On 09/06/2010 07:38 AM, Christian Lederer wrote:
Dear R-Users,
after the last upgrade from r-base 2.11.1-2hardy0 to 2.11.1-5hardy0
i lost the tlctk package.
I this a general problem under Ubuntu Hardy or should i search for
a configuration error on my system?
Assuming that you didn't
Thanks, it works fine now.
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Hi,
I have a vector who contents should look like this,
c d NULL e f etc
or
4 5 6 NULL 7 8 9
how can I represent the null value?
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
Perhaps you mean NA rather than NULL.
If NA is not what you want, then I
think you'll need to explain your
application.
On 06/09/2010 06:00, raje...@cse.iitm.ac.in wrote:
Hi,
I have a vector who contents should look like this,
c d NULL e f etc
or
4 5 6 NULL 7 8 9
how can I represent the
Hi Bill,
I didn't make the original post, but its pretty similar to some thing i
would have queried the list about. But, as an R dilatante i find more
curious your question-
...but why would you want to do so?
Is this because you'd typically use the given nine lines of explicit
code to
On 09/06/2010 03:00 PM, raje...@cse.iitm.ac.in wrote:
Hi,
I have a vector who contents should look like this,
c d NULL e f etc
or
4 5 6 NULL 7 8 9
how can I represent the null value?
Hi rajesh,
For character vectors, will probably suffice, but for numbers, you
are probably stuck with NA.
Dear All,
Is it possible to replace function with my own? I want to apply pca
clustering, but
to use some strange correlation function. I'm asking about replacing,
say, mean() with new content of mean() and use standard other functions,
which might use mean() as part.
karsar
Hi Karl,
The why do it like this is probably direct towards creating 9 new
objects for the arima results (Is this right Bill?). A better option
would be to create a list with nine entries. This is much easier for any
subsequent analyses. An example that uses lapply (an efficient syntax
for
Hi Karsar,
To replace mean you can make a new function with the same name:
l = runif(10)
mean(l)
mean = function(x) return(1)
mean(l)
But there must be a better way...
cheers,
Paul
On 09/06/2010 11:52 AM, Karen Sargsyan wrote:
Dear All,
Is it possible to replace function with my own? I
Hi Karl,
I think the question here is why would you want to create different
objects in the loop using assign().
Usually, using lists is better (more efficient?), although I sometimes
use assign() too in this context. I do it when I want to export the
object as separate files (xls, Rbin,
Dear R help,
I have the following data frame:
structure(list(prochi = c(ind_1, ind_1, ind_1,
ind_1, ind_1, ind_1, ind_1, ind_1,
ind_1, ind_1), date_1st_event = structure(c(14784,
14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784
), class = Date), bp_date = structure(c(12660,
lord12 wrote:
I have a data file with a given time series of price data and I would like
to split the time series into a test set and training set. I would then
like to build an ARIMA model on the training set and apply this model on
test set.
I had recently the same problem and, after
On 06/09/2010 1:00 AM, raje...@cse.iitm.ac.in wrote:
Hi,
I have a vector who contents should look like this,
c d NULL e f etc
or
4 5 6 NULL 7 8 9
how can I represent the null value?
As others have said, you probably want NA rather than NULL. If you
really want NULL, then use a list (a
Hi,
I have a loop as follows,
dataStr - character(0)
repeat{
fstr-read.socket(sockfd)
if(fstr==)
break
dataStr-paste(dataStr,fstr)
}
at what point does dataStr stop accepting(gets full)? I'm sending millions of
records over the socket and need to know if all of it can go into
Hi all,
I'm using optimize() to find the minimum of the following function f, and
minimize it (without
f-function(delta,P,U){
minimiz-P+delta*U
x-minimiz[1]
y-minimiz[2]
z-100*(y-x^2)^2+(1-x)^2
return(z)
}
result-optimize(f, interval=c(-1, 1), P=c(0.99,1.01),
NA is good.thanks
- Original Message -
From: Patrick Burns pbu...@pburns.seanet.com
To: r-help@r-project.org, raje...@cse.iitm.ac.in
Sent: Mon, 06 Sep 2010 13:55:34 +0530 (IST)
Subject: Re: [R] representing NULL values in a vector
Perhaps you mean NA rather than NULL.
If NA is not what
If I understand you correctly and each factor consists of binary data, you
may want to check out monothethic analysis, available in the package
'cluster'.
For a simple example and short description of the method to get you started,
just type in:
require(cluster)
?mona
As far as i know there's
Hi Friends,
I am new to R.
On R utility class pages, creating named.list is described with this command :
new(named.list,a=1,b=2)
For large matrix having two columns, such as :
row1 2334
row2 347
row3 379
...
I want to create a named.list like :
$row1
[1] 2334
$row2
[1] 347
...
Can
Hi Natalie,
By far the easiest thing to do is to convert the date to a special date
class. See as.POSIXct for example. I'm not sure that 14784 means, nor
what the data says in the bp_date column. Probably the two combine into
a specific date?
Once you've converted the columns into a POSIXct
Hello everyone.
I would like to ask you what happens when two functions with the same name
exist. I discovered this today when I wrote
?images (I was trying to understand how it works)
?images gave me the following output:
Help on topic 'image' was found in the following packages:
Image
(in
francogrex wrote:
out - glm(response~Var1+Var2+Var3..,family=binomial,data=mydata)
summary(out)
stepAIC(out)
anova(out, test='Chisq')
I understand that stepAIC is used to select the model with the lowest AIC
(the best model) but can someone explain what is the purpose of doing the
Is this what you want:
x
V1 V2
1 row1 2334
2 row2 347
3 row3 379
x.list - as.list(x$V2)
names(x.list) - x$V1
x.list
$row1
[1] 2334
$row2
[1] 347
$row3
[1] 379
On Mon, Sep 6, 2010 at 7:55 AM, Viki S is...@live.com wrote:
Hi Friends,
I am new to R.
On R utility class pages,
Hi,
I have a list which looks like this...
str(y)
List of 10
$ : chr [1:4] ABCD 5 0 1
$ : chr [1:4] DEF 15 1 16
$ : chr [1:4] AAA 2 17 8
$ : chr [1:4] SSS 15 25 1
$ : chr [1:4] III 15 26 4
$ : chr [1:4] OPQ 7 30 4
$ : chr [1:4] TYR 14 34 8
$ : chr [1:4] IRTS 15 42 1
$ : chr [1:4] LLL 15
Karen Sargsyan wrote:
Is it possible to replace function with my own? I want to apply pca
clustering, but to use some strange correlation function. I'm asking about
replacing, say, mean() with new content of mean() and use standard other
functions, which might use mean() as part.
The
Hi,
is it possible to insert a vector as a row in a data.frame?
[[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
Michael Bernsteiner wrote:
I'm using optimize() to find the minimum of the following function f, and
minimize it (without
.
But, when I choose a larger Interval in the optimization method:
The result gets worse (even though the old interval is included in the new
one).
In
Hi, this is more related to understanding some statistics while using R; I've
see such output in a paper:
out - glm(response~Var1+Var2+Var3..,family=binomial,data=mydata)
summary(out)
stepAIC(out)
anova(out, test='Chisq')
I understand that stepAIC is used to select the model with the lowest AIC
I am building a package say mypkg. Five months ago, when I built the package I
got the mypkg manual in pdf format.
Today, after making updates, I build the same package, same name, and steps;
unfortunately I do not get the manual in pdf format.
Rather I get the following message:
cd: can't
try this:
x - |1,ab,2.34|2,cd,3.44|
# split by the | and remove vectors of zero characters
x.sp - strsplit(x, '|', fixed = TRUE)[[1]]
x.sp - x.sp[nchar(x.sp) 0]
# now split by comma
x.comma - strsplit(x.sp, ',')
# you can now access you data
x.comma
[[1]]
[1] 1ab 2.34
[[2]]
[1] 2
Hi!
I'm sure there's an easier way, but that works for me:
test_list - list(c(ABC,5,0), c(DEF,10,1)) ##just a part of
your example, think about using dput() to create a copy/pastable example
test_df - t(as.data.frame(test_list)[-1,])
rownames(test_df) - t(as.data.frame(test_list)[1,])
HTH,
It is easy to store a list of that size:
x - list(1:1e6, 1:1e6, 1:1e6)
object.size(x)
12000112 bytes
str(x)
List of 3
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
Now it really depends on how you are
Hi again,
see ?rbind
Ivan
Le 9/6/2010 14:11, raje...@cse.iitm.ac.in a écrit :
Hi,
is it possible to insert a vector as a row in a data.frame?
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
On Sep 6, 2010, at 4:03 AM, B W wrote:
Snipped out formatting detritus and added back many missing speces.
-Hello,I am trying to take the information from the summary of my
best fit logisticregression model for the occurrence of a high
elevation plant spp. and create the appropriate
Thank you, David:
I obviously didn't look hard enough. This is exactly what I need.
Charles Annis, P.E.
charles.an...@statisticalengineering.com
561-352-9699
http://www.StatisticalEngineering.com
-Original Message-
From: r-help-boun...@r-project.org
Everyone -
What do the NaN's mean here? Is this analysis a problem?
Linear mixed-effects model fit by maximum likelihood
Data: tmp.dat
AIC BIClogLik
1611.251 1638.363 -797.6253
Random effects:
Formula: ~1 | group_id
(Intercept) Residual
StdDev: 0.0003077668
Hi Adrian,
dat=data.frame(matrix(0,3,3))
write.xlsx(dat,z:/dat.xlsx,sheetName=sheet1,append=F)
write.xlsx(dat,z:/dat.xlsx,sheetName=sheet2,append=F)
The above code works and creates new worksheets. But if I want to append
to an existing worksheet I seem to get an error.
Hi Paul, Ivan,
Hartstikke bedankt and thanks alot for sharing these thoughts. I can see
'listing up' multiple symmetrical data sets makes a lot of sense. As
does using lapply() on them which i understand to be more
efficient/faster than for().
Goodo- with your concensus (and helpful
Dear Edward,
You have no degrees of freedom left to estimate those p-values. Your
design does not allows for the model your implemented. We need a brief
summary of your design in order to help you further.
HTH,
Thierry
Hello everyone.
Say we have the following:
a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010,
c(ES, PT, Z , CF, GX, ST, EO)))
b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT,
CF, AT, EM, ST)))
d - cbind(a, b)
I want to calculate sums of the columns that have similar
Dear all,
Thanks very much for the replies and for the help.
This whole data set consists of about 7000 individuals who have had
multiple blood pressure measures taken over time so I just used one
individual as an example. I'm sorry if it looked like homework...it isn't.
Jim your solution
On Mon, Sep 6, 2010 at 9:56 AM, Sergey Goriatchev serg...@gmail.com wrote:
Hello everyone.
Say we have the following:
a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010,
c(ES, PT, Z , CF, GX, ST, EO)))
b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT,
CF, AT, EM,
On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote:
Hello everyone.
Say we have the following:
a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010,
c(ES, PT, Z , CF, GX, ST, EO)))
b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT,
CF, AT, EM, ST)))
d - cbind(a, b)
I
Hi
How would I analyse time series with
- different lengths (i.e. one has 9 entries and the other has 14 entries)
- different frequency (i.e. dates are random - no repeated length)
- multiple values for the same time entry (e.g. 2009-10-23 below)
i.e. my data takes the form:
1st time series
Gabor, David, thank you.
David, your last suggestion is what I need.
Regards,
Sergey
On Mon, Sep 6, 2010 at 16:12, David Winsemius dwinsem...@comcast.net wrote:
On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote:
Hello everyone.
Say we have the following:
a - matrix(c(-75, 3, 5, 9, 2,
On Sep 5, 2010, at 10:32 AM, Daniele Sluijters wrote:
Hello,
I'm sorry to just pop-up on the mailing list like this and ask a
relatively non-R related question but I had no idea whom else to
contact on this matter.
I'm working on a completely different port of an application to OS X
Hi,
I have a matrix that looks like this
a - c(1,1,1,1,2,2,3,3,3,3)
b - c(2,2,2,3,4,4,4,5,5,6)
c - c(1,2,3,4,5,6,7,8,9,10)
M - matrix(nr=10,nc=3)
M[,1] - a
M[,2] - b
M[,3] - c
M
[,1] [,2] [,3]
[1,]121
[2,]122
[3,]123
[4,]13
Hi everyone. I have these data:
probClass-seq(0,0.9,0.1)
prob1-c(0.0070,0.0911,0.1973,0.2949,0.3936,0.5030,0.5985,0.6869,0.7820,0.8822)
prob2-c(0.0066,0.0791,0.2358,0.3478,0.3714,0.3860,0.6667,0.6400,0.7000,1.)
# which I'm plotting as follows:
one way is the following:
M - cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6),
c(1,2,3,4,5,6,7,8,9,10))
ind - do.call(paste, c(as.data.frame(M[, 1:2], sep = \r)))
M[, 3] - ave(M[, 3], ind, FUN = sum)
unique(M)
I hope it helps.
Best,
Dimitris
On 9/6/2010 4:29 PM, Kennedy wrote:
On Mon, Sep 6, 2010 at 3:29 PM, Kennedy henrik.aldb...@gmail.com wrote:
I want to reduce the matrix according to the following: If the values of the
two first columns are the same in two or more rows the values in the third
column of the corresponding rows should be added and only one of the
Hello everyone.
I would kindly request your help concerning how R converts data between
different structrures.
In the following example please keep attention on the following two
1)
I create
f - GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean, variance, nugget,
scale, alpha))
with
On 06/09/2010 9:19 AM, Juliet Ndukum wrote:
I am building a package say mypkg. Five months ago, when I built the package I
got the mypkg manual in pdf format.
Today, after making updates, I build the same package, same name, and steps;
unfortunately I do not get the manual in pdf format.
On Mon, Sep 6, 2010 at 10:24 AM, trb1 thomasrbol...@yahoo.co.uk wrote:
Hi
How would I analyse time series with
- different lengths (i.e. one has 9 entries and the other has 14 entries)
- different frequency (i.e. dates are random - no repeated length)
- multiple values for the same time
Hi,
first off, I wanna ask how do I declare a data.frame of 0 rows and n
columns?
Coming to my problem,
I have a data.frame of 22 columns by dynamic rows which I insert using
rbind. The total number of rows could go upto 2,00,000. The problem is that
after about 800 or 900 get inserted rbind
Hi again!
I'm trying to follow your general goal from your questions today but
it's not easy.
First, declaring a data.frame of 0 rows is a bad idea. It is much faster
to define the length and number of rows from the beginning and to fill
it then.
Second, I don't know how to do it! What I
On Sep 6, 2010, at 10:47 AM, Dimitris Rizopoulos wrote:
one way is the following:
M - cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6),
c(1,2,3,4,5,6,7,8,9,10))
ind - do.call(paste, c(as.data.frame(M[, 1:2], sep = \r)))
M[, 3] - ave(M[, 3], ind, FUN = sum)
unique(M)
I had been
This will give a matrix with 0 rows.
data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))
But you should avoid growing dataframes is the final dataframe is going
to be large. You are very likely to get memory problems. It is much to
better to create a large enough
that was my first idea as well, but as the result shows, the minimized function
value of the wider interval is greater.
In
addidtion, the problem also exists, if the minimized parameter in the
case of the larger interval also already lies within the smaller
interval:
f-function(delta,P,U){
Hi.
which package i need to install to be able to run Path analysis using r?
many thanks, Guy
--
Guy Rotem
Department of Life Sciences
The Spatial Ecology Lab
Ben Gurion University of the Negev
P.O.B. 653 Beer-Sheva 84105
ISRAEL
+972-52-3354485 (mobile)
+972-8-6461350 (lab)
Dear list,
I am using a external program that outputs Q1, Q3, median, upper and lower
whisker values for various datasets simultaneously in a tab delimited format.
After importing this text file into R, I would like to plot a boxplot using
these given values and not the original series of
But If I do that how will I resize later?
On Mon, Sep 6, 2010 at 8:54 PM, ONKELINX, Thierry
thierry.onkel...@inbo.bewrote:
This will give a matrix with 0 rows.
data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))
But you should avoid growing dataframes is the
Thank you very much for your post.
Your answer has been very helpful.
Is it possible to merge 2 time series?
--
View this message in context:
http://r.789695.n4.nabble.com/Time-Series-tp2528444p2528584.html
Sent from the R help mailing list archive at Nabble.com.
Also, when I create the data.frame with matrix and try to rbind, I get
warnings..
Warning messages:
1: In `[-.factor`(`*tmp*`, ri, value = 4) :
invalid factor level, NAs generated
2: In `[-.factor`(`*tmp*`, ri, value = 5) :
invalid factor level, NAs generated
3: In `[-.factor`(`*tmp*`, ri,
Hi R users:
I don't know if you have had the following problem trying to
export to an xls format file in a non windows platform.
I try to use the following packages:
1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1)
2. WriteXLS (version 1.9.0) (with perl and testPerl working)
Even
Hi,
Are you sure you used the correct syntax and object names? It might just
be because of that...(reading the error messages)
There is another function, xlsReadWrite::write.xls(), that I like a lot:
it is really easy to use and does not require Perl or Python.
HTH,
Ivan
Le 9/6/2010 18:03,
There are lots of options for path analysis in R.
If you go to http://www.rseek.org and type path analysis into the search box,
you will get lots of information on functions/packages, and more general
info as well.
Beyond that, we'd need more specifics about your task.
Sarah
On Mon, Sep 6,
Hi Dave,
You can look at the function ?bxp it might work for you. Alternately,
create a meaningless boxplot object, and then just edit that data, in
which case I know it will work with bxp().
# Create a boxplot, the data does not matter
x - boxplot(1:10)
x # view the data for the boxplot
Hi folks,
Debian 504 64-bit
I found following document;
http://www.biostat.wisc.edu/~kbroman/Rintro/
Whether it is the right document for installing Emacs+ESS and R so that R can
run on Emacs?
TIA
B.R.
Stephen L
__
R-help@r-project.org mailing
Thank you Ivan for you answer:
El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:
Hi,
Are you sure you used the correct syntax and object names? It might just
be because of that...(reading the error messages)
Im sure, because it works with write.csv or write.table.
There is
On 6 September 2010 at 09:18, Stephen Liu wrote:
| Hi folks,
|
| Debian 504 64-bit
Good. All you need is
sudo apt-get install ess
| I found following document;
| http://www.biostat.wisc.edu/~kbroman/Rintro/
|
| Whether it is the right document for installing Emacs+ESS and R so that R can
On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote:
Thank you Ivan for you answer:
El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:
Hi,
Are you sure you used the correct syntax and object names? It might
just
be because of that...(reading the error messages)
Im
I use the following sintaxis for the packages:
For WriteXLS I use:
writeXLS(todo2009,todo2009.xls)
And for dataframes2xls I use:
dataframe2xls::write.xls(todo2009,todo2009.xls)
El lun, 06-09-2010 a las 12:34 -0400, David Winsemius escribió:
On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera
[forwarding back to r-help for archiving/further discussion]
On 10-09-05 08:48 PM, Sally Luo wrote:
Prof. Bolker,
Thanks for your reply and the helpful info.
I still have a few questions.
1. I also tried to use different methods other than BFGS without
changing their default
Hello!
I need some help.
How I know it to draw the formula of the poisson distribution?
expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda) ---
not good
on the screen the k! not the Poisson Formula, but factorial(k)
Thanx!
[[alternative HTML version deleted]]
I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt) column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.
n.h1 = sqldf(select distinct h, count(*) from x group by h)
Error in sqliteExecStatement(con, statement, bind.data) :
RS-DBI driver:
Hi,
I am not sure if this make sense at all. I'd like to sample a matrix, which
follows a wishart / inverted wishart distribution. However, the (1,1)
element of this matrix should always be equal to 1. How can I handle it in
R? Any suggestion is greatly appreciated. Thanks a lot.
Sonia
Hi Jim,
Thanks,
That´s right. But the problem is that it
introduces unnecessary quotes, perhaps due to the format of first
column data in this case :
x-cbind(c(row:1, row:2, row:3), c(4889, 9987, 494))
x1-as.list(x[,2])
names(x1)-x[,1]
x1
$`row:1`
[1] 4889
$`row:2`
[1] 9987
$`row:3`
[1] 494
Hi Everyone,
I have two different data set in 2 different scale.
I want to plot these two data in the same plot
in their respective scale. So the plot will have 2 different scale.
I have added an image below to show how it should look.
does any bode has any idea how this can be done.
2
Hi,
I am not sure if this make sense at all. I'd like to sample a matrix, which
follows a wishart / inverted wishart distribution. However, the (1,1)
element of this matrix should always be equal to 1. How can I handle it in
R? Any suggestion is greatly appreciated. Thanks a lot.
Sonia
On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:
Hello!
I need some help.
How I know it to draw the formula of the poisson distribution?
expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-
lambda) ---
not good
?plotmath
(Do not see factorial as a plotmath function
Try:
Successful!
Thank you!
2010/9/6 David Winsemius dwinsem...@comcast.net
On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:
Hello!
I need some help.
How I know it to draw the formula of the poisson distribution?
expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda)
---
Dear useRs,
the deadline for submission of abstracts is approaching for ERCIM'10.
Please upload your abstract until 2010-09-08 if you would like to give
a presentation at our track on Statistical Algorithms and Software at the
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On Sep 6, 2010, at 12:15 PM, Dimitri Shvorob wrote:
I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt)
column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.
n.h1 = sqldf(select distinct h, count(*) from x group by h)
Error in
Hi,
Looking at the picture, I think you are just talking about plotting
two datasets. Here is an example I made up, that looks sort of like
your picture:
# make a barplot
barplot(-50:50)
# add points into the existing plot at the coordinates set by x and y
# and use a line to connect them
On Mon, Sep 6, 2010 at 11:56 AM, trb1 thomasrbol...@yahoo.co.uk wrote:
Thank you very much for your post.
Your answer has been very helpful.
Is it possible to merge 2 time series?
zz is my posted code was formed by merging two univariate and one
multivariate series.
--
Statistics Software
Hi
This question is far less simple than the title suggests, please read
carefully, thanks.
I have 2 sets of data, both read into R
data1-read.table (1.txt, header=T, sep=\t)
data2-read.table (2.txt, header=T, sep=\t)
data1
Taxon stage1 stage2 stage3 stage4
T1 0 0
I simply put, plot(density(), main=, + xlab = XXX), it says that
I have an unexpected = in it.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
On Mon, Sep 6, 2010 at 12:15 PM, Dimitri Shvorob
dimitri.shvo...@gmail.com wrote:
I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt) column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.
n.h1 = sqldf(select distinct h, count(*) from x group
On Mon, Sep 6, 2010 at 11:07 AM, tooblue kai...@berkeley.edu wrote:
I simply put, plot(density(), main=, + xlab = XXX), it says that
I have an unexpected = in it.
You just have an extra ' + ' before the xlab argument:
plot(density(rnorm(100)), main = , xlab = XXX)
ought to do
On Sep 6, 2010, at 2:07 PM, tooblue wrote:
I simply put, plot(density(), main=, + xlab = XXX), it
says that
I have an unexpected = in it.
It may be a case of a confused parser. You have an extraneous + in
there:
= rnorm(100)
plot(density(), main=, xlab = XXX)
Dear community,
I am currently trying to fit an ordinal logistic regression model with the
polr function. I often get the same error message :
attempt to find suitable starting values failed, for example with :
require(MASS)
data(iris)
Dear friends, two questions
(1) does anyone know if there are any non-parametric equivalents of the two-way
ANOVA in R? I have an ordinal non-normally distributed dependent variable and
two factors (gender and city of birth). Normally, one would try a two-way
anova, but if R has any
Hi
I have got a long script which will not run for me as i keep getting errors :
source(clusterfixV1_4.r)
Error in source(clusterfixV1_4.r) :
clusterfixV1_4.r: unexpected symbol at
158: eck[k,2] - as.numeric(1)
159: #ClusterInfo[k,2] - Clustered
I have sorted all the ones i
The usual least-squares methods are fairly robust to departures from
normality. Furthermore, it is the residuals that are assumed to be
normally distributed (not the marginal distributions that you are
probably looking at) , so it does not sound as though you have yet
examined the data
I am using the standard phonetic font Doulos SIL in a graph
(http://scripts.sil.org/cms/scripts/page.php?site_id=nrsiid=DoulosSILfont)
This is an example:
windowsFonts(IPA=TT Doulos SIL)
barplot(c(1,2,3,4,5),names=c(\u{0251},\u{0252},\u{0253},\u{0254},\u{0255}),family=IPA)
However, I am unable
On 09/06/2010 09:41 PM, raje...@cse.iitm.ac.in wrote:
Hi,
I have a list which looks like this...
str(y)
List of 10
$ : chr [1:4] ABCD 5 0 1
$ : chr [1:4] DEF 15 1 16
$ : chr [1:4] AAA 2 17 8
$ : chr [1:4] SSS 15 25 1
$ : chr [1:4] III 15 26 4
$ : chr [1:4] OPQ 7 30 4
$ : chr
On Sat, 4 Sep 2010, raje...@cse.iitm.ac.in wrote:
Hi,
I have the following piece of code,
repeat{
ss-read.socket(sockfd);
if(ss==) break
output-paste(output,ss)
}
but somehow, output is not receiving all the data that is coming through the
socket.My suspicion is on the if statement. what
We would need to at least see 10 lines either side of the error to
understand its context. Also take a look with you editor (hex editor
would be handy) to see if there is some unprintable character around
it. Can you isolate just that portion of the code? Try putting it
inside a function to see
Try this (after making sure that Col_1 in data2 matches your column
names in data1
data1 - read.table(textConnection(Taxon stage1 stage2 stage3 stage4
+ T1 0 0 1 1
+ T2 0 1 1 0
+ T3 0 0 0
The design is a repeated measures with 3 instances. There are 3 groups:
Controls, Heavy Cocaine Users, Light Cocaine Users.
I reshaped the data so that there was one variable for the 3 instances
called AvgTrials. Time is the indicator of each instance.
Here is the model call:
mod5 -
Hi Folks:
I found the following paper to be an interesting example of how even
well designed and conducted studies -- randomized trials, even -- can
be rendered problematic by systematic effects beyond the control of
the investigators:
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