong? I was informed that MtGox does not use
 the proper websocket protocol but something called Socket-io.
Is it the reason why it doesn't work? If so, is there any other way to handle 
socket-io in R? I searched all websockets packages but did not find any.

 

Many thanks for help.

Stepan

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Re: [R] Select the column from the data.frame?

Hi,
Try:
set.seed(24)
dat1- as.data.frame(matrix(sample(1:60,15*5,replace=TRUE),ncol=15))
colnames(dat1)- paste0(a,c(3,1,5,7,2,8,11,14:15,10,9,6,12:13,4))
 
subDat1-dat1[colnames(dat1)[as.numeric(gsub([A-Za-z],,colnames(dat1)))=10]]
 subDat1
#  a3 a1 a5 a7 a2 a8 a10 a9 a6 a4
#1 18 56 37 55  6 42   5 28 60  9
#2 14 17 23 12 34  5  24 52 28 13
#3 43 46 41  3 44 36  22 28 18  1
#4 32 49 41 31  9 38  16 43 11 60
#5 40 16 20  9 14  3  21 33  9 39
A.K.


Hi, 
    I have one data set with 15 variable like a3, a1, a5, a7,a2, a8,
 a11, a14, a15, a10, a9, a6, a12, a13, a4. From this variable i want to 
select variables from a1:a10. Please helpe to to get the result..

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[R] apcluster webinar: Thursday, June 13, 2013, 7:00pm CEST

2013-05-14 Thread Ulrich Bodenhofer


Dear colleagues,

This is to inform you that I will be giving a webinar on the apcluster 
package on Thursday, June 13, 2013, 7:00pm CEST (10:00am PDT).


The outline of the one-hour webinar is as follows:

- Introduction to affinity propagation (AP) clustering
- The apcluster package, its algorithms, and visualization tools
- Live apcluster demonstration
- Question and Answer period

To register for the webinar, please visit the following URL:
https://www3.gotomeeting.com/register/503109182

The webinar is kindly brought to you by the Orange County R User Group, 
and will be moderated by its president, Ray DiGiacomo, Jr.


We are aware that the time may be inconvenient for users from the 
Asia-Pacific region. If there is demand (please let us know!), we are 
willing to organize a second webinar at a time that is more suitable for 
Asian-Pacific users.


Best regards,
Ulrich Bodenhofer



*Dr. Ulrich Bodenhofer*
Associate Professor
Institute of Bioinformatics

*Johannes Kepler University*
Altenberger Str. 69
4040 Linz, Austria

Tel. +43 732 2468 4526
Fax +43 732 2468 4539
bodenho...@bioinf.jku.at
http://www.bioinf.jku.at/

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[R] need help for Imbalanced classification problems!!!

2013-05-14 Thread Kevin Hao

Hi all,

I am facing the imbalanced classification problems. That means I have a
dataset, in which the ratio of majority data to minority data is 100:1 (or
more).

In addition, the independent variables are many and this is a binary
classification questions.

The model I built give poor predictive power for minor data, but for the
majority data the predictivity seems to overfitting.


Could you give me a suggestions to solve the imbalanced classification
problems?


Thanks in advance!

Kevin

[[alternative HTML version deleted]]

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Re: [R] points overlay axis

2013-05-14 Thread David Carlson

Try this

set.seed(42)
dat - matrix(c(runif(48), 0, 0), 25, 2, byrow=TRUE)
# Complete plot symbol on axes, but axis on top
plot(dat, xaxs=i, yaxs=i, pch=16, col=red, xpd=TRUE)
# Complete plot symbol on axes with symbol on top
plot(dat, xaxs=i, yaxs=i, type=n)
points(dat, xaxs=i, yaxs=i, pch=16, col=red, xpd=TRUE)


David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of John Kane
Sent: Tuesday, May 14, 2013 7:47 AM
To: Jonathan Phillips; r-help@r-project.org
Subject: Re: [R] points overlay axis

Probably but since we don't know what you are doing, it is very hard to give
any advice.  

Please read this for a start
https://github.com/hadley/devtools/wiki/Reproducibility and give us a clear
statement of the problem 

Thanks

John Kane
Kingston ON Canada


 -Original Message-
 From: 994p...@gmail.com
 Sent: Tue, 14 May 2013 13:34:35 +0100
 To: r-help@r-project.org
 Subject: [R] points overlay axis
 
 Hi,
 I'm trying to do quite a simple task, but I'm stuck.
 
 I've set xaxs = 'i' as I want the origin to be (0,0), but 
 unfortunately I have points that are sat on the axis.  R draws the 
 axis over the points, which hides the points somewhat and looks unsightly.
 Is there any way of getting a point to be drawn over the axis?
 
 Thanks,
 Jon Phillips
 
   [[alternative HTML version deleted]]
 
 __
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FREE ONLINE PHOTOSHARING - Share your photos online with your friends and
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Re: [R] points overlay axis

2013-05-14 Thread David Carlson

Let's try again after restraining Outlook's desire to use html.

set.seed(42)
dat - matrix(c(runif(48), 0, 0), 25, 2, byrow=TRUE)

# Complete plot symbol on axes, but axis on top
plot(dat, xaxs=i, yaxs=i, pch=16, col=red, xpd=TRUE)

# Complete plot symbol on axes with symbol on top
plot(dat, xaxs=i, yaxs=i, type=n)
points(dat, xaxs=i, yaxs=i, pch=16, col=red, xpd=TRUE)


David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of John Kane
Sent: Tuesday, May 14, 2013 7:47 AM
To: Jonathan Phillips; r-help@r-project.org
Subject: Re: [R] points overlay axis

Probably but since we don't know what you are doing, it is very hard to give
any advice.  

Please read this for a start
https://github.com/hadley/devtools/wiki/Reproducibility and give us a clear
statement of the problem 

Thanks

John Kane
Kingston ON Canada


 -Original Message-
 From: 994p...@gmail.com
 Sent: Tue, 14 May 2013 13:34:35 +0100
 To: r-help@r-project.org
 Subject: [R] points overlay axis
 
 Hi,
 I'm trying to do quite a simple task, but I'm stuck.
 
 I've set xaxs = 'i' as I want the origin to be (0,0), but 
 unfortunately I have points that are sat on the axis.  R draws the 
 axis over the points, which hides the points somewhat and looks unsightly.
 Is there any way of getting a point to be drawn over the axis?
 
 Thanks,
 Jon Phillips
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
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Re: [R] Possible bug in 'data.table'

2013-05-14 Thread Ben Bolker

Manta mantino84 at libero.it writes:

 
 Dear R users,
 
 I may have found a bug in the function 'data.table'. I have a similar
 question as the one in this post:
 
 http://stackoverflow.com/questions/3367190/
  aggregate-and-weighted-mean-in-r
 

[snip]


 While I run the command 
 
 temp[,list(weighted.mean(TIME3,QTY)),by=DATE2]
 
 R crashes. This behavior occurs in both R Studio and R GUI, regardless of
 the version (2.14, 2.15, 3.0). Is it really a bug or am I doing something
 wrong?
 

  If by crash you mean that the program actually terminates (i.e.
not just an error), then this is by definition a bug in data.table --
there is a bug tracker at 
https://r-forge.r-project.org/tracker/?group_id=240 , or you
could try contacting the package maintainer (help(package=data.table)
or maintainer(data.table)

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Re: [R] Need help to building R package with devtools

2013-05-14 Thread Hadley Wickham

 Can someone help what I need to do to make 'devtools' work?

 A quick asking around indicates that Rtools 3.0 should work fine for 2.15.3 
 maintenance. Thus, the issue is probably a purely formal bug in devtools's 
 version comparison logic, and you need to pester its maintainer. Unless it 
 has been fixed already, in which case you need to update devtools.

I'm pretty sure it has been fixed already, otherwise please file a bug
(with a copy of your sessionInfo()) at
https://github.com/hadley/devtools/issues

--
RStudio / Rice University
http://had.co.nz/

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Re: [R] Dataframe and conditions

2013-05-14 Thread Frédéric Grelier

Thanks a lot Guys
That works
I appreciate your helps

Best regards
Frederic

FREDERIC GRELIER  DIRECTEUR DATA
T +33 (0) 1 83 94 04 08 - P +33 (0) 6 70 50  01 05 - F +33 (0) 1 53 19 41
84
WWW.WEBORAMA.COM http://www.weborama.com/  - 15, RUE CLAVEL - 75019
PARIS




-Message d'origine-
De : arun [mailto:smartpink...@yahoo.com]
Envoyé : mardi 14 mai 2013 15:19
À : Pascal Oettli
Cc : R help; fgrelier
Objet : Re: [R] Dataframe and conditions



#this should also work

 within(X,a- ifelse(b,c,a))
#  a b c
#1 2  TRUE 2
#2 2  TRUE 2
#3 1 FALSE 2
#4 1 FALSE 2
#5 1 FALSE 2
#6 2  TRUE 2
A.K.

- Original Message -
From: Pascal Oettli kri...@ymail.com
To: fgrelier fgrel...@weborama.com
Cc: r-help@r-project.org
Sent: Tuesday, May 14, 2013 4:47 AM
Subject: Re: [R] Dataframe and conditions

Hello,

One approach is using ifelse:

 X - data.frame(a=c(1,1,1,1,1,1),
b=c(TRUE,TRUE,FALSE,FALSE,FALSE,TRUE), c=c(2,2,2,2,2,2))
 X
   a     b c
1 1  TRUE 2
2 1  TRUE 2
3 1 FALSE 2
4 1 FALSE 2
5 1 FALSE 2
6 1  TRUE 2

 X - within(X, a - ifelse(b==TRUE, c, a)) X
   a     b c
1 2  TRUE 2
2 2  TRUE 2
3 1 FALSE 2
4 1 FALSE 2
5 1 FALSE 2
6 2  TRUE 2

Hope this helps,
Pascal


On 05/14/2013 05:06 PM, fgrelier wrote:

 I have in a dataframe X  :  3 Variables X$a , X$b, X$c I would like to
 replace in X the values of X$a by the values of  X$c but only when
 X$b==TRUE
 I have tried to put in place a loop but as I have a lot of rows it is
 very very long to run.

 Thanks for your help





 --
 View this message in context:
 http://r.789695.n4.nabble.com/Dataframe-and-conditions-tp4667012.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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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
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[R] Problem with constrained nlsList model

2013-05-14 Thread Rob Forsyth

I have some longitudinal data I'm fitting with an asymptotic growth function 
constrained to the origin. Visual inspection of the dataset suggests this is 
reasonable!

 library(nlme)
 results - groupedData(y~Days | ID, data=results)
 m1 - nlsList(SSasympOrig, na.omit(results))
 m1

Call:
 Model: y ~ SSasympOrig(Days, Asym, lrc) | ID 
  Data: na.omit(results) 

Coefficients:
  Asym   lrc
1765.259815 -5.048464
186   13.833871 -4.337267
300   16.007862 -4.783315
159   21.509863 -4.241172
144  103.505322 -7.040130
161   49.746946 -5.682995
170   50.964702 -5.739456
293   63.060652 -5.928408
142   49.851672 -4.707905
261   48.612882 -4.568154
184   63.633909 -5.361407
162   49.478622 -4.088146
202  714.579639 -9.128026
303   53.688201 -3.750039
275   63.298936 -4.757936
264   60.868979 -5.029757
199   64.776418 -5.163388
1368 107.784023 -6.478387
212   99.200012 -6.566332
204   67.462536 -4.994269
177  120.435299 -6.125227
92   125.594800 -6.699715
210   63.899406 -4.055854
268   68.053434 -4.037159
201  133.915623 -6.786009
255   78.692978 -4.262876
273   80.826720 -4.481234
249   90.353950 -4.188984
238   97.372830 -3.969175
248   84.202629 -3.894503
208  104.325471 -4.209675
258  265.856493 -5.741356


So this appears to work reasonably well. For theoretical reasons in this 
context I also want to look at an asymptotic growth function additionally 
constrained with asymptote a constant (=100), i.e. with only the 
rate-constant/half-life free. I'm not sure how to modify the built-in 
SSasympOrig function to achieve this however thought this would work:

 m2 - nlsList(y~100*(1-exp(-exp(lrc*Days))), start=list(lrc=-5), 
 data=na.omit(results)) 

but this results in the error message

Error in nlsModel(formula, mf, start, wts) : 
 singular gradient matrix at initial parameter estimates

for every estimation

I realise that constraining the asymptote will force large changes in the 
estimate of lrc (e.g. particularly in the first case ID == 176) but many of the 
free-floating estimates of Asym are of the order of  ~ 100 so I'm not sure why 
the constrained nls model doesn't converge on at least some occasions? Am I 
doing something else wrong?

Thanks

Rob
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[R] where clauses - help

2013-05-14 Thread martinizza

hello, I wrote a foreach loop containing where clauses. R indicates an
error in the compilation of the where clause.
Could you please tell me why not right?

foreach (series. combine = c)%:% when (mydata [3] = u [1]) % Dopar% (mydata
[5] / sum (mydata [5]))

L 'error reported by R is:
Error: unexpected '=' in foreach (series. Combine = c)%:% when (mydata [3]
=


many thanks in advance
Martina

[[alternative HTML version deleted]]

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Re: [R] Dataframe and conditions



#this should also work

 within(X,a- ifelse(b,c,a))
#  a b c
#1 2  TRUE 2
#2 2  TRUE 2
#3 1 FALSE 2
#4 1 FALSE 2
#5 1 FALSE 2
#6 2  TRUE 2
A.K.

- Original Message -
From: Pascal Oettli kri...@ymail.com
To: fgrelier fgrel...@weborama.com
Cc: r-help@r-project.org
Sent: Tuesday, May 14, 2013 4:47 AM
Subject: Re: [R] Dataframe and conditions

Hello,

One approach is using ifelse:

 X - data.frame(a=c(1,1,1,1,1,1), 
b=c(TRUE,TRUE,FALSE,FALSE,FALSE,TRUE), c=c(2,2,2,2,2,2))
 X
   a     b c
1 1  TRUE 2
2 1  TRUE 2
3 1 FALSE 2
4 1 FALSE 2
5 1 FALSE 2
6 1  TRUE 2

 X - within(X, a - ifelse(b==TRUE, c, a))
 X
   a     b c
1 2  TRUE 2
2 2  TRUE 2
3 1 FALSE 2
4 1 FALSE 2
5 1 FALSE 2
6 2  TRUE 2

Hope this helps,
Pascal


On 05/14/2013 05:06 PM, fgrelier wrote:

 I have in a dataframe X  :  3 Variables X$a , X$b, X$c
 I would like to replace in X the values of X$a by the values of  X$c but
 only when X$b==TRUE
 I have tried to put in place a loop but as I have a lot of rows it is very
 very long to run.

 Thanks for your help





 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Dataframe-and-conditions-tp4667012.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@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.


__
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Re: [R] where clauses - help

2013-05-14 Thread peter dalgaard


On May 14, 2013, at 17:06 , martinizza wrote:

 hello, I wrote a foreach loop containing where clauses. R indicates an
 error in the compilation of the where clause.
 Could you please tell me why not right?
 
 foreach (series. combine = c)%:% when (mydata [3] = u [1]) % Dopar% (mydata
 [5] / sum (mydata [5]))
 
 L 'error reported by R is:
 Error: unexpected '=' in foreach (series. Combine = c)%:% when (mydata [3]
 =
 

This comes from a package, the package has a maintainer, the maintainer is 
associated with a commercial company...

Anyways, offhand:The comparison operator is == . You likely also need %dopar% 
(case and spaces do matter). And remove a space and add a comma within the 
first parenthesis.


-- 
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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Re: [R] where clauses - help


On 14/05/2013 11:06 AM, martinizza wrote:

hello, I wrote a foreach loop containing where clauses. R indicates an
error in the compilation of the where clause.
Could you please tell me why not right?

foreach (series. combine = c)%:% when (mydata [3] = u [1]) % Dopar% (mydata
[5] / sum (mydata [5]))

L 'error reported by R is:
Error: unexpected '=' in foreach (series. Combine = c)%:% when (mydata [3]
=


I don't know foreach, but it looks as though your = sign should be 
== to test for equality.


The first part looks funny too:  shouldn't that be a comma after series, 
rather than a period?


Duncan Murdoch

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[R] query re plot(confint(lmList...

2013-05-14 Thread Michelle Morters

Hi -

My sample size is about 190, consequently the plot output (below) is 
quite squashed up and the id numbers down the L axis overlay each other 
and are not legible

plot(confint(lmList(x ~ slope | id, data), pooled = TRUE), order = 1)

Is it possible to either reduce the size of the id numbers down the L 
axis or spread the plots out over a couple of pages so the text is less 
squashed up? (- I need to retain the ordering). I've tried the usual 
plot function arguments (amongst other things) but alas to no avail.

Thanks!

M




[[alternative HTML version deleted]]

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[R] query in plot(intervals....

2013-05-14 Thread Michelle Morters

Hi -

I would like the plot ordered by intercept.

Ordering is doable if the intervals function is substituted with the 
confint function and order=1 included. Is ordering doable with intervals 
function, please?

Thanks!

M

results-lmList(x~slope|id,data)

plot(intervals(results))


[[alternative HTML version deleted]]

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[R] 3D plot

2013-05-14 Thread Amir


Dear all,

I need to plot more than one surface in 3D using R.
I tried with persp function, it produce only one surface.
Could you please help me how can I do it? The X and Y axis are the same 
but I have Z1 and Z2. I want to have both results in one graph.


Thanks
Amir


--
__
 Amir Darehshoorzadeh |  Computer Engineering Department
 PostDoc Fellow   |  University of Ottawa, Paradise LAb
 Email: adare...@uottawa.ca   |  800 King Edward Ave
 Tel: -   |  ON K1N 6N5, Ottawa - CANADA
 http://personals.ac.upc.edu/amir

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[R] Changing Order of Factor Levels in Mixed Model (nlme)

2013-05-14 Thread Edward Patzelt

R Help -

Why is that in the results below, changing the order of the factor
(trialType2: levels - DD, SD, DS, SS) changes the estimates in the fixed
effects tests?

 tmp.dat4$trialType2 - sort(tmp.dat4$trialType, decreasing = TRUE)
 mod2c - lme(proportion.down ~ trialType2, data = tmp.dat4, random = ~ 1
| subject, na.action = na.omit, method = ML)
 summary(mod2c)
Linear mixed-effects model fit by maximum likelihood
 Data: tmp.dat4
   AIC  BIClogLik
  27.92306 48.23003 -7.961531

Random effects:
 Formula: ~1 | subject
(Intercept)  Residual
StdDev:   0.3800017 0.1530272

Fixed effects: proportion.down ~ trialType2
  Value  Std.Error  DF   t-value p-value
(Intercept)   0.6788875 0.08613476 141  7.881690  0.
trialType2DS  0.0287062 0.11267357 141  0.254773  0.7993
trialType2SD -0.0197194 0.12142018 141 -0.162406  0.8712
trialType2SS -0.0941918 0.12204707 141 -0.771766  0.4415
 Correlation:
 (Intr) trT2DS trT2SD
trialType2DS -0.658
trialType2SD -0.709  0.467
trialType2SS -0.706  0.464  0.571

Standardized Within-Group Residuals:
Min  Q1 Med  Q3 Max
-3.67176824 -0.22090663  0.08677971  0.14331603  2.86301670

Number of Observations: 218
Number of Groups: 74
 tmp.dat4$trialType2 - sort(tmp.dat4$trialType, decreasing = FALSE)
 mod2c - lme(proportion.down ~ trialType2, data = tmp.dat4, random = ~ 1
| subject, na.action = na.omit, method = ML)
 summary(mod2c)
Linear mixed-effects model fit by maximum likelihood
 Data: tmp.dat4
   AIC  BIClogLik
  27.92306 48.23003 -7.961531

Random effects:
 Formula: ~1 | subject
(Intercept)  Residual
StdDev:   0.3800017 0.1530272

Fixed effects: proportion.down ~ trialType2
 Value  Std.Error  DF  t-value p-value
(Intercept)  0.5846957 0.08646554 141 6.762181  0.
trialType2DS 0.0744724 0.1123 141 0.660347  0.5101
trialType2SD 0.1228981 0.12175741 141 1.009368  0.3145
trialType2SS 0.0941918 0.12204707 141 0.771766  0.4415
 Correlation:
 (Intr) trT2DS trT2SD
trialType2DS -0.660
trialType2SD -0.710  0.469
trialType2SS -0.708  0.468  0.573

Standardized Within-Group Residuals:
Min  Q1 Med  Q3 Max
-3.67176824 -0.22090663  0.08677971  0.14331603  2.86301670

Number of Observations: 218
Number of Groups: 74

-- 
*Edward H Patzelt | Research Assistant
Psychology | University of Minnesota  | Elliott Hall, 75 East River Road |
Minneapolis, MN 55455
Email: patze...@umn.edu  |
Main: 612.626.0072
|
Mobile: 651.315.3410
| Office: S355
**

*

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Re: [R] query re plot(confint(lmList...

Please don't post in html. The list strips it out and we, now, have no idea of 
what you are doing.  

Have a look at https://github.com/hadley/devtools/wiki/Reproducibility for 
suggestions on how to ask a qustion here.

Sample code and sample data (see ?dupt) are usually desirable.

John Kane
Kingston ON Canada


 -Original Message-
 From: mm...@hermes.cam.ac.uk
 Sent: Tue, 14 May 2013 16:54:36 +0100
 To: r-help@r-project.org
 Subject: [R] query re plot(confint(lmList...
 
 Hi -
 
 My sample size is about 190, consequently the plot output (below) is
 quite squashed up and the id numbers down the L axis overlay each other
 and are not legible
 
 plot(confint(lmList(x ~ slope | id, data), pooled = TRUE), order = 1)
 
 Is it possible to either reduce the size of the id numbers down the L
 axis or spread the plots out over a couple of pages so the text is less
 squashed up? (- I need to retain the ordering). I've tried the usual
 plot function arguments (amongst other things) but alas to no avail.
 
 Thanks!
 
 M
 
 
 
 
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Re: [R] 3D plot


On 14/05/2013 1:10 PM, Amir wrote:

Dear all,

  I need to plot more than one surface in 3D using R.
I tried with persp function, it produce only one surface.
Could you please help me how can I do it? The X and Y axis are the same
but I have Z1 and Z2. I want to have both results in one graph.


That's quite hard to do with persp because it needs to plot the segments 
in a certain order.  You can do it in the rgl package using persp3d for 
the first surface, then persp3d( ..., add=TRUE) for the second.   Note 
that persp3d and persp use slightly different conventions for how to 
specify colours; see ?persp3d for details.


Duncan Murdoch

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Re: [R] query re plot(confint(lmList...


On 14/05/2013 11:54 AM, Michelle Morters wrote:

Hi -

My sample size is about 190, consequently the plot output (below) is
quite squashed up and the id numbers down the L axis overlay each other
and are not legible

plot(confint(lmList(x ~ slope | id, data), pooled = TRUE), order = 1)

Is it possible to either reduce the size of the id numbers down the L
axis or spread the plots out over a couple of pages so the text is less
squashed up? (- I need to retain the ordering). I've tried the usual
plot function arguments (amongst other things) but alas to no avail.


You don't give a reproducible example or say what package lmList came 
from.  However, if you want two plots, presumably you could use the 
subset() function to select out only some of the id values, and then do 
it again to get the rest.


Duncan Murdoch

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Re: [R] query re plot(confint(lmList...

Or better yet see ?dput, sorry.

John Kane
Kingston ON Canada

 -Original Message-
 From: jrkrid...@inbox.com
 Sent: Tue, 14 May 2013 09:20:59 -0800
 To: mm...@hermes.cam.ac.uk, r-help@r-project.org
 Subject: Re: [R] query re plot(confint(lmList...

 Please don't post in html. The list strips it out and we, now, have no
 idea of what you are doing.

 Have a look at https://github.com/hadley/devtools/wiki/Reproducibility
 for suggestions on how to ask a qustion here.

 Sample code and sample data (see ?dupt) are usually desirable.

 John Kane
 Kingston ON Canada

 -Original Message-
 From: mm...@hermes.cam.ac.uk
 Sent: Tue, 14 May 2013 16:54:36 +0100
 To: r-help@r-project.org
 Subject: [R] query re plot(confint(lmList...

 Hi -

 My sample size is about 190, consequently the plot output (below) is
 quite squashed up and the id numbers down the L axis overlay each other
 and are not legible

 plot(confint(lmList(x ~ slope | id, data), pooled = TRUE), order = 1)

 Is it possible to either reduce the size of the id numbers down the L
 axis or spread the plots out over a couple of pages so the text is less
 squashed up? (- I need to retain the ordering). I've tried the usual
 plot function arguments (amongst other things) but alas to no avail.

 Thanks!

 M

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 PLEASE do read the posting guide
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Re: [R] query in plot(intervals....

2013-05-14 Thread ilai

On Tue, May 14, 2013 at 10:05 AM, Michelle Morters
mm...@hermes.cam.ac.ukwrote:

 Hi -

 I would like the plot ordered by intercept.


One way will be to tweak the ?intervals.lmList object

require(nlme)
fm1 - intervals(lmList(distance ~ age | Subject, Orthodont))
fm2 - fm1[order(fm1[,2,1]),,]
class(fm2) - class(fm1)
plot(fm2)

 Next time please, *you* provide the reproducible example...



 Ordering is doable if the intervals function is substituted with the
 confint function and order=1 included. Is ordering doable with intervals
 function, please?

 Thanks!

 M

 results-lmList(x~slope|id,data)

 plot(intervals(results))


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Re: [R] ?on-consequitve # of lags in VAR (package 'vars)?

2013-05-14 Thread Dimitri Liakhovitski

My question has been answered elsewhere, but here is the answer:
?restrict


On Mon, May 13, 2013 at 8:49 PM, Dimitri Liakhovitski 
dimitri.liakhovit...@gmail.com wrote:

 Hello!

 I was wondering if it is at all possible (in vars or maybe outside of
 it?) to include non-consequite lags into the model, vor example lags 1 and
 3.

 So far, it looks like when I set p = 3 under VAR, it includes all lags
 (1:3).

 Thank you very much!

 --
 Dimitri Liakhovitski



 --
 Dimitri Liakhovitski




-- 
Dimitri Liakhovitski

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[R] Specifying Correlation Structures in Linear Multivariate State Space Models

2013-05-14 Thread Katharina Best

I am investigating various R packages that facilitate estimation of linear
Gaussian multivariate state space models. I stumbled across the MARSS
package (http://cran.r-project.org/web/packages/MARSS/index.html), which I
believe is very well done, but am finding one missing feature that I cannot
live without. I am trying to estimate a Gaussian multiuvatiate local level
model

The MARSS package seems to require that *w *and *v *have the same
dimensions as the state and observation equations, respectively. For
example, a model with 4 states in the state equation would need to have
four error terms and a 4x4 covariance matrix Q. My questions is whether
there is any way to get around this restriction either by

   - adding a *p x m *coefficient matrix in front of *w*, where *p *is the
   number of error terms and *m *is the number of state equations, thus
   allowing the error term of each of the *m *companies to be a linear
   combination of *p *independent error terms, OR
   - by specifying the correlation structure in *Q *appropriately. When I
   attempt to restrict *Q* in a way that mimics the model that would be
   created by a coefficient matrix like one described above I run into the
   problem that expressions defining values of entries in *Q* must be
   linear in parameters. This prevents me from e.g. writing expressions that
   are products of estimated correlation coefficients.

I found a similar issue when investigating the dlm package, but am also
looking for suggestions on other approaches to try.

*Some more detail:*
Currently in MARSS, the equations look like this:

   - X_t = B_t X_{t-1} + u_t + w_t  where X is mx1, B is mxm, u is mx1, w
   is mx1
   - Y_t = Z_t X_t + a_t + v_t where Y is nx1, Z is nxm, X is mx1, a is
   nx1, v is nx1

My goal is to write error terms as linear combinations, yielding something
like:

   - X_t = B_t X_{t-1} + u_t + R_t w_t  where X is mx1, B is mxm, u is mx1,
   R is pxm, w is px1
   - Y_t = Z_t X_t + a_t + S_t v_t  where Y is nx1, Z is nxm, X is mx1, a
   is nx1, S is qxn, v is qx1

*Thanks so much for the help!*

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Re: [R] Unexpected behavior of apply when FUN=sample

2013-05-14 Thread Patrick Burns


This is Circle 8.1.47 of 'The R Inferno'.

http://www.burns-stat.com/documents/books/the-r-inferno/

Pat


On 14/05/2013 09:52, Luca Nanetti wrote:

Dear experts,

I wanted to signal a peculiar, unexpected behaviour of 'apply'. It is not a
bug, it is per spec, but it is so counterintuitive that I thought it could
be interesting.

I have an array, let's say test, dim=c(7,5).


test - array(1:35, dim=c(7, 5))
test


  [,1] [,2] [,3] [,4] [,5]
[1,]18   15   22   29
[2,]29   16   23   30
[3,]3   10   17   24   31
[4,]4   11   18   25   32
[5,]5   12   19   26   33
[6,]6   13   20   27   34
[7,]7   14   21   28   35

I want a new array where the content of the rows (columns) are permuted,
differently per row (per column)

Let's start with the columns, i.e. the second MARGIN of the array:

test.m2 - apply(test, 2, sample)
test.m2


  [,1] [,2] [,3] [,4] [,5]
[1,]1   10   18   23   32
[2,]79   16   25   30
[3,]6   14   17   22   33
[4,]4   11   15   24   34
[5,]2   12   21   28   31
[6,]58   20   26   29
[7,]3   13   19   27   35

perfect. That was exactly what I wanted: the content of each column is
shuffled, and differently for each column.
However, if I use the same with the rows (MARGIIN = 1), the output is
transposed!


test.m1 - apply(test, 1, sample)
test.m1


  [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]12345   13   21
[2,]   22   30   17   18   19   20   35
[3,]   15   23   24   32   26   27   14
[4,]   29   16   31   25   33   34   28
[5,]89   10   11   1267

In other words, I wanted to permute the content of the rows of test, and
I expected to see in the output, well, the shuffled rows as rows, not as
column!

I would respectfully suggest to make this behavior more explicit in the
documentation.

Kind regards,
Luca Nanetti



--
Patrick Burns
pbu...@pburns.seanet.com
twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
 'Impatient R'
 'The R Inferno'
 'Tao Te Programming')

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[R] Help me please: gplot, facets_wrap and ordering of x axis dates

2013-05-14 Thread David Lyon

I have a text file of data as below and doing a ggplot line plot of all the ids 
as separate mini line plots which works with the following code.

Problem how do I order the dates for each id plot on the x axis so that the 
dates are going from oldest to most recent


Thanks in advance

Dave





CODE:

a-read.table(DATA,header=TRUE);
b-a[order(as.Date(a$date, format=%m/%d/%Y)),]

ggplot(data=b) + geom_line(aes(x=date, y=value, group=id, colour= factor(id))) 
+ facet_wrap(~id, scales = free)






DATA:


id  date    value
001 8/16/10 0.16
001 10/25/10    0.16
001 11/8/10 0.42
001 11/22/10    0.81
001 12/6/10 0.16
002 8/18/10 2.93
002 10/25/10    2.4
002 11/8/10 1.36
002 11/22/10    1.22
004 8/20/10 0.77
004 10/25/10    0.85
004 11/8/10 1.22
004 11/22/10    0.21
004 12/6/10 1.81
004 10/26/11    0.54
005 6/4/09  1.33
005 6/18/10 1.32
005 8/25/10 2.5
005 9/15/10 1.3
005 10/25/10    1.1
005 11/8/10 0.66
005 11/22/10    0.84
005 12/6/10 7.42
005 11/4/11 1.21
006 9/23/10 0.97
006 10/25/10    2.25
006 11/8/10 0.51
006 11/22/10    0.53
006 12/6/10 0.41
008 8/25/10 3.14
008 10/25/10    3.58
008 11/8/10 2.41
008 12/6/10 2.08
008 6/11/12 3.2
009 10/6/10 0.24
009 11/22/10    0.34
009 2/2/11  0.58
009 2/16/11 0.54
009 3/2/11  0.25
009 3/16/11 0.39
010 9/14/10 0.28
010 2/2/11  0.19
010 2/16/11 0.42
010 3/2/11  0.39
010 3/16/11 0.26
011 8/20/10 0.16
011 2/2/11  0.16
011 2/16/11 0.16
011 3/2/11  0.16
011 3/16/11 1.76
011 10/26/11    0.16
012 12/14/10    0.48
012 2/2/11  1.2
012 2/16/11 0.44
012 3/2/11  0.32
012 3/16/11 0.34
013 11/13/09    0.73
013 8/19/10 3.32
013 2/2/11  13.7
014 2/6/13  1.35
014 4/24/13 0.85
014 8/18/10 0.66
014 10/5/10 0.68
014 10/27/10    0.53
014 2/2/11  0.54
014 2/16/11 0.49
014 3/2/11  0.31
014 3/16/11 0.4
014 6/29/11 0.53
014 8/15/11 0.55
014 8/15/12 0.94
014 10/31/12    0.74
015 12/10/10    0.4
015 2/2/11  0.44
015 3/2/11  0.38
015 3/16/11 0.43
016 12/17/10    0.18
016 1/25/11 0.16
016 2/2/11  0.53
016 2/2/11  0.54
016 2/16/11 0.46
016 3/2/11  0.29
016 3/16/11 0.2
016 3/20/12 0.18
016 3/26/12 0.23
016 3/30/12 0.52
016 4/2/12  0.33
016 4/23/12 0.3
017 11/17/11    3.35
017 12/9/11 2.56
018 2/25/13 18.1
018 3/11/13 14.9
018 3/25/13 11.1
018 4/10/13 8.47
018 4/22/13 15.9


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Re: [R] Help me please: gplot, facets_wrap and ordering of x axis dates

Thank you for supplying the code. It would be easier to help you  if we also 
had some data to work with.  ?dput

https://github.com/hadley/devtools/wiki/Reproducibility

I think reorder() is likely to do the trick but I don't have enough time to 
mock up some data and check at the moment.

Have a look at  
http://stackoverflow.com/questions/3744178/ggplot2-sorting-a-plot

Good luck.
John Kane
Kingston ON Canada


 -Original Message-
 From: david_ly...@yahoo.com
 Sent: Tue, 14 May 2013 12:59:07 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] Help me please: gplot, facets_wrap and ordering of x axis
 dates
 
 I have a text file of data as below and doing a ggplot line plot of all
 the ids as separate mini line plots which works with the following code.
 
 Problem how do I order the dates for each id plot on the x axis so that
 the dates are going from oldest to most recent
 
 
 Thanks in advance
 
 Dave
 
 
 
 
 
 CODE:
 
 a-read.table(DATA,header=TRUE);
 b-a[order(as.Date(a$date, format=%m/%d/%Y)),]
 
 ggplot(data=b) + geom_line(aes(x=date, y=value, group=id, colour=
 factor(id))) + facet_wrap(~id, scales = free)
 
 
 
 
 
 
 DATA:
 
 
 id  date    value
 001 8/16/10 0.16
 001 10/25/10    0.16
 001 11/8/10 0.42
 001 11/22/10    0.81
 001 12/6/10 0.16
 002 8/18/10 2.93
 002 10/25/10    2.4
 002 11/8/10 1.36
 002 11/22/10    1.22
 004 8/20/10 0.77
 004 10/25/10    0.85
 004 11/8/10 1.22
 004 11/22/10    0.21
 004 12/6/10 1.81
 004 10/26/11    0.54
 005 6/4/09  1.33
 005 6/18/10 1.32
 005 8/25/10 2.5
 005 9/15/10 1.3
 005 10/25/10    1.1
 005 11/8/10 0.66
 005 11/22/10    0.84
 005 12/6/10 7.42
 005 11/4/11 1.21
 006 9/23/10 0.97
 006 10/25/10    2.25
 006 11/8/10 0.51
 006 11/22/10    0.53
 006 12/6/10 0.41
 008 8/25/10 3.14
 008 10/25/10    3.58
 008 11/8/10 2.41
 008 12/6/10 2.08
 008 6/11/12 3.2
 009 10/6/10 0.24
 009 11/22/10    0.34
 009 2/2/11  0.58
 009 2/16/11 0.54
 009 3/2/11  0.25
 009 3/16/11 0.39
 010 9/14/10 0.28
 010 2/2/11  0.19
 010 2/16/11 0.42
 010 3/2/11  0.39
 010 3/16/11 0.26
 011 8/20/10 0.16
 011 2/2/11  0.16
 011 2/16/11 0.16
 011 3/2/11  0.16
 011 3/16/11 1.76
 011 10/26/11    0.16
 012 12/14/10    0.48
 012 2/2/11  1.2
 012 2/16/11 0.44
 012 3/2/11  0.32
 012 3/16/11 0.34
 013 11/13/09    0.73
 013 8/19/10 3.32
 013 2/2/11  13.7
 014 2/6/13  1.35
 014 4/24/13 0.85
 014 8/18/10 0.66
 014 10/5/10 0.68
 014 10/27/10    0.53
 014 2/2/11  0.54
 014 2/16/11 0.49
 014 3/2/11  0.31
 014 3/16/11 0.4
 014 6/29/11 0.53
 014 8/15/11 0.55
 014 8/15/12 0.94
 014 10/31/12    0.74
 015 12/10/10    0.4
 015 2/2/11  0.44
 015 3/2/11  0.38
 015 3/16/11 0.43
 016 12/17/10    0.18
 016 1/25/11 0.16
 016 2/2/11  0.53
 016 2/2/11  0.54
 016 2/16/11 0.46
 016 3/2/11  0.29
 016 3/16/11 0.2
 016 3/20/12 0.18
 016 3/26/12 0.23
 016 3/30/12 0.52
 016 4/2/12  0.33
 016 4/23/12 0.3
 017 11/17/11    3.35
 017 12/9/11 2.56
 018 2/25/13 18.1
 018 3/11/13 14.9
 018 3/25/13 11.1
 018 4/10/13 8.47
 018 4/22/13 15.9
 
 
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[R] Post hoc test for GLM with poisson distribution

2013-05-14 Thread Bel Braz

Hi R-people,

 I performed controlled experiments to evaluated the seeds germination of
two palms under four levels of water treatments. I conducted a generalized
linear model (GLM) with a Poisson distribution to verify whether there were
significant differences in the number of seed germination (NS-count
variable)  between treatments and species (explanatory variables). Thus, my
model and output were:

model1-glm(NS~Treatments*Species, family=poisson)



Coefficients:

   Estimate  Std. Error  z value  Pr(|z|)

(Intercept)   2.562470.57544   4.4538.46e-06 ***

Treatments  -2.072670.35065  -5.911   3.40e-09 ***

Species -0.003120.30527  -0.010  0.992

Treatments:Species 0.903970.17896   5.051  4.39e-07 ***

 Null deviance: 379.870  on 98  degrees of freedom

Residual deviance:  68.302  on 95  degrees of freedom



There is a significant interaction between treatments:species. Which is the
post hoc test appropriate for this model?



Thanks,

Maria Isabel

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Re: [R] need means on all boxplots, but only half of them got that

2013-05-14 Thread S Ellison



boxplot( Daten$weight~interaction(Daten$Dosis,Daten$sex, drop=TRUE))
means-tapply( Daten$weight, Daten$Dosis, mean)
points(means, pch=5, col=red, lwd=5)

 ...only the boxplots for male got that point on them, 
 its really weird because I don't think that I separated the sex in the codes 
 above
Look again; the boxplot is clearly distinguished by sex.

Try 
means-tapply( Daten$weight, interaction(Daten$Dosis,Daten$sex, drop=TRUE), 
mean)

S Ellison

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Re: [R] Help me please: gplot, facets_wrap and ordering of x axis dates

2013-05-14 Thread David Winsemius

Your value column will be input as character because of the  signs and you 
need to convert the dates in %m/%d/%y format to real R dates.

-- 
David
Sent from my iPhone

On May 14, 2013, at 3:59 PM, David Lyon david_ly...@yahoo.com wrote:

 I have a text file of data as below and doing a ggplot line plot of all the 
 ids as separate mini line plots which works with the following code.
 
 Problem how do I order the dates for each id plot on the x axis so that the 
 dates are going from oldest to most recent
 
 
 Thanks in advance
 
 Dave
 
 
 
 
 
 CODE:
 
 a-read.table(DATA,header=TRUE);
 b-a[order(as.Date(a$date, format=%m/%d/%Y)),]
 
 ggplot(data=b) + geom_line(aes(x=date, y=value, group=id, colour= 
 factor(id))) + facet_wrap(~id, scales = free)
 
 
 
 
 
 
 DATA:
 
 
 id  datevalue
 001 8/16/10 0.16
 001 10/25/100.16
 001 11/8/10 0.42
 001 11/22/100.81
 001 12/6/10 0.16
 002 8/18/10 2.93
 002 10/25/102.4
 002 11/8/10 1.36
 002 11/22/101.22
 004 8/20/10 0.77
 004 10/25/100.85
 004 11/8/10 1.22
 004 11/22/100.21
 004 12/6/10 1.81
 004 10/26/110.54
 005 6/4/09  1.33
 005 6/18/10 1.32
 005 8/25/10 2.5
 005 9/15/10 1.3
 005 10/25/101.1
 005 11/8/10 0.66
 005 11/22/100.84
 005 12/6/10 7.42
 005 11/4/11 1.21
 006 9/23/10 0.97
 006 10/25/102.25
 006 11/8/10 0.51
 006 11/22/100.53
 006 12/6/10 0.41
 008 8/25/10 3.14
 008 10/25/103.58
 008 11/8/10 2.41
 008 12/6/10 2.08
 008 6/11/12 3.2
 009 10/6/10 0.24
 009 11/22/100.34
 009 2/2/11  0.58
 009 2/16/11 0.54
 009 3/2/11  0.25
 009 3/16/11 0.39
 010 9/14/10 0.28
 010 2/2/11  0.19
 010 2/16/11 0.42
 010 3/2/11  0.39
 010 3/16/11 0.26
 011 8/20/10 0.16
 011 2/2/11  0.16
 011 2/16/11 0.16
 011 3/2/11  0.16
 011 3/16/11 1.76
 011 10/26/110.16
 012 12/14/100.48
 012 2/2/11  1.2
 012 2/16/11 0.44
 012 3/2/11  0.32
 012 3/16/11 0.34
 013 11/13/090.73
 013 8/19/10 3.32
 013 2/2/11  13.7
 014 2/6/13  1.35
 014 4/24/13 0.85
 014 8/18/10 0.66
 014 10/5/10 0.68
 014 10/27/100.53
 014 2/2/11  0.54
 014 2/16/11 0.49
 014 3/2/11  0.31
 014 3/16/11 0.4
 014 6/29/11 0.53
 014 8/15/11 0.55
 014 8/15/12 0.94
 014 10/31/120.74
 015 12/10/100.4
 015 2/2/11  0.44
 015 3/2/11  0.38
 015 3/16/11 0.43
 016 12/17/100.18
 016 1/25/11 0.16
 016 2/2/11  0.53
 016 2/2/11  0.54
 016 2/16/11 0.46
 016 3/2/11  0.29
 016 3/16/11 0.2
 016 3/20/12 0.18
 016 3/26/12 0.23
 016 3/30/12 0.52
 016 4/2/12  0.33
 016 4/23/12 0.3
 017 11/17/113.35
 017 12/9/11 2.56
 018 2/25/13 18.1
 018 3/11/13 14.9
 018 3/25/13 11.1
 018 4/10/13 8.47
 018 4/22/13 15.9
 
 
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[R] empirical and GPD for time series simulation

2013-05-14 Thread Al Ehan

Hi,

Does anyone know how to simulate a long time series (say 1000 daily series)
or generally a series, with inverse empirical distribution and generalized
pareto distribution (meaning to say the time series has a marginal
distribution of empirical and GPD distribution.) using the R package?

Does anybody know if there is a package in R that do this thing?

Thanks in advance,

B

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Re: [R] How to capture the expression corresponding to the i param in the [ function

2013-05-14 Thread David Winsemius

I think you need to read ?setClass and ?setMethod. There is an example of 
defining a [ method for a class that inherits from 'data.frame'. I suspect 
you need to capture the various possibilities for the arguments being present 
or missing.

-- 
David

Sent from my iPhone

On May 14, 2013, at 5:45 AM, Nhan Vu Lam Chi nhani...@adatao.com wrote:

 Dear David,
 
 First, I would like to say thank you for your very soon reply. Second, I want 
 to clarify the question because it seems to not carrying exactly what I want 
 to ask. Let take an example on R data.frame:
 V1 - 1
 df2 - df[V1== 1,] # df is a data.frame, this command is correct, right?
 
 The evaluation steps for the above command are:
 1. R evaluate V1  1 to get TRUE
 2. The command becomes df2 - df[TRUE,] which copies all rows of df to df2
 
 What I want is to capture the V1  1 expression instead of letting R do the 
 evaluation in case of the custom [ function. Assume my class is mydf, the S4 
 function should be:
 setMethod([, signature(x=mydf), function(x,i,j,...,drop=TRUE) {
  e - substitute(i)
  // do parsing and custom-evaluation tasks
 })
 
 Currently, i is always a vector of type character, numeric or logic due to R 
 evaluation.
 
 I am a newbie to R, so please tolerate my mistakes or misunderstanding. 
 Thanks!
 Nhan Vu
 
 
 
 On Tue, May 14, 2013 at 12:14 PM, David Winsemius dwinsem...@comcast.net 
 wrote:
 
 On May 13, 2013, at 6:38 PM, Nhan Vu Lam Chi wrote:
 
  Hi everyone,
  I currently work on a S4 class that has the [ function. I want to capture
  the unevaluated expression corresponding to the i param using substitute()
  function and do a non-standard evaluation. However R automatically
  evaluates the expression and give me its value.
  For example:
  Given mydf[mydf$V1  1,] with mydf is an object of my custom S4 dataframe
  class and V1 is one of its columns, I want to get the unevaluated
  expression mydf$V1  1.
 
  My questions are:
  1. Is it possible to do that in R?
  2. If yes, how to do?
 
 Doesn't this cry out for the S4 class definition of [ to be answerable?. 
 Because [ is generic, it could have almost any definition at the whim of 
 the package author.
 
  My R version and OS info are:
  R version 2.15.3 (2013-03-01) -- Security Blanket
  Copyright (C) 2013 The R Foundation for Statistical Computing
  ISBN 3-900051-07-0
  Platform: x86_64-pc-linux-gnu (64-bit)
 
  This is the first time I post to the mailing list, so please forgive any
  mistakes and/or advise me if possible.
  Regards,
  Nhan Vu
 
[[alternative HTML version deleted]]
 
 You are forgiven, but  this once, for posting in HTML.
 
 --
 David Winsemius
 Alameda, CA, USA
 

[[alternative HTML version deleted]]

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Re: [R] Changing Order of Factor Levels in Mixed Model (nlme)

2013-05-14 Thread Ben Bolker

Edward Patzelt patze003 at umn.edu writes:

 
 R Help -
 
 Why is that in the results below, changing the order of the factor
 (trialType2: levels - DD, SD, DS, SS) changes the estimates in the fixed
 effects tests?

  I think you're not doing what you expected.  By sorting the factor,
you are _not_ changing the order of the factor levels (which you might
have been trying to do in order to change the parameterization); rather,
you're changing the actual order of the observations of the factor,
which is scrambling their association with the other variables
(response=proportion.down and the grouping variable, subject).
  I can't think of a scenario under which sorting the order of only one
of the variables in the data frame is not a mistake, unless you're
trying to randomize the order to do a permutation test.

  What you might have meant to do is to change the order of
the _levels_ of the factor, which you can do via

tmp.dat4$trialType2 - factor(tmp.dat4$trialType2,
levels=c(DD,SD,DS,SS))

or perhaps

tmp.dat4 - transform(tmp.dat4,
   trialType2=factor(trialType2,levels=sort(levels(trialType2

(see also ?relevel and ?reorder)

  Changing the order of the factor levels will also change
the specific estimates of the fixed-effect parameters, in this
case by changing the parameterization (contrasts), which are
by default based on differences from the first factor level
(although see also ?contr.SAS), but not the overall meaning/fit
of the model.

  By the way, this isn't specifically a mixed-effects model
question -- the same issues would apply with just about any
statistical model in R (see e.g. Faraway's books on linear
and generalized models -- some early drafts are available in
the contributed documentation section).

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Re: [R] How to capture the expression corresponding to the i param in the [ function

2013-05-14 Thread David Winsemius



On May 14, 2013, at 8:02 PM, David Winsemius wrote:

I think you need to read ?setClass and ?setMethod. There is an  
example of defining a [ method for a class that inherits from  
'data.frame'. I suspect you need to capture the various  
possibilities for the arguments being present or missing.




setClass(myFrame, contains = data.frame,
representation(callexp = character))

df1 - data.frame(x = 1:10, y = rnorm(10), z = sample(letters,10))

mydf1 - new(myFrame, df1, callexp = )

setMethod([,
signature(x = myFrame),
function (x, i, j, ..., drop = TRUE)
{callexp - deparse(substitute(i))
S3Part(x) - callNextMethod()
x@callexp - callexp
x
}
)

mydf1[1:2, ]
#--
Object of class myFrame
  x  y z
1 1 -1.9119574 f
2 2  0.2719548 i
Slot callexp:
[1] 1:2

 mydf1[mydf1$x5, ]
Object of class myFrame
  x  y z
1 1 -0.1065694 u
2 2  0.8571367 l
3 3  1.7259175 z
4 4  0.3618450 x
Slot callexp:
[1] mydf1$x  5




--
David

Sent from my iPhone

On May 14, 2013, at 5:45 AM, Nhan Vu Lam Chi nhani...@adatao.com  
wrote:



Dear David,

First, I would like to say thank you for your very soon reply.  
Second, I want to clarify the question because it seems to not  
carrying exactly what I want to ask. Let take an example on R  
data.frame:

V1 - 1
df2 - df[V1== 1,] # df is a data.frame, this command is correct,  
right?


The evaluation steps for the above command are:
1. R evaluate V1  1 to get TRUE
2. The command becomes df2 - df[TRUE,] which copies all rows of df  
to df2


What I want is to capture the V1  1 expression instead of  
letting R do the evaluation in case of the custom [ function.  
Assume my class is mydf, the S4 function should be:

setMethod([, signature(x=mydf), function(x,i,j,...,drop=TRUE) {
e - substitute(i)
// do parsing and custom-evaluation tasks
})

Currently, i is always a vector of type character, numeric or logic  
due to R evaluation.


I am a newbie to R, so please tolerate my mistakes or  
misunderstanding. Thanks!

Nhan Vu



On Tue, May 14, 2013 at 12:14 PM, David Winsemius dwinsem...@comcast.net 
 wrote:


On May 13, 2013, at 6:38 PM, Nhan Vu Lam Chi wrote:


Hi everyone,
I currently work on a S4 class that has the [ function. I want to  
capture
the unevaluated expression corresponding to the i param using  
substitute()

function and do a non-standard evaluation. However R automatically
evaluates the expression and give me its value.
For example:
Given mydf[mydf$V1  1,] with mydf is an object of my custom S4  
dataframe

class and V1 is one of its columns, I want to get the unevaluated
expression mydf$V1  1.

My questions are:
1. Is it possible to do that in R?
2. If yes, how to do?


Doesn't this cry out for the S4 class definition of [ to be  
answerable?. Because [ is generic, it could have almost any  
definition at the whim of the package author.



My R version and OS info are:
R version 2.15.3 (2013-03-01) -- Security Blanket
Copyright (C) 2013 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-linux-gnu (64-bit)

This is the first time I post to the mailing list, so please  
forgive any

mistakes and/or advise me if possible.
Regards,
Nhan Vu

 [[alternative HTML version deleted]]


You are forgiven, but  this once, for posting in HTML.

--
David Winsemius
Alameda, CA, USA




[[alternative HTML version deleted]]

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David Winsemius, MD
Alameda, CA, USA

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[R] R-Help: nparLD Package Non-parametric Repeated Measures

2013-05-14 Thread James Casey

Hi,

I'm trying to analyze repeated measurements of body temperature data
collected from 7 randomly chosen subjects (e.g. turtles). I am using R,
along with the nparLD package to test for an effect of diel period (fixed
factor: day or night) and season (sub-plot fixed factor: spring, summer,
fall) on body temperature.

Based on this set-up (LD-F2), I am using the non-parametric nparLD
pacakge([url]http://www.inside-r.org/packages/cran/nparLD/docs/ld.f2[/url])
because data transformations were unsuccessful and I am randomly missing
some paired values.

Main issue/question: In R the nparLD ANOVA-type Test showed a significant
p-value for diel period, no effect of season, and no interaction between
diel period and season. But a post-hoc Wilcoxon Signed-Rank Test did NOT
find a significant difference (p = 0.054) for diel period (day vs night)
body temperature.

How is it possible to have a significant effect for day vs night, based on
the nparLD package, but NO significant difference between day and night for
the post-hoc Wilcoxon test?

Also, if I only have two levels of the fixed effect (day vs night), do I
need to run a post-hoc test or just look at the mean values after the
ANOVA-type test?

Data info:

The repeated measurements on the 7 subjects had 2 fixed effects:

1. Diel period (day or night)
2. Season (Spring, summer, and fall)(Subplot Factor)

Mean values for body temperature and for diel period are below. Diel column
(D=Day, N = Night). State column (RT=Spring, RF = Summer, PT = Fall).
Subject, N=7. NA = missing value.

All comments (good and bad) are greatly appreciated!

Thanks,
James



 -- output of sessionInfo():

[code]
 data=read.csv(file.choose(), header=TRUE)
 attach(data)
 data
  stp diel state subject
1  26.2DRT   1
2  26.4NRT   1
3  24.1DRT   2
4NANRT   2
5NADRT   3
6  25.2NRT   3
7  27.1DRT   4
8  26.5NRT   4
9  26.9DRT   5
10 27.1NRT   5
11 26.2DRT   6
12 26.0NRT   6
13 26.3DRT   7
14 26.7NRT   7
15 26.0DRF   1
16 26.6NRF   1
17 24.2DRF   2
18 25.6NRF   2
19 25.6DRF   3
20 26.6NRF   3
21 26.1DRF   4
22 26.9NRF   4
23 27.2DRF   5
24 27.4NRF   5
25 26.2DRF   6
26 26.7NRF   6
27 27.2DRF   7
28 27.5NRF   7
29 25.0DPT   1
30 24.8NPT   1
31   NADPT   2
32   NANPT   2
33   NADPT   3
34   NANPT   3
35 26.7DPT   4
36 26.9NPT   4
37 27.6DPT   5
38 27.5NPT   5
39 25.2DPT   6
40 24.9NPT   6
41 27.1DPT   7
42 27.0NPT   7


ex.f2-ld.f2(y=stp, time1=diel, time2=state, subject=subject,
time1.name=Diel, time2.name=State, description=FALSE)

 ex.f2$ANOVA.test
   Statistic   dfp-value
Diel   4.9028447 1.00 0.02681249
State  0.2332795 1.374320 0.70586274
Diel:State 2.1937783 1.062943 0.13717393
[/code]

[code]
 detach(data)
 data=read.csv(file.choose(), header=TRUE)
 attach(data)
 data
day night
1  26.2  26.4
2  26.0  26.6
3  25.0  24.8
4  24.2  25.6
5  25.6  26.6
6  27.1  26.5
7  26.1  26.9
8  26.7  26.9
9  26.9  27.1
10 27.2  27.4
11 27.6  27.5
12 26.2  26.0
13 26.2  26.7
14 25.2  24.9
15 26.3  26.7
16 27.2  27.5
17 27.1  27.0

 library(coin)

 wilcoxsign_test(day ~ night, distribution=exact)

Exact Wilcoxon-Signed-Rank Test

data:  y by x (neg, pos)
 stratified by block
Z = -1.9234, p-value = 0.05482
alternative hypothesis: true mu is not equal to 0

[/code]

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Re: [R] Broken line questions

Hi,
It is because of the unequal lengths of x and y.
I was able to plot without the errors.  But, not sure if this is what you 
wanted.

mydata - read.table(TestData.csv, header=TRUE, sep=\t)
reg1-lm(MW01~Year,data=mydata)
 
plot(MW01~Year,data=mydata[!is.na(mydata$MW01),],col=ifelse(D_MW01,black,red),ylab=END
 (mg/L),pch=ifelse(D_MW01,19,24),cex=1)
with(mydata[!is.na(mydata$MW01),],lines(Year,MW01,lty=c(1),col=black))

plx1-predict(loess(MW01 ~ Year, data=mydata[!is.na(mydata$MW01),]), se=T)
#rough  ready CI by adding and subtracting 2 times the standard error to the 
mean
lines(mydata[!is.na(mydata$MW01),Year],plx1$fit+2*plx1$s, lty=2) 
lines(mydata[!is.na(mydata$MW01),Year],plx1$fit-2*plx1$s, lty=2) 
mydata1-mydata[!is.na(mydata$MW01),2:3]
y.loess - loess(y ~ x, span=0.8, data.frame(x=mydata1$Year,y=mydata1$MW01))

# Compute loess smoothed values for all points along the curve
y.predict - predict(y.loess, data.frame(x=mydata1$Year))

# Plots the curve.
lines(mydata1$Year,y.predict, lty=2, lwd=2)

BTW,  Non-detect in the legend is not clear.
A.K.

#Add line between the points
lines(mydata1$Year,mydata1$MW01)







From: David Doyle kydaviddo...@gmail.com
To: arun smartpink...@yahoo.com 
Sent: Tuesday, May 14, 2013 11:10 PM
Subject: Re: [R] Broken line questions



Hi,

I'm trying to do something very similar to the graph before and I got the 
broken line problem fixed with your example.

On this one I'm trying to add 2 times the standard error (approxmating a 10% 
confidence) onto the LOESS curve.

The data is at 
https://docs.google.com/file/d/0B0CqEkyKWiq9SjFwemRhZnV4aTA/edit?usp=sharing 
(sorry I don't have access to my FTP right now.)

When I use the code below, I get 
Error in xy.coords(x, y) : 'x' and 'y' lengths differ
I assume this is because there are blanks in the data.  

Any suggestions???


The code I'm  using is:

#set the dir to where you data is

setwd(c:/R)

#Load your data.  The data is in a spreadsheet named nd-spreadsheet and we are 
going to call it data in R
mydata - read.table(TestData.csv, header=TRUE, sep=,,)
attach(mydata)

reg1 - lm(MW01~Year)
par(cex=1)


#Plots the data but makes nondetects a different color and type based on column 
D_MW01 being a 0 for ND and 1 for detect.

plot(MW01~Year,data=mydata[!is.na(mydata$MW01),], col=ifelse (D_MW01, black, 
red),ylab = EMD (mg/L), pch=ifelse(D_MW01, 19, 24), cex = 1)
with(mydata[!is.na(mydata$MW01),],lines(Year,MW01,lty=c(1),col=black))
plx-predict(loess(MW01 ~ Year, data=mydata), se=T)

#rough  ready CI by adding and subtracting 2 times the standard error to the 
mean
lines(mydata$Year,plx$fit+2*plx$s, lty=2) 
lines(mydata$Year,plx$fit-2*plx$s, lty=2)

# Apply loess smoothing using the default span value of 0.8.  You can change 
the curve by changing the span value.
y.loess - loess(y ~ x, span=0.8, data.frame(x=Year, y=MW01))

# Compute loess smoothed values for all points along the curve
y.predict - predict(y.loess, data.frame(x=Year))

# Plots the curve.
lines(Year,y.predict, lty=2, lwd=2)


#Add line between the points
lines(Year,MW01)


#Add Legend to MW01.  You can change the size of the box by changing cex = 0.75 
 Large # makes it larger.
legend (topleft, 
c(Smoothing Curve,Detect,Non-Detect), 
col = c(1,1,1),
cex = 1, 
text.col = black, 
lty = c(2,-1,-1), 
lwd = c(2,-1,-1),
pch = c(-1,19,24), 
merge = TRUE, bg = 'gray90')

#Add title
title(main=MW01)

# Done

Thanks again
David




On Sun, May 12, 2013 at 10:11 AM, arun smartpink...@yahoo.com wrote:

No problem,
Regards,
Arun








From: David Doyle kydaviddo...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Sunday, May 12, 2013 8:03 AM
Subject: Re: [R] Broken line questions




Thanks!

That took care of it.



On Sat, May 11, 2013 at 10:01 PM, arun smartpink...@yahoo.com wrote:

Hi,
May be this helps:
plot(dataset1~Date,data=mydata[!is.na(mydata$dataset1),],ylim=range(5.7,8),pch=10,cex=0.8,col=black,xlab=Date,ylab=pH)
 
with(mydata[!is.na(mydata$dataset2),],points(Date,dataset2,col=blue,pch=2,cex=0.8))
 
with(mydata[!is.na(mydata$dataset3),],points(Date,dataset3,col=red,pch=2,cex=0.8))
 
with(mydata[!is.na(mydata$dataset1),],lines(Date,dataset1,lty=c(1),col=black))
 
with(mydata[!is.na(mydata$dataset2),],lines(Date,dataset2,lty=c(1),col=blue))
 
with(mydata[!is.na(mydata$dataset3),],lines(Date,dataset3,lty=c(1),col=red))



A.K.




- Original Message -
From: David Doyle kydaviddo...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Saturday, May 11, 2013 10:22 PM
Subject: [R] Broken line questions

Hello Everyone,

I have some data that like most real world data isn't complete.  I'm trying
to plot all of it together with lines connecting the data points.  Because
I have breaks in the data I have breaks in the lines.

Is there a way that the lines will connect all the markers???   Below is
the code.

Thanks in advance.
David

#Load your data.  The data is in a spreadsheet named KW-spreadsheet and we

Re: [R] Sampling Weights and lmer() update?

2013-05-14 Thread Thomas Lumley

Arguably you are looking in the wrong place (there's a special mixed-models
mailing list for R), but I can answer the question.

No.

At least, there's nothing in lme4, and I haven't done anything (since I
want a more general solution than Stata and MLWiN implement) and I'd be
surprised if someone else had done it.

-thomas

On Tue, May 14, 2013 at 3:35 PM, Richard Blissett rsl.bl...@gmail.comwrote:

Perhaps I am not looking in the right place, but I am looking for a way to
use lmer() to run a multilevel model that incorporates sampling weights. I
have used the Lumley survey package to use sampling weights in the past,
but according to post I found online from Thomas Lumley in mid-2012, R is
currently not equipped to be able to do this.

His post is here:

http://r.789695.n4.nabble.com/sampling-weights-for-multilevel-models-tp4632947p4632955.html

Does anyone know if there has been an update since then to be able to do
this, or if there's another way to go about doing this in R? Otherwise, I
am thinking that I will have to move my data over to Stata and try to run
the multilevel models there.

Richard

--
Richard Blissett

Eco-Tip: Before printing, please consider whether you really need to have
this email on paper.

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and provide commented, minimal, self-contained, reproducible code.

--
Thomas Lumley
Professor of Biostatistics
University of Auckland

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Post hoc test for GLM with poisson distribution

2013-05-14 Thread peter dalgaard


On May 14, 2013, at 21:04 , Bel Braz wrote:

 Hi R-people,
 
 I performed controlled experiments to evaluated the seeds germination of
 two palms under four levels of water treatments. I conducted a generalized
 linear model (GLM) with a Poisson distribution to verify whether there were
 significant differences in the number of seed germination (NS-count
 variable)  between treatments and species (explanatory variables). Thus, my
 model and output were:
 
 model1-glm(NS~Treatments*Species, family=poisson)
 
 
 
 Coefficients:
 
   Estimate  Std. Error  z value  Pr(|z|)
 
 (Intercept)   2.562470.57544   4.4538.46e-06 ***
 
 Treatments  -2.072670.35065  -5.911   3.40e-09 ***
 
 Species -0.003120.30527  -0.010  0.992
 
 Treatments:Species 0.903970.17896   5.051  4.39e-07 ***
 
 Null deviance: 379.870  on 98  degrees of freedom
 
 Residual deviance:  68.302  on 95  degrees of freedom
 
 
 
 There is a significant interaction between treatments:species. Which is the
 post hoc test appropriate for this model?

There's not much post hoc testing to do if the effect is described by a single 
coefficient. Did you forget to code Treatments as a factor variable?

Apart from that, it depends on what you want to do. Do you want to know where 
the interaction comes from, or just within which treatment(s) there is a 
species effect?

Since there is only two species, the easiest way forward is to compute the four 
species effects, one for each treatment. You can then compare the effects 
pairwise (6 comparisons)  or compare each effect to zero (4 comparison). I 
don't think you can do much better than simple Bonferroni corrections in either 
case. 

 
 
 
 Thanks,
 
 Maria Isabel
 
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 R-help@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@cbs.dk  Priv: pda...@gmail.com

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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Broken line questions