date:20140121

[R] how to get the numbers of factors in a matrix

2014-01-21 Thread 张以春

Dear friends,


I have a question do not know how to resolve.


I have a big matrix composed of different columns (I use N here). A column is 
species and another one is latitudes. Now, I want to know how I can get the 
number of different latitudes for every species. I have tried to split the 
matrix according to species (X-split(N, N$species) and then use sapply(X, 
function(m){nlevels(m$latitudes)}) to get that. But the result shows the total 
factor numbers of latitudes but not the factor numbers of every species I 
splitted. Also, I have tried to use tapply(N$latitudes, N$species, nlevels) to 
do this. The result is the same. I am confused about this. Can someone help me 
with that? Thank you very much!


Best regards,
Yichun






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[R] My problem with R

2014-01-21 Thread Lee Marine

Hi, I am a student in trouble with R.
please help me!

I have a 16 huge data(.csv).

data1-read.csv(D://data1.csv, header=FALSE)
.
.
.
data16-read.csv(D://data16.csv, header=FALSE)

and then, I tried to seperated data w/ column.

a1-cbind(data1[1:36568,1])
b1-cbind(data1[1:36568,2])
.
.
.
ac16-cbind(data16[1:85207,31])

a column is a Date(-mm-dd)
b column is a Time(hh:mm:ss)
c to ac is a numeric data.

finished this work, use rbind function.

date-rbind(a1,a2,a3,a4,.,a16)
time-rbind(b1,b2,b3,..,b16)
.
.
.
.
ac-rbind(ac1,ac2,ac3,...,ac16)

I want to see -mm-dd format, but It does not work.
b
  [,1]
[1,] 17753
[2,] 17754
[3,] 17755
[4,] 17756
[5,] 17757
[6,] 17758

a
 [,1]
[1,]1
[2,]1
[3,]1
[4,]1
[5,]1
[6,]1

I don't know why...
Question 1.
I want to paste a and b  -mm-dd hh:mm:ss

Question 2.
I want to averaging mean, median for every each minute groups.
for example)
 Time  c  de f  . .
.
2014-01-20 00:00:00 3.1428 1.99 0.123455 7.5526
2014-01-20 00:00:01 3.1887 1.03 0.176545 7.5234
2014-01-20 00:00:02 2.1876 1.27 0.987455 7.5996
...
2014-01-20 00:01:00 5.4428 1.96 0.164455 7.8666
2014-01-20 00:01:01 1.5658 1.39 0.176545 7.7786
2014-01-20 00:01:02 3.1776 1.67 0.254655 7.4567

then,
 Time   average_C c  de
f  . . .
2014-01-20 00:00:00 here   3.1428 1.99 0.123455 7.5526
2014-01-20 00:00:01  3.1887 1.03 0.176545 7.5234
2014-01-20 00:00:02  2.1876 1.27 0.987455 7.5996
...
2014-01-20 00:01:00  here  5.4428 1.96 0.164455 7.8666
2014-01-20 00:01:01  1.5658 1.39 0.176545 7.7786
2014-01-20 00:01:02  3.1776 1.67 0.254655 7.4567

PLEASE HELP ME, R-help!!!

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Re: [R] how to replace NA values

2014-01-21 Thread kingsly

First of all I thank all the friends.

Though IÂ  haveÂ written zero to replace NA inÂ  previous email, I will replace 
with extreme value (9).



On Tuesday, 21 January 2014 7:05 AM, jwd [via R] 
ml-node+s789695n4683894...@n4.nabble.com wrote:
  
On Sun, 19 Jan 2014 11:39:43 -0800 (PST) 

kingsly [hidden email] wrote: 



The age variable is being read in as a factor because of the 
. Â If you were to replace it with NA, the type becomes numerical: 

Before replacement: 

str(Elder1) 
'data.frame': Â  5 obs. of Â 2 variables: 
Â $ ID : Factor w/ 5 levels ID1,ID2,ID3,..: 1 2 3 4 5 
Â $ age: Factor w/ 3 levels ,35,38: 3 2 1 NA NA 

Notice that the  is treated as a factor level. 

After: 

str(Elder1) 
'data.frame': Â  5 obs. of Â 2 variables: 
Â $ ID : Factor w/ 5 levels ID1,ID2,ID3,..: 1 2 3 4 5 
Â $ age: num Â 38 35 NA NA NA 

SO, the question, is what do you want to do with that column? Â An NA
value tells you honestly that the information is missing. Â Replacing it 
with a zero can be misleading and can bias some basic parameter 
estimates. 

After you know how you want to treat the data in that field, you may 
have a better idea of how to handle the missing data. 

JWD 

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To unsubscribe from R help, click here.
NAML



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[R] lattice

2014-01-21 Thread Said Filahi

hello

 If I have a a program



trend - read.table(txx.csv, header=T, sep=,)
library(lattice)
trellis.par.set(theme = canonical.theme(postscript, col=FALSE))
trellis.par.set(list(fontsize=list(text=8),
 par.xlab.text=list(cex=1.5),
 add.text=list(cex=3.5),
 superpose.symbol=list(cex=1.0)))
key - simpleKey(levels(trend$Season), space = right)
key$text$cex - 1.5
print(
 dotplot(Ville ~ Slope , data = trend, groups = Season,
 key = key,
 xlab = NULL,
 aspect=0.8, layout = c(1,1), ylab=NULL)
)

and my dat det is

 Ville Season Slope significance  Alh DJF -0.3 0.02  Ben DJF 0.13 0.7  Cas
MAM 0.1 0.1  Ess JJA 0.4 0.03  Fes SON 0.9 0.02
how can i change the color of symbole of slope value if the significance is
below 0.05

thank you

FILAHI

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Re: [R] how to get the numbers of factors in a matrix

2014-01-21 Thread Liu , Jun Yi

Dear YZ,

I guess this is what you want:

1.
http://stackoverflow.com/questions/3418128/how-to-convert-a-factor-to-an-integer-numeric-without-a-loss-of-information
2.
http://stackoverflow.com/questions/6979625/arithmetic-operations-on-r-factors/6980780#6980780
3.
http://toddjobe.blogspot.jp/2010/08/converting-r-contingency-tables-to-data.html

but beware of that “If you really want the levels of the factor to be used,
you're either doing something very wrong or too clever for its own good.

All the best
JY

--
From: 张以春 yczh...@nigpas.ac.cn
Reply: 张以春 yczh...@nigpas.ac.cn
Date: 21 January 2014 at 16:57:38
To: r-help@r-project.org r-help@r-project.org
Subject: [R] how to get the numbers of factors in a matrix

Dear friends,

I have a question do not know how to resolve.

I have a big matrix composed of different columns (I use N here).
A column is species and another one is latitudes. Now, I want
to know how I can get the number of different latitudes for every
species. I have tried to split the matrix according to species
(X-split(N, N$species) and then use sapply(X,
function(m){nlevels(m$latitudes)})
to get that. But the result shows the total factor numbers of latitudes
but not the factor numbers of every species I splitted. Also,
I have tried to use tapply(N$latitudes, N$species, nlevels)
to do this. The result is the same. I am confused about this. Can
someone help me with that? Thank you very much!

Best regards,
Yichun

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code.

[R] (no subject)

2014-01-21 Thread Mariacarla Memeo

Hello to everyone!

I'm a new to R and I don't succed in updating a variable (SR) in a for
loop. The script created is the following one:

path=c(C:/Users/memeo/Documents/Dottorato/soggetti visus danneggiato/)
subjects=list.files(path)
# subjects=subjects[c(1:35,37)]

n_gradini=rep(c(1,2,4),length(subjects))
allsubjects=c(rep(subjects,each=3))


for (i in length(subjects))
{ path_each_subj=paste0(path,subjects[i])
  no_thumbs-grep(Thumbs,subjects)
  n=1:length(subjects)
  clear_thumbs=n!=no_thumbs

  if(clear_thumbs[i])
   {
  each_subject=list.files(path_each_subj)
  find_gradini-grep(gradini,each_subject)
  file_gradini=each_subject[find_gradini]
  path_file_gradini=paste0(path_each_subj,/,file_gradini)
  SR-(rep(0,3))
#   SR-matrix(data=NA,nrow=36*3,ncol=1)
for(j in length(path_file_gradini))
   {
gradini-read.table(file=path_file_gradini[j],header=T,quote=\,skip=1)
 data-data.frame(gradini)
 names(data)=c(Id,Time,X,Y,H,Sound)
 #Calculate derivatives for isolate the positive
contributes
 Hdiff=diff(data$H)
 Height=Hdiff[Hdiff0]
 SR[j]-length(Height)/tSR
 }
  blind=data.frame(Names=allsubjects,Gradini=n_gradini,StimuliRate=SR)

  print(SR)  }
else i=i+1
}
write.table(blind,file=10Blind,sep= , eol=\r\n)


CAN ANYONE HELP ME? Thank you very much!



-- 

*M. Memeo*

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Re: [R] how to replace NA values

2014-01-21 Thread PIKAL Petr

Hi

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of kingsly
 Sent: Tuesday, January 21, 2014 7:22 AM
 To: r-help@r-project.org
 Subject: Re: [R] how to replace NA values

 First of all I thank all the friends.

 Though I  have written zero to replace NA in  previous email, I will
 replace with extreme value (9).

Why? Many R functions has ability to work smoothly with NA values but AFAIK 
none is able to work with 9 values.

It is the same as if you wanted to change bicycle tyres to wooden rim. You 
could ride but not with the same comfort.

Regards
Petr

 On Tuesday, 21 January 2014 7:05 AM, jwd [via R] ml-
 node+s789695n4683894...@n4.nabble.com wrote:

 On Sun, 19 Jan 2014 11:39:43 -0800 (PST)

 kingsly [hidden email] wrote:

 The age variable is being read in as a factor because of the .  If
 you were to replace it with NA, the type becomes numerical:

 Before replacement:

 str(Elder1)
 'data.frame':   5 obs. of  2 variables:
  $ ID : Factor w/ 5 levels ID1,ID2,ID3,..: 1 2 3 4 5
  $ age: Factor w/ 3 levels ,35,38: 3 2 1 NA NA

 Notice that the  is treated as a factor level.

 After:

 str(Elder1)
 'data.frame':   5 obs. of  2 variables:
  $ ID : Factor w/ 5 levels ID1,ID2,ID3,..: 1 2 3 4 5
  $ age: num  38 35 NA NA NA

 SO, the question, is what do you want to do with that column?  An NA
 value tells you honestly that the information is missing.  Replacing it
 with a zero can be misleading and can bias some basic parameter
 estimates.

 After you know how you want to treat the data in that field, you may
 have a better idea of how to handle the missing data.

 JWD

 __
 [hidden email] 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.

 If you reply to this email, your message will be added to the
 discussion below: http://r.789695.n4.nabble.com/how-to-replace-NA-
 values-tp4683831p4683894.html
 To start a new topic under R help, email ml-
 node+s789695n78969...@n4.nabble.com
 To unsubscribe from R help, click here.
 NAML

 --
 View this message in context: http://r.789695.n4.nabble.com/how-to-
 replace-NA-values-tp4683831p4683896.html
 Sent from the R help mailing list archive at Nabble.com.
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- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným 
dosažením shody na všech jejích náležitostech.
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pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně 
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Re: [R] My problem with R

2014-01-21 Thread PIKAL Petr

Hi

see in line

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Lee Marine
 Sent: Tuesday, January 21, 2014 7:17 AM
 To: r-help@r-project.org
 Subject: [R] My problem with R

 Hi, I am a student in trouble with R.
 please help me!

 I have a 16 huge data(.csv).

 data1-read.csv(D://data1.csv, header=FALSE) .
 .
 .
 data16-read.csv(D://data16.csv, header=FALSE)

 and then, I tried to seperated data w/ column.

 a1-cbind(data1[1:36568,1])
 b1-cbind(data1[1:36568,2])
 .
 .
 .
 ac16-cbind(data16[1:85207,31])

 a column is a Date(-mm-dd)
 b column is a Time(hh:mm:ss)
 c to ac is a numeric data.

 finished this work, use rbind function.

 date-rbind(a1,a2,a3,a4,.,a16)
 time-rbind(b1,b2,b3,..,b16)
 .
 .
 .
 .
 ac-rbind(ac1,ac2,ac3,...,ac16)

 I want to see -mm-dd format, but It does not work.
 b
   [,1]
 [1,] 17753
 [2,] 17754
 [3,] 17755
 [4,] 17756
 [5,] 17757
 [6,] 17758

 a
  [,1]
 [1,]1
 [2,]1
 [3,]1
 [4,]1
 [5,]1
 [6,]1

 I don't know why...

It is because you did not bother to make a glimpse to R-Intro to learn about 
data types and how to work with them.

1. Reading such data to R means that they are transformed to character or 
factor. See

?str
?factor

2. Functions rbind and cbind produce matrices if input is not a data frame. 
Matrices can not hold different types of data as data frames.

 Question 1.
 I want to paste a and b  -mm-dd hh:mm:ss

timestamp -strptime(paste(data1[,1], data1[,2], sep= ), format = )

see
?paste
?strptime
?as.POSIXct

how to do it exactly

 Question 2.
 I want to averaging mean, median for every each minute groups.
 for example)

see
?aggregate

for averaging and

?format

how to specify minute chunks.

Final remarks.

If you prototype the code for one file, you can do all computation in cycle.
Do not post in HTML
Post some data eg. by dput
Do your homework and read help pages. They are available, usually well written 
and much effort is invested to them to be simply overlooked.

Regards
Petr

  Time  c  de f
 . .
 .
 2014-01-20 00:00:00 3.1428 1.99 0.123455 7.5526 2014-01-20 00:00:01
 3.1887 1.03 0.176545 7.5234 2014-01-20 00:00:02 2.1876 1.27 0.987455
 7.5996 ...
 2014-01-20 00:01:00 5.4428 1.96 0.164455 7.8666 2014-01-20 00:01:01
 1.5658 1.39 0.176545 7.7786 2014-01-20 00:01:02 3.1776 1.67 0.254655
 7.4567

 then,
  Time   average_C c  de
 f  . . .
 2014-01-20 00:00:00 here   3.1428 1.99 0.123455 7.5526
 2014-01-20 00:00:01  3.1887 1.03 0.176545 7.5234
 2014-01-20 00:00:02  2.1876 1.27 0.987455 7.5996
 ...
 2014-01-20 00:01:00  here  5.4428 1.96 0.164455 7.8666
 2014-01-20 00:01:01  1.5658 1.39 0.176545 7.7786
 2014-01-20 00:01:02  3.1776 1.67 0.254655 7.4567

 PLEASE HELP ME, R-help!!!

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 PLEASE do read the posting guide http://www.R-project.org/posting-
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jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či 
zpožděním přenosu e-mailu.

V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a 
to z jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; 
Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce 
s dodatkem či odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným 
dosažením shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost 
žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně 
pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně 
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či osobě jím zastoupené známá.

This e-mail and any documents attached to it may be confidential and are 
intended only for its intended recipients.
If you received this e-mail by mistake, please immediately inform its sender. 
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If you are not the

Re: [R] how to get the numbers of factors in a matrix

2014-01-21 Thread PIKAL Petr

Hi

It is rather difficult to understand what problem you have.

post some data e.g. by

dput(head(bigmatrix))

Maybe your problem is in a factor feature that it preserves also empty levels 
until you specifically drop them.

 ff-factor(letters[1:5])
 levels(ff[1:2])
[1] a b c d e
 fff-ff[1:2]
 nlevels(fff)
[1] 5

 fff
[1] a b
Levels: a b c d e

Regards
Petr

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of ???
 Sent: Tuesday, January 21, 2014 7:36 AM
 To: r-help@r-project.org
 Subject: [R] how to get the numbers of factors in a matrix

 Dear friends,


 I have a question do not know how to resolve.


 I have a big matrix composed of different columns (I use N here). A
 column is species and another one is latitudes. Now, I want to know
 how I can get the number of different latitudes for every species.
 I have tried to split the matrix according to species (X-split(N,
 N$species) and then use sapply(X, function(m){nlevels(m$latitudes)}) to
 get that. But the result shows the total factor numbers of latitudes
 but not the factor numbers of every species I splitted. Also, I have
 tried to use tapply(N$latitudes, N$species, nlevels) to do this. The
 result is the same. I am confused about this. Can someone help me with
 that? Thank you very much!


 Best regards,
 Yichun






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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.


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Re: [R] lattice

2014-01-21 Thread David Winsemius



On Jan 21, 2014, at 2:02 AM, Said Filahi wrote:


hello

If I have a a program



trend - read.table(txx.csv, header=T, sep=,)
library(lattice)
trellis.par.set(theme = canonical.theme(postscript, col=FALSE))
trellis.par.set(list(fontsize=list(text=8),
par.xlab.text=list(cex=1.5),
add.text=list(cex=3.5),
superpose.symbol=list(cex=1.0)))
key - simpleKey(levels(trend$Season), space = right)
key$text$cex - 1.5
print(
dotplot(Ville ~ Slope , data = trend, groups = Season,
key = key,
xlab = NULL,
aspect=0.8, layout = c(1,1), ylab=NULL)
)

and my dat det is

Ville Season Slope significance  Alh DJF -0.3 0.02  Ben DJF 0.13  
0.7  Cas

MAM 0.1 0.1  Ess JJA 0.4 0.03  Fes SON 0.9 0.02
how can i change the color of symbole of slope value if the  
significance is

below 0.05


My early efforts were sabotaged by the groups argument. Removing it  
allowed the expected behavior to occur.



dotplot(Ville ~ Slope , data = trend,
 key = key, col=c(red,blue)[1+(trend$Slope  0.05)],
 xlab = NULL,
 aspect=0.8, layout = c(1,1), ylab=NULL)

--
David Winsemius, MD
Alameda, CA, USA

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Re: [R] how to get the numbers of factors in a matrix

2014-01-21 Thread 张以春

Dear Pikal,


Thank you very much for your answer. 


I think your example is just the problem I have. 


In the following example you gave to me, 


  ff-factor(letters[1:5])
  levels(ff[1:2])
 [1] a b c d e
  fff-ff[1:2]
  nlevels(fff)
 [1] 5
 
  fff
 [1] a b
 Levels: a b c d e

In my understanding, fff is a subset of ff. Why fff's levels is not a, b but 
a,b,c,d,e.


My problem is quite similar to the example. I just want to split the matrix 
into many subsets and calculate the levels of every subset. Can you tell me how 
to do? Thank you very much!


Best regards,
Yichun

 -åå§é®ä»¶-
 åä»¶äºº: PIKAL Petr petr.pi...@precheza.cz
 åéæ¶é´: 2014å¹´1æ21æ¥ ææäº
 æ¶ä»¶äºº: å¼ ä»¥æ¥ yczh...@nigpas.ac.cn, r-help@r-project.org 
 r-help@r-project.org
 æé: 
 ä¸»é¢: RE: [R] how to get the numbers of factors in a matrix
 
 Hi
 
 It is rather difficult to understand what problem you have.
 
 post some data e.g. by
 
 dput(head(bigmatrix))
 
 Maybe your problem is in a factor feature that it preserves also empty levels 
 until you specifically drop them.
 
  ff-factor(letters[1:5])
  levels(ff[1:2])
 [1] a b c d e
  fff-ff[1:2]
  nlevels(fff)
 [1] 5
 
  fff
 [1] a b
 Levels: a b c d e
 
 Regards
 Petr
 
  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
  project.org] On Behalf Of ???
  Sent: Tuesday, January 21, 2014 7:36 AM
  To: r-help@r-project.org
  Subject: [R] how to get the numbers of factors in a matrix
 
  Dear friends,
 
 
  I have a question do not know how to resolve.
 
 
  I have a big matrix composed of different columns (I use N here). A
  column is species and another one is latitudes. Now, I want to know
  how I can get the number of different latitudes for every species.
  I have tried to split the matrix according to species (X-split(N,
  N$species) and then use sapply(X, function(m){nlevels(m$latitudes)}) to
  get that. But the result shows the total factor numbers of latitudes
  but not the factor numbers of every species I splitted. Also, I have
  tried to use tapply(N$latitudes, N$species, nlevels) to do this. The
  result is the same. I am confused about this. Can someone help me with
  that? Thank you very much!
 
 
  Best regards,
  Yichun
 
 
 
 
 
 
[[alternative HTML version deleted]]
 
  __
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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.
 
 
 Tento e-mail a jakÃ©koliv k nÄmu pÅipojenÃ© dokumenty jsou dÅ¯vÄrnÃ© a 
 jsou urÄeny pouze jeho adresÃ¡tÅ¯m.
 JestliÅ¾e jste obdrÅ¾el(a) tento e-mail omylem, informujte laskavÄ 
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 modifikacemi Äi zpoÅ¾dÄnÃm pÅenosu e-mailu.
 
 V pÅÃpadÄ, Å¾e je tento e-mail souÄÃ¡stÃ obchodnÃho jednÃ¡nÃ:
 - vyhrazuje si odesÃlatel prÃ¡vo ukonÄit kdykoliv jednÃ¡nÃ o uzavÅenÃ 
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 - a obsahuje-li nabÃdku, je adresÃ¡t oprÃ¡vnÄn nabÃdku bezodkladnÄ 
 pÅijmout; OdesÃlatel tohoto e-mailu (nabÃdky) vyluÄuje pÅijetÃ nabÃdky 
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 - odesÃlatel tohoto emailu informuje, Å¾e nenÃ oprÃ¡vnÄn uzavÃrat za 
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 zmocnÄn nebo pÃsemnÄ povÄÅen a takovÃ© povÄÅenÃ nebo plnÃ¡ moc byly 
 adresÃ¡tovi tohoto emailu pÅÃpadnÄ osobÄ, kterou adresÃ¡t zastupuje, 
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 Delete the contents of this e-mail with all attachments and its copies from 
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Re: [R] how to get the numbers of factors in a matrix

2014-01-21 Thread 张以春

Dear Jun Yi,

Thank you very much for your help. But, the link you attached can not resolve
my question.

What I want is to get the numbers of factor levels for every subsets of a
vector. When I do this, it always gives me the total numbers of levels for the
whole vector. It really makes me confused.

Best,
Yichun

-åå§é®ä»¶-
åä»¶äºº: Liu, Jun Yi 7yu...@gmail.com
åéæ¶é´: 2014å¹´1æ21æ¥ ææäº
æ¶ä»¶äºº: å¼ ä»¥æ¥ yczh...@nigpas.ac.cn, r-help@r-project.org
æé:
ä¸»é¢: Re: [R] how to get the numbers of factors in a matrix

Dear YZ,

I guess this is what you want:

but beware of that âIf you really want the levels of the factor to be used,
you're either doing something very wrong or too clever for its own good.

All the best
JY

--
From: å¼ ä»¥æ¥ yczh...@nigpas.ac.cn
Reply: å¼ ä»¥æ¥ yczh...@nigpas.ac.cn
Date: 21 January 2014 at 16:57:38
To: r-help@r-project.org r-help@r-project.org
Subject: [R] how to get the numbers of factors in a matrix

Dear friends,

I have a question do not know how to resolve.

Best regards,
Yichun

[[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.

--
Dr Yichun Zhang
State Key Laboratory of Palaeobiology and Stratigraphy, Nanjing Institute of
Geology and Palaeontology
39 East Beijing Road, Nanjing, China, 210008

[[alternative HTML version deleted]]

Re: [R] how to get the numbers of factors in a matrix

2014-01-21 Thread Jeff Newmiller

You are mis-using factors. Would you pretend that women do not exist just 
because your sample happens to include only men?

If you really need to ignore those other levels, convert your factor to 
character and summarize those values.
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

张以春 yczh...@nigpas.ac.cn wrote:
Dear Pikal,


Thank you very much for your answer. 


I think your example is just the problem I have. 


In the following example you gave to me, 


  ff-factor(letters[1:5])
  levels(ff[1:2])
 [1] a b c d e
  fff-ff[1:2]
  nlevels(fff)
 [1] 5
 
  fff
 [1] a b
 Levels: a b c d e

In my understanding, fff is a subset of ff. Why fff's levels is not a,
b but a,b,c,d,e.


My problem is quite similar to the example. I just want to split the
matrix into many subsets and calculate the levels of every subset. Can
you tell me how to do? Thank you very much!


Best regards,
Yichun

 --
 �: PIKAL Petr petr.pi...@precheza.cz
 : 2014���1���21��� �
 �: � yczh...@nigpas.ac.cn, r-help@r-project.org
r-help@r-project.org
 ��: 
 ��: RE: [R] how to get the numbers of factors in a matrix
 
 Hi
 
 It is rather difficult to understand what problem you have.
 
 post some data e.g. by
 
 dput(head(bigmatrix))
 
 Maybe your problem is in a factor feature that it preserves also
empty levels until you specifically drop them.
 
  ff-factor(letters[1:5])
  levels(ff[1:2])
 [1] a b c d e
  fff-ff[1:2]
  nlevels(fff)
 [1] 5
 
  fff
 [1] a b
 Levels: a b c d e
 
 Regards
 Petr
 
  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
  project.org] On Behalf Of ???
  Sent: Tuesday, January 21, 2014 7:36 AM
  To: r-help@r-project.org
  Subject: [R] how to get the numbers of factors in a matrix
 
  Dear friends,
 
 
  I have a question do not know how to resolve.
 
 
  I have a big matrix composed of different columns (I use N here). A
  column is species and another one is latitudes. Now, I want to
know
  how I can get the number of different latitudes for every
species.
  I have tried to split the matrix according to species (X-split(N,
  N$species) and then use sapply(X,
function(m){nlevels(m$latitudes)}) to
  get that. But the result shows the total factor numbers of
latitudes
  but not the factor numbers of every species I splitted. Also, I
have
  tried to use tapply(N$latitudes, N$species, nlevels) to do this.
The
  result is the same. I am confused about this. Can someone help me
with
  that? Thank you very much!
 
 
  Best regards,
  Yichun
 
 
 
 
 
 
[[alternative HTML version deleted]]
 
  __
  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.
 
 
 Tento e-mail a jak��koliv k n��mu p��ipojen�� dokumenty jsou
d��v��rn�� a jsou ur��eny pouze jeho adres��t��m.
 Jestli��e jste obdr��el(a) tento e-mail omylem, informujte laskav��
neprodlen�� jeho odes��latele. Obsah tohoto emailu i s plohami a
jeho kopie vyma��te ze sv��ho syst��mu.
 Nejste-li zamlen��m adres��tem tohoto emailu, nejste opr��vn��ni
tento email jakkoliv uvat, roz��i��ovat, kop��rovat ��i
zve��ej��ovat.
 Odes��latel e-mailu neodpov��d�� za eventu��ln�� ��kodu zp��sobenou
modifikacemi ��i zpo��d��n��m p��enosu e-mailu.
 
 V ppad��, ��e je tento e-mail soust�� obchodn��ho jedn��n��:
 - vyhrazuje si odes��latel pr��vo ukon��it kdykoliv jedn��n�� o
uzav��en�� smlouvy, a to z jak��hokoliv d��vodu i bez uveden�� d��vodu.
 - a obsahuje-li nab��dku, je adres��t opr��vn��n nab��dku
bezodkladn�� p��ijmout; Odes��latel tohoto e-mailu (nab��dky) vylu��uje
p��ijet�� nab��dky ze strany pjemce s dodatkem ��i odchylkou.
 - trv�� odes��latel na tom, ��e pslu��n�� smlouva je uzav��ena
teprve v��slovn��m dosa��en��m shody na v��ech jej��ch n��le��itostech.
 - odes��latel tohoto emailu informuje, ��e nen�� opr��vn��n uzav��rat
za spole��nost dn�� smlouvy s v��jimkou ppad��, kdy k tomu byl
p��semn�� zmocn��n nebo p��semn�� poven a takov�� poven�� nebo
pln�� moc byly adres��tovi tohoto emailu ppadn�� osob��, kterou
adres��t zastupuje, p��edlo��eny nebo jejich existence je adres��tovi
��i osob�� j��m zastoupen�� zn��m��.
 
 This e-mail and any documents

[R] repeated measures ANOVA using Anova in car library

2014-01-21 Thread Rüdiger Wolf

Dear all,

I am trying to do a repeated-measures ANOVA using Anova of the car package.

I have a quite basic design with 24 subjects and only 2 within-subject
factors (condition (2 levels: legal an illegal) and day (1,2,3)).

I found the Examples (see
https://stat.ethz.ch/pipermail/r-help/2008-December/181981.html) in
?Anova, but I don't know what to do if there are only within-subject
factors. What should I do with the mod.ok model object?

--snip-

name_test_fribbles-read.csv('Naming_test_Fribbles.csv') # Import
cond-factor(rep(c(legal,illegal),c(3,3)),levels=c(legal,illegal))
day-ordered(rep(1:3,2))
idata-data.frame(cond,day)
mod.ok-lm(cbind(legTag1,legTag2,legTag3,illegTag1,illegTag2,illegTag3)
   ~ THIS IS MISSING

,data=name_test_fribbles)


---snip-

I would be glad if anyone could help me! Thanks a lot!

Rudi

__
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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] how to replace NA values

2014-01-21 Thread John Kane

I'm pretty sure that what you have is not what you think you have.

Do a str(Elder1) on the data set and you will see that age is a factor not a 
numerical variable. Bascally you have a mangled data set from the look of it.

Why would you want to change NA's to 0's?  

John Kane
Kingston ON Canada


 -Original Message-
 From: ecoking...@yahoo.co.in
 Sent: Sun, 19 Jan 2014 11:39:43 -0800 (PST)
 To: r-help@r-project.org
 Subject: [R] how to replace NA values
 
 Dear R community
 B
 I have a large data set contain some empty cells. BecauseB of that,B  may
 be I am wrong, NA values are produced.
 Now I want replace both empty and NA values with zero.
 B
 Elder1 - data.frame(
 B  ID=c(ID1,ID2,ID3,ID6,ID8),
 B  age=c(38,35,,NA,NA))
 Output I am expecting
 B
 IDB B  age
 ID1B  38
 ID2B  35
 ID3B  0
 ID6B  0
 ID8B  0
 B
 In advance I thank your help.
 
 
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/how-to-replace-NA-values-tp4683831.html
 Sent from the R help mailing list archive at Nabble.com.
   [[alternative HTML version deleted]]
 
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 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] repeated measures ANOVA using Anova in car library

2014-01-21 Thread Bert Gunter

You should post this instead on r-sig-mixed-models, rather than here.
They are specifically concerned with such statistical issues; here the
concern is primarily R programming.

Cheers,
Bert

Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch




On Tue, Jan 21, 2014 at 9:56 AM, Rüdiger Wolf r.h.w...@gmx.net wrote:
 Dear all,

 I am trying to do a repeated-measures ANOVA using Anova of the car package.

 I have a quite basic design with 24 subjects and only 2 within-subject
 factors (condition (2 levels: legal an illegal) and day (1,2,3)).

 I found the Examples (see
 https://stat.ethz.ch/pipermail/r-help/2008-December/181981.html) in
 ?Anova, but I don't know what to do if there are only within-subject
 factors. What should I do with the mod.ok model object?

 --snip-

 name_test_fribbles-read.csv('Naming_test_Fribbles.csv') # Import
 cond-factor(rep(c(legal,illegal),c(3,3)),levels=c(legal,illegal))
 day-ordered(rep(1:3,2))
 idata-data.frame(cond,day)
 mod.ok-lm(cbind(legTag1,legTag2,legTag3,illegTag1,illegTag2,illegTag3)
~ THIS IS MISSING

 ,data=name_test_fribbles)


 ---snip-

 I would be glad if anyone could help me! Thanks a lot!

 Rudi

 __
 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] how to get the numbers of factors in a matrix

2014-01-21 Thread peter dalgaard


On 21 Jan 2014, at 15:58 , Jeff Newmiller jdnew...@dcn.davis.ca.us wrote:

 You are mis-using factors. Would you pretend that women do not exist just 
 because your sample happens to include only men?
 
 If you really need to ignore those other levels, convert your factor to 
 character and summarize those values.

Actually, just factor(fff) will do. Or droplevels(fff).


-- 
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] Strange gcfit output in R package Grofit

2014-01-21 Thread Lucy Rayner

Hi all,

I'm trying to use the R growth curve analysis package Grofit. I've used it
successfully before but the protocol which previously worked for me is now
not giving me the expected output.

Entering my data and time info and running Test-grofit(time,data,FALSE)
gives me the expected growth curves as plots for my consideration. However,
when I try to access the output of gcfit using either summary(Test,gcfit) or
summary(Test$gcfit) I don't get the raw growth curve data which I expect, i.e. 
the 'hard' quantified info from the fitted growth curves, but instead seem to 
get information about the data that I entered and the running of the process: 
summary(TestRun2,gcfit)

Length
Class  Mode   

time55 data.frame list 
 

data58 data.frame list  


gcFit7 gcFit  list  


drFit1 -none- logical

control 16
grofit.control list   
I've tried using the in-package generated data and get the same output. I
can't see why the same protocol now doesn't work, except that the package is
a later version than the one I previously used. Also, the package is now
'orphaned'.

Can anyone give any advice? Would loading an earlier version of the package
from the tar.gz file be an option? I'm a bit shaky on getting this to work
(I had a bit of a go using the instructions here): 

http://stat.ethz.ch/pipermail/r-help/2008-August/169599.html 

but wasn't entirely successful

Thanks for any suggestions!   
[[alternative HTML version deleted]]

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[R] PCA factominer package, question about changing labels in individuals factor map

2014-01-21 Thread Andy.P

I am looking to do a PCA with the factominer package on a dataset of mine,
named dat. The individuals in my dataset have character names, represented
in the first column of my dataset, but since they aren't quantitative I
can't include that column in my PCA analysis. Leading to the command 
res.pca = PCA(dat[,3:8], scale.unit=TRUE, ncp=5, graph=T)

This is all well and good, except when my individuals factor map shows up,
each data point is labeled as a number and not as their actual character
names.
http://r.789695.n4.nabble.com/file/n4683924/Screen_Shot_2014-01-21_at_12.19.23_PM.png
 



 How do I go about getting the map to display the character names? If I
include the column with the names I get an error because it's not
quantitative. I know it can be done because I've seen it on the decathlon
dataset, I just don't know what they're doing differently to get those
character names on the map.



--
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Re: [R] how do I extract all possible combinations of rows from a column

2014-01-21 Thread arun

Hi,
May be this helps:
library(gtools) 

as.data.frame(t(combinations(5,3,1:5)))
#  V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
#1  1  1  1  1  1  1  2  2  2   3
#2  2  2  2  3  3  4  3  3  4   4
#3  3  4  5  4  5  5  4  5  5   5
A.K.



For a column of data, I need to extract all possible unique 
combinations of 3 rows and combine the results into a data frame (as 
columns). So, for example, if I had a column called test that had five rows 
(the contents of which were the numbers 1:5) all possible 
combinations of 3 rows would be 
1,2,3 
1,2,4 
1,2,5 
1,3,4 
1,3,5 
1,4,5 
2,3,4 
2,4,5 
2,4,5 
3,4,5 

So for this example, R would extract all the combinations shown, and I would 
get a data frame that looked like this 
1     1     1     1     1     1     2     2     2      3       
2     2     2     3     3     4     3     3     4      4   
3     4     5     4     5     5     4     5     5      5     

Note: I am not considering rearrangements of the same three rows
 to be unique. In other words, the combinations 123, 132, 213, 231, 312,
 321 are all the same as far as I am concerned. If it is not possible to
 do it that way, it will still be helpful, but not quite as helpful. 
Thanks for the help!

__
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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] how do I extract all possible combinations of rows from a column

2014-01-21 Thread Bert Gunter

... or combn() in utils.

which was the first hit when I googled on finding combinations in R.

So moral: Use search engines before you post.

Cheers,
Bert

Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch




On Tue, Jan 21, 2014 at 12:57 PM, arun smartpink...@yahoo.com wrote:
 Hi,
 May be this helps:
 library(gtools)

 as.data.frame(t(combinations(5,3,1:5)))
 #  V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
 #1  1  1  1  1  1  1  2  2  2   3
 #2  2  2  2  3  3  4  3  3  4   4
 #3  3  4  5  4  5  5  4  5  5   5
 A.K.



 For a column of data, I need to extract all possible unique
 combinations of 3 rows and combine the results into a data frame (as
 columns). So, for example, if I had a column called test that had five rows 
 (the contents of which were the numbers 1:5) all possible
 combinations of 3 rows would be
 1,2,3
 1,2,4
 1,2,5
 1,3,4
 1,3,5
 1,4,5
 2,3,4
 2,4,5
 2,4,5
 3,4,5

 So for this example, R would extract all the combinations shown, and I would 
 get a data frame that looked like this
 1 1 1 1 1 1 2 2 2  3
 2 2 2 3 3 4 3 3 4  4
 3 4 5 4 5 5 4 5 5  5

 Note: I am not considering rearrangements of the same three rows
  to be unique. In other words, the combinations 123, 132, 213, 231, 312,
  321 are all the same as far as I am concerned. If it is not possible to
  do it that way, it will still be helpful, but not quite as helpful.
 Thanks for the help!

 __
 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] repeated measures ANOVA using Anova in car library

2014-01-21 Thread John Fox

Dear Rudiger,

See the following paper in the R Journal, which, among other things, shows how 
to set up a repeated-measures ANOVA or MANOVA using Anova(): 
http://journal.r-project.org/archive/2013-1/fox-friendly-weisberg.pdf.

I hope this helps,
 John


John Fox
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/


On Tue, 21 Jan 2014 18:56:56 +0100
 Rüdiger Wolf r.h.w...@gmx.net wrote:
 Dear all,
 
 I am trying to do a repeated-measures ANOVA using Anova of the car package.
 
 I have a quite basic design with 24 subjects and only 2 within-subject
 factors (condition (2 levels: legal an illegal) and day (1,2,3)).
 
 I found the Examples (see
 https://stat.ethz.ch/pipermail/r-help/2008-December/181981.html) in
 ?Anova, but I don't know what to do if there are only within-subject
 factors. What should I do with the mod.ok model object?
 
 --snip-
 
 name_test_fribbles-read.csv('Naming_test_Fribbles.csv') # Import
 cond-factor(rep(c(legal,illegal),c(3,3)),levels=c(legal,illegal))
 day-ordered(rep(1:3,2))
 idata-data.frame(cond,day)
 mod.ok-lm(cbind(legTag1,legTag2,legTag3,illegTag1,illegTag2,illegTag3)
~ THIS IS MISSING
 
   ,data=name_test_fribbles)
 
 
 ---snip-
 
 I would be glad if anyone could help me! Thanks a lot!
 
 Rudi
 
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Re: [R] PCA factominer package, question about changing labels in individuals factor map

2014-01-21 Thread David Carlson

Try

rownames(dat) - dat[,1]

before running PCA.

-
David L Carlson
Department 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 Andy.P
Sent: Tuesday, January 21, 2014 12:18 PM
To: r-help@r-project.org
Subject: [R] PCA factominer package, question about changing
labels in individuals factor map

I am looking to do a PCA with the factominer package on a
dataset of mine,
named dat. The individuals in my dataset have character names,
represented
in the first column of my dataset, but since they aren't
quantitative I
can't include that column in my PCA analysis. Leading to the
command 
res.pca = PCA(dat[,3:8], scale.unit=TRUE, ncp=5, graph=T)

This is all well and good, except when my individuals factor map
shows up,
each data point is labeled as a number and not as their actual
character
names.
http://r.789695.n4.nabble.com/file/n4683924/Screen_Shot_2014-01
-21_at_12.19.23_PM.png 



 How do I go about getting the map to display the character
names? If I
include the column with the names I get an error because it's
not
quantitative. I know it can be done because I've seen it on the
decathlon
dataset, I just don't know what they're doing differently to get
those
character names on the map.



--
View this message in context:
http://r.789695.n4.nabble.com/PCA-factominer-package-question-ab
out-changing-labels-in-individuals-factor-map-tp4683924.html
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__
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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.

__
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Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : dims [product 1] do not match the length of object [0]

2014-01-21 Thread arun

Hi,
Couldn't reproduce the error after running your code:
 d - dcast(d, date ~ id, value.var=simple_return)
 dim(d)
#[1] 356   9
 sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-unknown-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_CA.UTF-8   LC_NUMERIC=C  
 [3] LC_TIME=en_CA.UTF-8    LC_COLLATE=en_CA.UTF-8    
 [5] LC_MONETARY=en_CA.UTF-8    LC_MESSAGES=en_CA.UTF-8   
 [7] LC_PAPER=en_CA.UTF-8   LC_NAME=C 
 [9] LC_ADDRESS=C   LC_TELEPHONE=C    
[11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C   

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base 

other attached packages:
[1] plyr_1.8   quantmod_0.4-0 TTR_0.22-0 xts_0.9-7  zoo_1.7-10    
[6] Defaults_1.1-1 stringr_0.6.2  reshape2_1.2.2

loaded via a namespace (and not attached):
[1] grid_3.0.2  lattice_0.20-23


A.K.


On Tuesday, January 21, 2014 2:57 PM, rcse2006 rcse2...@gmail.com wrote:
Trying to run below code.

library(quantmod)
symbols - c(AAPL, DELL, GOOG, MSFT, AMZN, BIDU, EBAY, YHOO)
d - list()
for(s in symbols) {
  tmp - getSymbols(s, auto.assign=FALSE, verbose=TRUE)
  tmp - Ad(tmp)
  names(tmp) - price
  tmp - data.frame( date=index(tmp), id=s, price=coredata(tmp) )
  d[[s]] - tmp
}
d - do.call(rbind, d)
d - d[ d$date = as.Date(2007-01-01), ]
rownames(d) - NULL

# Weekly returns
library(plyr)
library(reshape2)
d$next_friday - d$date - as.numeric(format(d$date, %u)) + 5
d - subset(d, date==next_friday)
d - ddply(d, id, mutate,
           previous_price = lag(xts(price,date)),
           log_return    = log(price / previous_price),
           simple_return = price / previous_price - 1
)
d - dcast(d, date ~ id, value.var=simple_return)

Getting error

 d - dcast(d, date ~ id, value.var=simple_return)
Error in structure(ordered, dim = ns) : 
  dims [product 1] do not match the length of object [0]

Please help me how to use ddply and dcast or using other similar function to
get same data.



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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] obtain mean trait values by haplotype in haplo.stats package

2014-01-21 Thread ylwgto

Hello,

I am using the haplo.stats package to calculate haplotype frequencies and
test for associations of haplotype with a continuous phenotype (in this case
BMI) in my population. My analysis is rather simple, as only 2 SNPs are
included. 

I have a few questions related to the package:

-How exactly is the hap-score calculated and what does it signify?
-Is there a simple script I can use to obtain mean trait (BMI) values BY
haplotype 


Thanks!



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

[R] Windows7, R-Studio, can't find package rJava,

2014-01-21 Thread Rainer K. SACHS

example:

 getwd() # with RStudio you can read it off the console bar[1] 
 C:/Users/ra`-/Documents/1ray/plans/learnR/week3liats library(xlsx)Loading 
 required package: rJavaError : .onLoad failed in loadNamespace() for 'rJava', 
 details:
  call: fun(libname, pkgname)
  error: JAVA_HOME cannot be determined from the RegistryIn addition:
Warning messages:1: package xlsx was built under R version 3.0.2 2:
package rJava was built under R version 3.0.2 Error: package rJava
could not be loaded install.packages(rJava)Installing package into
C:/Users/ra`-/Documents/R/win-library/3.0
(as lib is unspecified)trying URL
'http://cran.rstudio.com/bin/windows/contrib/3.0/rJava_0.9-6.zip'Content
type 'application/zip' length 758820 bytes (741 Kb)opened
URLdownloaded 741 Kbpackage rJava successfully unpacked and MD5 sums
checked
The downloaded binary packages are in
C:\Users\ra`-\AppData\Local\Temp\Rtmpq0N2f2\downloaded_packages
library(rJava,
lib.loc=C:/Users/ra`-/Documents/R/win-library/3.0)Error : .onLoad
failed in loadNamespace() for 'rJava', details:
  call: fun(libname, pkgname)
  error: JAVA_HOME cannot be determined from the RegistryIn addition:
Warning message:package rJava was built under R version 3.0.2 Error:
package or namespace load failed for rJava

I tried various things with searhpath, but can't figure out if this a
windows 7 problem, an R problem, an R-studio problem, an R4 vs R3
problem, an rJava problem, a firewall problem, or some clever
combination of such problems. Any tips greatly appreciated even if all
they tell me is in which direction to look (e.g. give Windows a
different searchpath?)

TIA Ray Sachs

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

Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : dims [product 1] do not match the length of object [0]

2014-01-21 Thread William Dunlap

When I use R-3.0.2 on Windows 7 the %u descriptor for format.Date() always
gives , while on Linux in gives as.character(day-of-the-week).  The resulting 
NA's
on Windows could be the source of your problem.

On Linux I get:
format(as.Date(c(2014-01-21, 2014-01-22, 2014-01-28)), %u)
   [1] 2 3 2
as.numeric(.Last.value)
   [1] 2 3 2
cat(version$version.string, on, version$platform, \n)
   R version 3.0.2 (2013-09-25) on x86_64-unknown-linux-gnu

while on Windows:
   format(as.Date(c(2014-01-21, 2014-01-22, 2014-01-28)), %u)
  [1]   
   as.numeric(.Last.value)
  [1] NA NA NA
   cat(version$version.string, on, version$platform, \n)
  R version 3.0.2 (2013-09-25) on x86_64-w64-mingw32

 d - subset(d, date==next_friday)
 d - ddply(d, id, mutate,
previous_price = lag(xts(price,date)),
log_return= log(price / previous_price),
simple_return = price / previous_price - 1
 )
 d - dcast(d, date ~ id, value.var=simple_return)

I you didn't reuse the same name, d, for the result of all these
steps it would be easier to poke through the intermediate
results to see where the trouble began (the output of subset()
is a 0-row data.frame and dcast() dies when its input has
zero rows).

Bill Dunlap
TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of arun
 Sent: Tuesday, January 21, 2014 12:48 PM
 To: R help
 Subject: Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : 
 dims [product 1]
 do not match the length of object [0]
 
 Hi,
 Couldn't reproduce the error after running your code:
  d - dcast(d, date ~ id, value.var=simple_return)
  dim(d)
 #[1] 356   9
  sessionInfo()
 R version 3.0.2 (2013-09-25)
 Platform: x86_64-unknown-linux-gnu (64-bit)
 
 locale:
  [1] LC_CTYPE=en_CA.UTF-8   LC_NUMERIC=C
  [3] LC_TIME=en_CA.UTF-8    LC_COLLATE=en_CA.UTF-8
  [5] LC_MONETARY=en_CA.UTF-8    LC_MESSAGES=en_CA.UTF-8
  [7] LC_PAPER=en_CA.UTF-8   LC_NAME=C
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods   base
 
 other attached packages:
 [1] plyr_1.8   quantmod_0.4-0 TTR_0.22-0 xts_0.9-7  zoo_1.7-10
 [6] Defaults_1.1-1 stringr_0.6.2  reshape2_1.2.2
 
 loaded via a namespace (and not attached):
 [1] grid_3.0.2  lattice_0.20-23
 
 
 A.K.
 
 
 On Tuesday, January 21, 2014 2:57 PM, rcse2006 rcse2...@gmail.com wrote:
 Trying to run below code.
 
 library(quantmod)
 symbols - c(AAPL, DELL, GOOG, MSFT, AMZN, BIDU, EBAY, YHOO)
 d - list()
 for(s in symbols) {
   tmp - getSymbols(s, auto.assign=FALSE, verbose=TRUE)
   tmp - Ad(tmp)
   names(tmp) - price
   tmp - data.frame( date=index(tmp), id=s, price=coredata(tmp) )
   d[[s]] - tmp
 }
 d - do.call(rbind, d)
 d - d[ d$date = as.Date(2007-01-01), ]
 rownames(d) - NULL
 
 # Weekly returns
 library(plyr)
 library(reshape2)
 d$next_friday - d$date - as.numeric(format(d$date, %u)) + 5
 d - subset(d, date==next_friday)
 d - ddply(d, id, mutate,
            previous_price = lag(xts(price,date)),
            log_return    = log(price / previous_price),
            simple_return = price / previous_price - 1
 )
 d - dcast(d, date ~ id, value.var=simple_return)
 
 Getting error
 
  d - dcast(d, date ~ id, value.var=simple_return)
 Error in structure(ordered, dim = ns) :
   dims [product 1] do not match the length of object [0]
 
 Please help me how to use ddply and dcast or using other similar function to
 get same data.
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Error-in-structure-
 ordered-dim-ns-dims-product-1-do-not-match-the-length-of-object-0-tp4683923.html
 Sent from the datatable-help mailing list archive at Nabble.com.
 ___
 datatable-help mailing list
 datatable-h...@lists.r-forge.r-project.org
 https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/datatable-help
 
 
 __
 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : dims [product 1] do not match the length of object [0]

2014-01-21 Thread William Dunlap

You probably want to use %w instead of %u.  On Linux 'man strftime' says
   %w The day of the week as a decimal, range 0 to 6, Sunday being  0.
  See also %u.
   %u The day of the week as a decimal, range 1 to 7, Monday being  1.
  See also %w.  (SU)
where (SU) means according to the Single Unix specification and the lack
of (...) after %w means it is in some year's ANSI C standard.   I assume 
Windows
does not attempt to subscribe to the Single Unix standard.

Bill Dunlap
TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of William Dunlap
 Sent: Tuesday, January 21, 2014 2:33 PM
 To: arun; R help
 Subject: Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : 
 dims [product 1]
 do not match the length of object [0]
 
 When I use R-3.0.2 on Windows 7 the %u descriptor for format.Date() always
 gives , while on Linux in gives as.character(day-of-the-week).  The 
 resulting NA's
 on Windows could be the source of your problem.
 
 On Linux I get:
 format(as.Date(c(2014-01-21, 2014-01-22, 2014-01-28)), %u)
[1] 2 3 2
 as.numeric(.Last.value)
[1] 2 3 2
 cat(version$version.string, on, version$platform, \n)
R version 3.0.2 (2013-09-25) on x86_64-unknown-linux-gnu
 
 while on Windows:
format(as.Date(c(2014-01-21, 2014-01-22, 2014-01-28)), %u)
   [1]   
as.numeric(.Last.value)
   [1] NA NA NA
cat(version$version.string, on, version$platform, \n)
   R version 3.0.2 (2013-09-25) on x86_64-w64-mingw32
 
  d - subset(d, date==next_friday)
  d - ddply(d, id, mutate,
 previous_price = lag(xts(price,date)),
 log_return= log(price / previous_price),
 simple_return = price / previous_price - 1
  )
  d - dcast(d, date ~ id, value.var=simple_return)
 
 I you didn't reuse the same name, d, for the result of all these
 steps it would be easier to poke through the intermediate
 results to see where the trouble began (the output of subset()
 is a 0-row data.frame and dcast() dies when its input has
 zero rows).
 
 Bill Dunlap
 TIBCO Software
 wdunlap tibco.com
 
 
  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
  Behalf
  Of arun
  Sent: Tuesday, January 21, 2014 12:48 PM
  To: R help
  Subject: Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : 
  dims [product 1]
  do not match the length of object [0]
 
  Hi,
  Couldn't reproduce the error after running your code:
   d - dcast(d, date ~ id, value.var=simple_return)
   dim(d)
  #[1] 356   9
   sessionInfo()
  R version 3.0.2 (2013-09-25)
  Platform: x86_64-unknown-linux-gnu (64-bit)
 
  locale:
   [1] LC_CTYPE=en_CA.UTF-8   LC_NUMERIC=C
   [3] LC_TIME=en_CA.UTF-8    LC_COLLATE=en_CA.UTF-8
   [5] LC_MONETARY=en_CA.UTF-8    LC_MESSAGES=en_CA.UTF-8
   [7] LC_PAPER=en_CA.UTF-8   LC_NAME=C
   [9] LC_ADDRESS=C   LC_TELEPHONE=C
  [11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C
 
  attached base packages:
  [1] stats graphics  grDevices utils datasets  methods   base
 
  other attached packages:
  [1] plyr_1.8   quantmod_0.4-0 TTR_0.22-0 xts_0.9-7  zoo_1.7-10
  [6] Defaults_1.1-1 stringr_0.6.2  reshape2_1.2.2
 
  loaded via a namespace (and not attached):
  [1] grid_3.0.2  lattice_0.20-23
 
 
  A.K.
 
 
  On Tuesday, January 21, 2014 2:57 PM, rcse2006 rcse2...@gmail.com wrote:
  Trying to run below code.
 
  library(quantmod)
  symbols - c(AAPL, DELL, GOOG, MSFT, AMZN, BIDU, EBAY, YHOO)
  d - list()
  for(s in symbols) {
    tmp - getSymbols(s, auto.assign=FALSE, verbose=TRUE)
    tmp - Ad(tmp)
    names(tmp) - price
    tmp - data.frame( date=index(tmp), id=s, price=coredata(tmp) )
    d[[s]] - tmp
  }
  d - do.call(rbind, d)
  d - d[ d$date = as.Date(2007-01-01), ]
  rownames(d) - NULL
 
  # Weekly returns
  library(plyr)
  library(reshape2)
  d$next_friday - d$date - as.numeric(format(d$date, %u)) + 5
  d - subset(d, date==next_friday)
  d - ddply(d, id, mutate,
             previous_price = lag(xts(price,date)),
             log_return    = log(price / previous_price),
             simple_return = price / previous_price - 1
  )
  d - dcast(d, date ~ id, value.var=simple_return)
 
  Getting error
 
   d - dcast(d, date ~ id, value.var=simple_return)
  Error in structure(ordered, dim = ns) :
    dims [product 1] do not match the length of object [0]
 
  Please help me how to use ddply and dcast or using other similar function to
  get same data.
 
 
 
  --
  View this message in context: 
  http://r.789695.n4.nabble.com/Error-in-structure-
  ordered-dim-ns-dims-product-1-do-not-match-the-length-of-object-0-tp4683923.html
  Sent from the datatable-help mailing list archive at Nabble.com.
  ___
  datatable-help mailing list
  datatable-h...@lists.r-forge.r-project.org

[R] Problems downloading and installing nlme under Linux Mint 15

2014-01-21 Thread John Sorkin

I am running R under Linux Mint 15 and trying to download and install
nlme. During the download I get an error message:
Warning message:
In install.packages(nlme) :
  installation of package ‘nlme’ had non-zero exit status

When I try to load the package, I get an error message 
 library(nlme)
Error: package ‘nlme’ was built before R 3.0.0: please re-install it
Why am I getting the error message, and what can I do to fix the
problem? Please see output of R session below.
Thank you,
John


version
platform   i486-pc-linux-gnu   
arch   i486
os linux-gnu   
system i486, linux-gnu 
status 
major  3   
minor  0.1 
year   2013
month  05  
day16  
svn rev62743   
language   R   
version.string R version 3.0.1 (2013-05-16)
nickname   Good Sport


 install.packages(nlme)
Installing package into ‘/home/john/R/i486-pc-linux-gnu-library/3.0’
(as ‘lib’ is unspecified)
--- Please select a CRAN mirror for use in this session ---
trying URL
'http://watson.nci.nih.gov/cran_mirror/src/contrib/nlme_3.1-113.tar.gz'
Content type 'application/octet-stream' length 752188 bytes (734 Kb)
opened URL
==
downloaded 734 Kb

* installing *source* package ‘nlme’ ...
** package ‘nlme’ successfully unpacked and MD5 sums checked
** libs
gfortran   -fpic  -O2 -pipe -g  -c chol.f -o chol.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c corStruct.c -o corStruct.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c gnls.c -o gnls.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c init.c -o init.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c matrix.c -o matrix.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nlOptimizer.c -o nlOptimizer.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nlme.c -o nlme.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nlmefit.c -o nlmefit.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nls.c -o nls.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c pdMat.c -o pdMat.o
gfortran   -fpic  -O2 -pipe -g  -c rs.f -o rs.o
gcc -std=gnu99 -shared -o nlme.so chol.o corStruct.o gnls.o init.o
matrix.o nlOptimizer.o nlme.o nlmefit.o nls.o pdMat.o rs.o -lgfortran
-lm -lquadmath -L/usr/lib/R/lib -lR
installing to /home/john/R/i486-pc-linux-gnu-library/3.0/nlme/libs
** R
** data
*** moving datasets to lazyload DB
** inst
** byte-compile and prepare package for lazy loading
Error : package ‘lattice’ was built before R 3.0.0: please re-install it
ERROR: lazy loading failed for package ‘nlme’
* removing ‘/home/john/R/i486-pc-linux-gnu-library/3.0/nlme’

The downloaded source packages are in
‘/tmp/Rtmpo7eFZf/downloaded_packages’
Warning message:
In install.packages(nlme) :
  installation of package ‘nlme’ had non-zero exit status
 library(nlme)
Error: package ‘nlme’ was built before R 3.0.0: please re-install it




John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing) 

CThis email message, including any attachments, is for the sole use of
the intended recipient(s) and may contain confidential and privileged
information.  Any unauthorized use, disclosure or distribution is
prohibited.  If you are not the intended recipient, please contact the
sender by reply email and destroy all copies of the original message. 
__
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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] PCA factominer package, question about changing labels in individuals factor map

2014-01-21 Thread Andy.P

That worked! Wow easy fix, i feel dumb! Thanks!



--
View this message in context: 
http://r.789695.n4.nabble.com/PCA-factominer-package-question-about-changing-labels-in-individuals-factor-map-tp4683924p4683941.html
Sent from the R help mailing list archive at Nabble.com.

__
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] Problems downloading and installing nlme under Linux Mint 15

2014-01-21 Thread Clint Bowman


There is the error message:

Error : package ‘lattice’ was built before R 3.0.0: please re-install it

Perhaps the problem lies there.

Clint BowmanINTERNET:   cl...@ecy.wa.gov
Air Quality Modeler INTERNET:   cl...@math.utah.edu
Department of Ecology   VOICE:  (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600

USPS:   PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274

On Tue, 21 Jan 2014, John Sorkin wrote:


I am running R under Linux Mint 15 and trying to download and install
nlme. During the download I get an error message:
Warning message:
In install.packages(nlme) :
 installation of package ‘nlme’ had non-zero exit status

When I try to load the package, I get an error message

library(nlme)

Error: package ‘nlme’ was built before R 3.0.0: please re-install it
Why am I getting the error message, and what can I do to fix the
problem? Please see output of R session below.
Thank you,
John



version

platform   i486-pc-linux-gnu
arch   i486
os linux-gnu
system i486, linux-gnu
status
major  3
minor  0.1
year   2013
month  05
day16
svn rev62743
language   R
version.string R version 3.0.1 (2013-05-16)
nickname   Good Sport



install.packages(nlme)

Installing package into ‘/home/john/R/i486-pc-linux-gnu-library/3.0’
(as ‘lib’ is unspecified)
--- Please select a CRAN mirror for use in this session ---
trying URL
'http://watson.nci.nih.gov/cran_mirror/src/contrib/nlme_3.1-113.tar.gz'
Content type 'application/octet-stream' length 752188 bytes (734 Kb)
opened URL
==
downloaded 734 Kb

* installing *source* package ‘nlme’ ...
** package ‘nlme’ successfully unpacked and MD5 sums checked
** libs
gfortran   -fpic  -O2 -pipe -g  -c chol.f -o chol.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c corStruct.c -o corStruct.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c gnls.c -o gnls.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c init.c -o init.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c matrix.c -o matrix.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nlOptimizer.c -o nlOptimizer.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nlme.c -o nlme.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nlmefit.c -o nlmefit.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c nls.c -o nls.o
gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG -fvisibility=hidden
-fpic  -O2 -pipe -g  -c pdMat.c -o pdMat.o
gfortran   -fpic  -O2 -pipe -g  -c rs.f -o rs.o
gcc -std=gnu99 -shared -o nlme.so chol.o corStruct.o gnls.o init.o
matrix.o nlOptimizer.o nlme.o nlmefit.o nls.o pdMat.o rs.o -lgfortran
-lm -lquadmath -L/usr/lib/R/lib -lR
installing to /home/john/R/i486-pc-linux-gnu-library/3.0/nlme/libs
** R
** data
*** moving datasets to lazyload DB
** inst
** byte-compile and prepare package for lazy loading
Error : package ‘lattice’ was built before R 3.0.0: please re-install it
ERROR: lazy loading failed for package ‘nlme’
* removing ‘/home/john/R/i486-pc-linux-gnu-library/3.0/nlme’

The downloaded source packages are in
   ‘/tmp/Rtmpo7eFZf/downloaded_packages’
Warning message:
In install.packages(nlme) :
 installation of package ‘nlme’ had non-zero exit status

library(nlme)

Error: package ‘nlme’ was built before R 3.0.0: please re-install it




John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)

CThis email message, including any attachments, is for the sole use of
the intended recipient(s) and may contain confidential and privileged
information.  Any unauthorized use, disclosure or distribution is
prohibited.  If you are not the intended recipient, please contact the
sender by reply email and destroy all copies of the original message.
__
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.

Re: [R] Problems downloading and installing nlme under Linux Mint 15

2014-01-21 Thread John Sorkin

Clint,
Thank you!
Problem solved.
John

 
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing) 
 Clint Bowman cl...@ecy.wa.gov 1/21/2014 6:57 PM 
There is the error message:

Error : package ‘lattice’ was built before R 3.0.0: please re-install
it

Perhaps the problem lies there.

Clint BowmanINTERNET:cl...@ecy.wa.gov
Air Quality ModelerINTERNET:cl...@math.utah.edu
Department of EcologyVOICE:(360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600

 USPS:   PO Box 47600, Olympia, WA 98504-7600
 Parcels:300 Desmond Drive, Lacey, WA 98503-1274

On Tue, 21 Jan 2014, John Sorkin wrote:

 I am running R under Linux Mint 15 and trying to download and
install
 nlme. During the download I get an error message:
 Warning message:
 In install.packages(nlme) :
  installation of package ‘nlme’ had non-zero exit status

 When I try to load the package, I get an error message
 library(nlme)
 Error: package ‘nlme’ was built before R 3.0.0: please re-install it
 Why am I getting the error message, and what can I do to fix the
 problem? Please see output of R session below.
 Thank you,
 John


 version
 platform   i486-pc-linux-gnu
 arch   i486
 os linux-gnu
 system i486, linux-gnu
 status
 major  3
 minor  0.1
 year   2013
 month  05
 day16
 svn rev62743
 language   R
 version.string R version 3.0.1 (2013-05-16)
 nickname   Good Sport


 install.packages(nlme)
 Installing package into ‘/home/john/R/i486-pc-linux-gnu-library/3.0’
 (as ‘lib’ is unspecified)
 --- Please select a CRAN mirror for use in this session ---
 trying URL

'http://watson.nci.nih.gov/cran_mirror/src/contrib/nlme_3.1-113.tar.gz'
 Content type 'application/octet-stream' length 752188 bytes (734 Kb)
 opened URL
 ==
 downloaded 734 Kb

 * installing *source* package ‘nlme’ ...
 ** package ‘nlme’ successfully unpacked and MD5 sums checked
 ** libs
 gfortran   -fpic  -O2 -pipe -g  -c chol.f -o chol.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c corStruct.c -o corStruct.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c gnls.c -o gnls.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c init.c -o init.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c matrix.c -o matrix.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c nlOptimizer.c -o nlOptimizer.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c nlme.c -o nlme.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c nlmefit.c -o nlmefit.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c nls.c -o nls.o
 gcc -std=gnu99 -I/usr/share/R/include -DNDEBUG
-fvisibility=hidden
 -fpic  -O2 -pipe -g  -c pdMat.c -o pdMat.o
 gfortran   -fpic  -O2 -pipe -g  -c rs.f -o rs.o
 gcc -std=gnu99 -shared -o nlme.so chol.o corStruct.o gnls.o init.o
 matrix.o nlOptimizer.o nlme.o nlmefit.o nls.o pdMat.o rs.o
-lgfortran
 -lm -lquadmath -L/usr/lib/R/lib -lR
 installing to /home/john/R/i486-pc-linux-gnu-library/3.0/nlme/libs
 ** R
 ** data
 *** moving datasets to lazyload DB
 ** inst
 ** byte-compile and prepare package for lazy loading
 Error : package ‘lattice’ was built before R 3.0.0: please re-install
it
 ERROR: lazy loading failed for package ‘nlme’
 * removing ‘/home/john/R/i486-pc-linux-gnu-library/3.0/nlme’

 The downloaded source packages are in
‘/tmp/Rtmpo7eFZf/downloaded_packages’
 Warning message:
 In install.packages(nlme) :
  installation of package ‘nlme’ had non-zero exit status
 library(nlme)
 Error: package ‘nlme’ was built before R 3.0.0: please re-install it




 John David Sorkin M.D., Ph.D.
 Professor of Medicine
 Chief, Biostatistics and Informatics
 University of Maryland School of Medicine Division of Gerontology
and
 Geriatric Medicine
 Baltimore VA Medical Center
 10 North Greene Street
 GRECC (BT/18/GR)
 Baltimore, MD 21201-1524
 (Phone) 410-605-7119
 (Fax) 410-605-7913 (Please call phone number above prior to faxing)

 CThis email message, including any attachments, is for the sole use
of
 the intended recipient(s) and may contain confidential and
privileged
 information.  Any unauthorized use, disclosure or distribution is
 prohibited.  If you are not the intended recipient, please contact
the
 sender by reply email and destroy all copies

[R] Getting the current number of rows in windows R gui console (Carlos Arias)

2014-01-21 Thread Carlos Arias

Hello, I'm currently trying to set some default settings and I know how to
get the width of the console with getOption(width), this corresponds to
the number of columns of the console in the Rconsole file. I would also
like to get the current number of rows from inside R. Is there an R option
to get this value?

Thank you,

Carlos Arias

Statistician at ICFES Colombia

[[alternative HTML version deleted]]

__
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.

Re: [R] Getting the current number of rows in windows R gui console (Carlos Arias)

2014-01-21 Thread Duncan Murdoch


On 14-01-21 7:27 PM, Carlos Arias wrote:

Hello, I'm currently trying to set some default settings and I know how to
get the width of the console with getOption(width), this corresponds to
the number of columns of the console in the Rconsole file. I would also
like to get the current number of rows from inside R. Is there an R option
to get this value?



options(width) gives you the line length that various R functions use; 
it doesn't necessarily tell you the window width.  (Setting the window 
width does update the option value, but not vice versa.)


There aren't functions to get window size at the R level, though you 
could theoretically use getWindowsHandle() to get the handle of the R 
console Window, and use Windows API functions to get its measurements. 
You can use loadRconsole() to load new settings from a file.


Duncan Murdoch

__
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.

[R] Porting Legacy R Code to Different Language

2014-01-21 Thread toshiharu . saito

Hi,

I'm trying to port a quite big piece of legacy R code to a different
language ( Java or C# ).
The package I'm trying to port is quite large but I'm only interested
in one function in the 
package. 

I was just wondering if there's any tools that would allow me to get a
trace of all the code
that a R function executes when evaluating a given input : 

Example :

g 
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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] Windows7, R-Studio, can't find package rJava,

2014-01-21 Thread Prof Brian Ripley

You need to install Java.  That seems not to be in the rJava 
documentation: you get it from here:


http://www.oracle.com/technetwork/java/javase/downloads/jre7-downloads-1880261.html

and need the version for the version of R (32- or 64-bit, x86 or x64 in 
Oracle parlance) you want to run.


Please do follow the R-help posting guide and not send HTML, and give 
the 'at a minimum' information asked for (which includes that this is 
Windows i386 or x64).


On 21/01/2014 20:49, Rainer K. SACHS wrote:

example:


getwd() # with RStudio you can read it off the console bar[1] 
C:/Users/ra`-/Documents/1ray/plans/learnR/week3liats library(xlsx)Loading 
required package: rJavaError : .onLoad failed in loadNamespace() for 'rJava', details:

   call: fun(libname, pkgname)
   error: JAVA_HOME cannot be determined from the RegistryIn addition:
Warning messages:1: package ‘xlsx’ was built under R version 3.0.2 2:
package ‘rJava’ was built under R version 3.0.2 Error: package ‘rJava’
could not be loaded install.packages(rJava)Installing package into
‘C:/Users/ra`-/Documents/R/win-library/3.0’
(as ‘lib’ is unspecified)trying URL
'http://cran.rstudio.com/bin/windows/contrib/3.0/rJava_0.9-6.zip'Content
type 'application/zip' length 758820 bytes (741 Kb)opened
URLdownloaded 741 Kbpackage ‘rJava’ successfully unpacked and MD5 sums
checked
The downloaded binary packages are in
C:\Users\ra`-\AppData\Local\Temp\Rtmpq0N2f2\downloaded_packages
library(rJava,
lib.loc=C:/Users/ra`-/Documents/R/win-library/3.0)Error : .onLoad
failed in loadNamespace() for 'rJava', details:
   call: fun(libname, pkgname)
   error: JAVA_HOME cannot be determined from the RegistryIn addition:
Warning message:package ‘rJava’ was built under R version 3.0.2 Error:
package or namespace load failed for ‘rJava’

I tried various things with searhpath, but can't figure out if this a
windows 7 problem, an R problem, an R-studio problem, an R4 vs R3
problem, an rJava problem, a firewall problem, or some clever
combination of such problems. Any tips greatly appreciated even if all
they tell me is in which direction to look (e.g. give Windows a
different searchpath?)

TIA Ray Sachs

[[alternative HTML version deleted]]



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--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
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Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : dims [product 1] do not match the length of object [0]

2014-01-21 Thread Prof Brian Ripley


On 21/01/2014 22:45, William Dunlap wrote:

You probably want to use %w instead of %u.  On Linux 'man strftime' says
%w The day of the week as a decimal, range 0 to 6, Sunday being  0.
   See also %u.
%u The day of the week as a decimal, range 1 to 7, Monday being  1.
   See also %w.  (SU)
where (SU) means according to the Single Unix specification and the lack
of (...) after %w means it is in some year's ANSI C standard.   I assume 
Windows
does not attempt to subscribe to the Single Unix standard.


Maybe, but %u is in POSIX, standards Windows no long attempts to 
follow (and never did by default).


If you need %u on Windows, try R-devel.  strftime has been replaced 
there (by default on Windows, optionally on other platforms) by a 
POSIX-2008-compliant version.




Bill Dunlap
TIBCO Software
wdunlap tibco.com



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf
Of William Dunlap
Sent: Tuesday, January 21, 2014 2:33 PM
To: arun; R help
Subject: Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : dims 
[product 1]
do not match the length of object [0]

When I use R-3.0.2 on Windows 7 the %u descriptor for format.Date() always
gives , while on Linux in gives as.character(day-of-the-week).  The resulting 
NA's
on Windows could be the source of your problem.

On Linux I get:
 format(as.Date(c(2014-01-21, 2014-01-22, 2014-01-28)), %u)
[1] 2 3 2
 as.numeric(.Last.value)
[1] 2 3 2
 cat(version$version.string, on, version$platform, \n)
R version 3.0.2 (2013-09-25) on x86_64-unknown-linux-gnu

while on Windows:
format(as.Date(c(2014-01-21, 2014-01-22, 2014-01-28)), %u)
   [1]   
as.numeric(.Last.value)
   [1] NA NA NA
cat(version$version.string, on, version$platform, \n)
   R version 3.0.2 (2013-09-25) on x86_64-w64-mingw32


d - subset(d, date==next_friday)
d - ddply(d, id, mutate,
previous_price = lag(xts(price,date)),
log_return= log(price / previous_price),
simple_return = price / previous_price - 1
)
d - dcast(d, date ~ id, value.var=simple_return)


I you didn't reuse the same name, d, for the result of all these
steps it would be easier to poke through the intermediate
results to see where the trouble began (the output of subset()
is a 0-row data.frame and dcast() dies when its input has
zero rows).

Bill Dunlap
TIBCO Software
wdunlap tibco.com



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf
Of arun
Sent: Tuesday, January 21, 2014 12:48 PM
To: R help
Subject: Re: [R] [datatable-help] Error in structure(ordered, dim = ns) : dims 
[product 1]
do not match the length of object [0]

Hi,
Couldn't reproduce the error after running your code:
  d - dcast(d, date ~ id, value.var=simple_return)
  dim(d)
#[1] 356   9
  sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-unknown-linux-gnu (64-bit)

locale:
  [1] LC_CTYPE=en_CA.UTF-8   LC_NUMERIC=C
  [3] LC_TIME=en_CA.UTF-8LC_COLLATE=en_CA.UTF-8
  [5] LC_MONETARY=en_CA.UTF-8LC_MESSAGES=en_CA.UTF-8
  [7] LC_PAPER=en_CA.UTF-8   LC_NAME=C
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base

other attached packages:
[1] plyr_1.8   quantmod_0.4-0 TTR_0.22-0 xts_0.9-7  zoo_1.7-10
[6] Defaults_1.1-1 stringr_0.6.2  reshape2_1.2.2

loaded via a namespace (and not attached):
[1] grid_3.0.2  lattice_0.20-23


A.K.


On Tuesday, January 21, 2014 2:57 PM, rcse2006 rcse2...@gmail.com wrote:
Trying to run below code.

library(quantmod)
symbols - c(AAPL, DELL, GOOG, MSFT, AMZN, BIDU, EBAY, YHOO)
d - list()
for(s in symbols) {
   tmp - getSymbols(s, auto.assign=FALSE, verbose=TRUE)
   tmp - Ad(tmp)
   names(tmp) - price
   tmp - data.frame( date=index(tmp), id=s, price=coredata(tmp) )
   d[[s]] - tmp
}
d - do.call(rbind, d)
d - d[ d$date = as.Date(2007-01-01), ]
rownames(d) - NULL

# Weekly returns
library(plyr)
library(reshape2)
d$next_friday - d$date - as.numeric(format(d$date, %u)) + 5
d - subset(d, date==next_friday)
d - ddply(d, id, mutate,
previous_price = lag(xts(price,date)),
log_return= log(price / previous_price),
simple_return = price / previous_price - 1
)
d - dcast(d, date ~ id, value.var=simple_return)

Getting error


d - dcast(d, date ~ id, value.var=simple_return)

Error in structure(ordered, dim = ns) :
   dims [product 1] do not match the length of object [0]

Please help me how to use ddply and dcast or using other similar function to
get same data.



--
View this message in context: http://r.789695.n4.nabble.com/Error-in-structure-
ordered-dim-ns-dims-product-1-do-not-match-the-length-of-object-0-tp4683923.html
Sent from the datatable-help mailing list archive at Nabble.com.

[R] Strange error on text assignment to variable

2014-01-21 Thread Sumitrajit Dhar

Hello folks,

I am using 

R version 3.0.2 (2013-09-25) -- Frisbee Sailing
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: x86_64-apple-darwin10.8.0 (64-bit)


on Mac OS 10.9.1

When I try a simple assignment as below, the first For works but the second 
for gives me an error.

 P_for - expression(P[For])
 P_for - expression(P[for])
Error: unexpected ']' in P_for - expression(P[for]

I have tested the following to work just fine:
 P_for - expression(P[For])
 P_for - expression(P[F])
 P_for - expression(P[f])
 P_for - expression(P[fo])
 P_for - expression(P[Fo])
 P_for - expression(P[or])

Thanks for any help you can provide.

Regards,
Sumit

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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] Strange error on text assignment to variable

2014-01-21 Thread Frede Aakmann Tøgersen

Hi

Well, for is  a basic control-flow construct and as such a reserved word in R. 
See e.g.  ?for

Yours sincerely / Med venlig hilsen


Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance  Modeling

Technology  Service Solutions
T +45 9730 5135
M +45 2547 6050
fr...@vestas.com
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 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of Sumitrajit Dhar
 Sent: 22. januar 2014 08:01
 To: r-help@r-project.org
 Subject: [R] Strange error on text assignment to variable
 
 Hello folks,
 
 I am using
 
 R version 3.0.2 (2013-09-25) -- Frisbee Sailing
 Copyright (C) 2013 The R Foundation for Statistical Computing
 Platform: x86_64-apple-darwin10.8.0 (64-bit)
 
 
 on Mac OS 10.9.1
 
 When I try a simple assignment as below, the first For works but the
 second for gives me an error.
 
  P_for - expression(P[For])
  P_for - expression(P[for])
 Error: unexpected ']' in P_for - expression(P[for]
 
 I have tested the following to work just fine:
  P_for - expression(P[For])
  P_for - expression(P[F])
  P_for - expression(P[f])
  P_for - expression(P[fo])
  P_for - expression(P[Fo])
  P_for - expression(P[or])
 
 Thanks for any help you can provide.
 
 Regards,
 Sumit
 
 __
 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] My problem with R

2014-01-21 Thread PIKAL Petr

Hi

From: Lee Marine [mailto:marine1...@gmail.com]
Sent: Wednesday, January 22, 2014 2:01 AM
To: PIKAL Petr
Subject: Re: [R] My problem with R

ps.

str(timestamp3)
 POSIXlt[1:36568], format: 2013-10-04 04:42:00 2013-10-04 04:42:01 ...

POSIX format can have list structure or vector structure which is appropriate 
for your case.
see
?as.POSIXct

However did you consider to use merge which is probably more appropriate than 
rbind and does not have matrix side effect.

If I understand correctly you want to import several files with same structure 
and stack them together

 file1-data.frame(a=rnorm(10), b=1:10)
 file1
a  b
1   1.1507791  1
2   1.1080384  2
3   1.4140105  3
4   0.5001733  4
5  -0.2109147  5
6   0.4318809  6
7   0.7652300  7
8  -0.3629509  8
9  -0.1579185  9
10 -0.3710052 10
 file2-data.frame(a=rnorm(10), b=11:20)
 file2
 a  b
1   0.38297419 11
2  -0.77735905 12
3   1.50438015 13
4   0.30632757 14
5   0.30045189 15
6   0.66224030 16
7   1.64338395 17
8  -0.84172668 18
9  -0.33173291 19
10  0.0207 20
 merge(file1, file2, all=T)
 a  b
1  -0.84172668 18
2  -0.77735905 12
3  -0.37100523 10
4  -0.36295093  8
5  -0.33173291 19
6  -0.21091468  5
7  -0.15791848  9
8   0.0207 20
9   0.30045189 15
10  0.30632757 14
11  0.38297419 11
12  0.43188092  6
13  0.50017333  4
14  0.66224030 16
15  0.76522999  7
16  1.10803836  2
17  1.15077911  1
18  1.41401048  3
19  1.50438015 13
20  1.64338395 17

After merging you can convert date and time to POSIX and sort or aggregate your 
file . AFAIK merge can stack only two data frames but you can do it easily in 
cycle or by hand.

Assuming your objects are named data1-datan, you can do

final -merge(data1, data2, all=T)
final - merge(final, data3, all=T)


Regards
Petr

 mode(timestamp3)
[1] list

 Time-data.frame(rbind(timestamp3,timestamp4,timestamp5,timestamp6,timestamp7,timestamp8,timestamp9,
 
timestamp10,timestamp11,timestamp12,timestamp13,timestamp14,timestamp15,timestamp16))
 str(Time)
'data.frame':  14 obs. of  9 variables:
 $ sec  :List of 14
  ..$ timestamp3 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp4 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp5 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp6 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp7 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp8 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp9 : num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp10: num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp11: num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp12: num  0 1 2 4 5 6 7 8 9 10 ...
  ..$ timestamp13: num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp14: num  0 1 2 3 4 5 6 7 8 9 ...
  ..$ timestamp15: num  1 2 3 4 5 6 7 8 9 10 ...
  ..$ timestamp16: num  0 1 2 3 4 5 6 7 8 9 ...
 $ min  :List of 14
  ..$ timestamp3 : int  42 42 42 42 42 42 42 42 42 42 ...
  ..$ timestamp4 : int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp5 : int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp6 : int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp7 : int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp8 : int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp9 : int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp10: int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp11: int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp12: int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp13: int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp14: int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp15: int  0 0 0 0 0 0 0 0 0 0 ...
  ..$ timestamp16: int  0 0 0 0 0 0 0 0 0 0 ...
 $ hour :List of 14
  ..$ timestamp3 : int  4 4 4 4 4 4 4 4 4 4 ...
  ..$ timestamp4 : int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp5 : int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp6 : int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp7 : int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp8 : int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp9 : int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp10: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp11: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp12: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp13: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp14: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp15: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$ timestamp16: int  15 15 15 15 15 15 15 15 15 15 ...
 $ mday :List of 14
  ..$ timestamp3 : int  4 4 4 4 4 4 4 4 4 4 ...
  ..$ timestamp4 : int  4 4 4 4 4 4 4 4 4 4 ...
  ..$ timestamp5 : int  5 5 5 5 5 5 5 5 5 5 ...
  ..$ timestamp6 : int  6 6 6 6 6 6 6 6 6 6 ...
  ..$ timestamp7 : int  7 7 7 7 7 7 7 7 7 7 ...
  ..$ timestamp8 : int  8 8 8 8 8 8 8 8 8 8 ...
  ..$ timestamp9 : int  9 9 9 9 9 9 9 9 9 9 ...
  ..$ timestamp10: int  10 10 10 10 10 10 10 10 10 10 ...
  ..$ timestamp11: int  11 11 11 11 11 11 11 11 11 11 ...
  ..$ timestamp12: int  12 12 12 12 12 12 12 12 12 12 ...
  ..$ timestamp13: int  13 13 13 13 13 13 13 13 13 13 ...
  ..$ timestamp14: int  14 14 14 14 14 14 14 14 14 14 ...
  ..$ timestamp15: int  15 15 15 15 15 15 15 15 15 15 ...
  ..$

Re: [R] how to get the numbers of factors in a matrix

2014-01-21 Thread PIKAL Petr

Hi

elaborating answers you already got

sapply(X, function(m){nlevels(factor(m$latitudes))})
tapply(N$latitudes, N$species, function(x) nlevels(factor(x)))

shall do the trick

Petr

From: å¼ ä»¥æ¥ [mailto:yczh...@nigpas.ac.cn]
Sent: Tuesday, January 21, 2014 2:58 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: RE: [R] how to get the numbers of factors in a matrix

Dear Pikal,

Thank you very much for your answer.

I think your example is just the problem I have.

In the following example you gave to me,

  ff-factor(letters[1:5])
  levels(ff[1:2])
 [1] a b c d e
  fff-ff[1:2]
  nlevels(fff)
 [1] 5

  fff
 [1] a b
 Levels: a b c d e

In my understanding, fff is a subset of ff. Why fff's levels is not a, b but 
a,b,c,d,e.

My problem is quite similar to the example. I just want to split the matrix 
into many subsets and calculate the levels of every subset. Can you tell me how 
to do? Thank you very much!

Best regards,
Yichun

 -åå§é®ä»¶-
 åä»¶äºº: PIKAL Petr 
 petr.pi...@precheza.czmailto:petr.pi...@precheza.cz
 åéæ¶é´: 2014å¹´1æ21æ¥ ææäº
 æ¶ä»¶äºº: å¼ ä»¥æ¥ yczh...@nigpas.ac.cnmailto:yczh...@nigpas.ac.cn, 
 r-help@r-project.orgmailto:r-help@r-project.org 
 r-help@r-project.orgmailto:r-help@r-project.org
 æé:
 ä¸»é¢: RE: [R] how to get the numbers of factors in a matrix

 Hi

 It is rather difficult to understand what problem you have.

 post some data e.g. by

 dput(head(bigmatrix))

 Maybe your problem is in a factor feature that it preserves also empty levels 
 until you specifically drop them.

  ff-factor(letters[1:5])
  levels(ff[1:2])
 [1] a b c d e
  fff-ff[1:2]
  nlevels(fff)
 [1] 5

  fff
 [1] a b
 Levels: a b c d e

 Regards
 Petr

  -Original Message-
  From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org 
  [mailto:r-help-bounces@r-
  project.orgmailto:r-help-bounces@r-%0b%3eÂ %3eÂ project.org] On Behalf Of 
  ???
  Sent: Tuesday, January 21, 2014 7:36 AM
  To: r-help@r-project.orgmailto:r-help@r-project.org
  Subject: [R] how to get the numbers of factors in a matrix
 
  Dear friends,
 
 
  I have a question do not know how to resolve.
 
 
  I have a big matrix composed of different columns (I use N here). A
  column is species and another one is latitudes. Now, I want to know
  how I can get the number of different latitudes for every species.
  I have tried to split the matrix according to species (X-split(N,
  N$species) and then use sapply(X, function(m){nlevels(m$latitudes)}) to
  get that. But the result shows the total factor numbers of latitudes
  but not the factor numbers of every species I splitted. Also, I have
  tried to use tapply(N$latitudes, N$species, nlevels) to do this. The
  result is the same. I am confused about this. Can someone help me with
  that? Thank you very much!
 
 
  Best regards,
  Yichun
 
 
 
 
 
 
[[alternative HTML version deleted]]
 
  __
  R-help@r-project.orgmailto: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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