In my first email,I wrote:
summary(lm(extra~group))
Maybe the same as your "summary.lm"
2011/8/16 Jeff Newmiller
> ** Sorry, I missed the nature of the data.
>
> As to your original question, perhaps typing "summary.lm" at the R prompt
> can enlighten you?
>
> ---
By searching, i finally found the resolution.
We could set "par.settings = list( axis.line = list(col ='white') )" in
xyplot function, then the axis border lines will be white.
Thanks anyway.
Regards,
Black
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-change-axis-line-c
Hi all,
I tried and searched a lot to find a way of changing color of axis line in
applying xyplot() function.
Take following code as example:
library(lattice)
par(col='white')
par(col.axis='white')
par(col.lab='white')
z = data.frame(x = 1:5)
xyplot(x~x, z)
The par(col.axis='white') has no ef
Sorry, I missed the nature of the data.
As to your original question, perhaps typing "summary.lm" at the R prompt can
enlighten you?
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.
The mean of group1 is 0.75,just the same as intercept.
Acturally,R treated group1 as reference,so intercept is just group1.
2011/8/16 Jeff Newmiller
> Why would you assert that the mean is the same as the intercept? For data
> clustered far from the y axis this seems clearly unlikely to be tru
Why would you assert that the mean is the same as the intercept? For data
clustered far from the y axis this seems clearly unlikely to be true.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Hi all:
I have a question about lm on t-test.
data(sleep)
I wanna perform t-test to test the difference between the 2 groups:
I can use:
t.test(extra~group)
The t.test result shows that:t = -1.8608; mean1=0.75,mean2=2.33
But I still wanna use:
summary(lm(extra~group))
Intercept=0.75,which is
Hi:
As previously mentioned, the reshape package (in particular, melt())
is an alternative to make.groups(), although they serve the same
purpose in this example:
dat <- data.frame(`x1`=runif(6, 0, 125),
`x2`=runif(6, 50, 75),
`x3`=runif(6, 0, 100),
On Mon, Aug 15, 2011 at 10:34 PM, Mikhail Titov wrote:
> Hello!
>
> I'd like to have a function to draw correct grid while using log axis with
> xyplot from lattice package. Right now I have the following code inside of my
> panel function:
>
> lim <- current.panel.limits()
> v <- latticeExtra::
A combination of Predict (your newdata), cut2, and the plotting function of
your choice ought to suffice. But thought that cross-validation was an option.
Not at console at the moment (just off airplane.)
Sent from my iPhone
On Aug 15, 2011, at 5:26 PM, array chip wrote:
> is there a R functi
Hi All,
Is there anyway to combine vennDiagrams with venneuler commands so I can
have both of the diagrams as one? venneuler gives me proportionate circles
but without numbers and veenDiagrams gives me numbers without proportionate
circles.
Thanks,
Phil
[[alternative HTML version delete
Hello All,
I'm trying to load the rJava package and every time I do I receive this
error message:
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: fun(...)
error: JAVA_HOME cannot be determined from the Registry
Error: package/namespace load failed for 'rJava'
Can anyo
On Mon, 2011-08-15 at 10:41 -0500, Joseph Magagnoli wrote:
> Brian,
> I believe I had a similar problem. Make sure you have JDK (*sun-java6-jdk)
> * installed as well.
> Here is a link to a JDK install guide
> http://www.cyberciti.biz/faq/howto-ubuntu-linux-install-configure-jdk-jre/
>
Joseph
On 16/08/11 10:58, Bert Gunter wrote:
Well, of course that doesn't work for categorical covariates (duhhh!)
-- so I'll just stop at my first clause, "I don't know." Sorry.
I would suggest that a better specification of the model and the
constraints may elicit better and faster responses.
I thi
The labels can contain much more than just names. In my case, they are variable
descriptions (items from a questionnaire). I need to keep the names as they
are, hence the need for Hmisc's labels.
On Mon, Aug 15, 2011 at 3:53 PM, Monsieur Do wrote:
> I have a dataset and a list of labels. I simp
Hi everyone.
I'm using this following code:
pdf(file="Fig5.pdf", width = 4.86, height = 6.29, pointsize = 10, family
='Times')
par(mfcol = c(3,2), mai = c(0.4,0.8,0,0), omi = c(0.7, 0, 0.7, 0.1));
hist( rnorm(100) )
dev.off()
When I open this in any vectorial software (like Illustrator), it say
On 16/08/11 05:07, Mariana Varela wrote:
HI there, I have been trying to use a code posted on R help to be able to calculate area
under the curve for complicated data points and there seems to be an issue with the code:
no "b" object found. I am not a good R user and can't find were the problem
Here's a warning, but it sounds like it's at a deep (C, presumably) level,
xfig()
grid::grid.rect(gp=gpar(fill="red", alpha=0.5))
Warning message:
In grid.Call.graphics("L_rect", x$x, x$y, x$width, x$height,
resolveHJust(x$just, :
semi-transparency is not supported on this device: reported only
In ?par there is the following line:
"Semi-transparent colors are available for use on devices that support them."
I'm looking for a way to query for that support on the current device.
I'm pretty sure that this functionality exists at some level as I
remember seeing warnings to this effect but
Sorry about the incomplete post: gMail just vomited on me. Here's the
complete post:
On Mon, Aug 15, 2011 at 7:40 PM, Bert Gunter wrote:
> FWIW:
>
> 1. The best reference i know is the discussion in Venables's and
> Ripley's "S Programming" book
>
> 2. VERY BRIEF and INADEQUATE discussion. The ge
FWIW:
1. The best reference i know is the discussion in Venables's and
Ripley's "S Programming" book.
2. VERY BRIEF and INADEQUATE discussion. The general idea is that it
allows one to easily pass down "extra" arguments to the internals of a
function.
For example, suppose that you want to write
Inside your function that has a ... argument, you can get the optional
arguments with :
Arg.dot <- list(...)
Section 2.1.5 of the R Language Definition which should be accessible from the
help facility would appear to be sensible place to start. "Programming for Data
Analysis" could also be pu
Have you done any searching (with for instance the term "forecast")? If so then
you should describe what you found and why it doesn't meet your needs.
--
David
Sent from my iPhone
On Aug 15, 2011, at 7:54 PM, "Eduardo M. A. M.Mendes"
wrote:
> Many thanks but I am not looking for a remote
Hi,
Try this,
?capture.output
as in,
capture.output(cat("this is it"))
HTH,
baptiste
PS: Here's another example for fun,
#
library(textplot)
capture.output(txtplot(1:10))
library(gplots)
textplot(capture.output(txtplot(1:10)))
library(grid)
grid.cap()
# not sure how to get back to the orig
Many thanks but I am not looking for a remote consulting.
Actually it is not difficult but unfortunately I am a newbie as far as
writing functions on R.
Can I gather from your email that there is nothing available on R that deals
with dynamic models (k-step ahead and free-run)?
Cheers
Ed
-
Dear folks --
There are a number of functions -- I am thinking of str() as an example --
that produce text as a side-effect, rather then returning it. Is there any
way to send the text produced by such functions into a character variable?
Any suggestions would be greatly appreciated.
andrewH
?reshape
You have your data in a wide format, but you want it in a long format.
reshape can convert it both ways.
Mikhail
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Sam Albers
> Sent: Monday, August 15, 2011 6:58 PM
> T
Hello list,
## I have been doing the following process to convert data from one
form to another for a while but it occurs to me that there is probably
an easier way to do this. I am often given data that have column names
which are actually data and I much prefer dealing with data that are
sorted
I followed a couple threads from the archives and from
stackoverflow.com, and would like to know: just what is "..." ? What I
mean by this is,for example, from the point of view of a user running a
function in debug mode, is "..." an object, or does it exist in the
current environment as som
Hi Rolf:
Maybe. But I'm not sure whether the OP wants two levels of a single
variable to have the same coefficient, or two different categorical
variables in some way, or two different numeric variables, or...
Maybe it's obvious, but I thought it fairer to the OP to make clear
that I was not a re
Well, of course that doesn't work for categorical covariates (duhhh!)
-- so I'll just stop at my first clause, "I don't know." Sorry.
I would suggest that a better specification of the model and the
constraints may elicit better and faster responses.
-- Bert
On Mon, Aug 15, 2011 at 3:53 PM, Ber
Hi:
Here's one way, using the following reproducible example.
# Method 1: the variable names are the same in each data frame
# Create two separate data frames
ds1 <- data.frame(x1 = rnorm(10), x2 = rnorm(10), x3 = rnorm(10))
ds2 <- data.frame(x1 = rnorm(10), x2 = rnorm(10), x3 = rnorm(10))
# Mel
I don't know the answer in general, but for the specific constraint of
two coefficients being the same, I would assume that you should create
a new covariate which is the sum of the two individual ones and fit
this single combined covariate instead of the two separate ones.
Cheers,
Bert
On Mon, A
This may not be helpful, but this sounds difficult enough that you
should work with a local statistician rather than trying to get remote
consulting here.
-- Bert
On Mon, Aug 15, 2011 at 3:00 PM, Eduardo M. A. M.Mendes
wrote:
> Dear R-users
>
>
>
> I need to fit a nonlinear model to a piece of d
Hello,
I'm pretty new to R. Basically, how do I speed up the for loop below. Or
better yet, get rid of the for loop all together.
objective: plot two data sets column against column by index. These data
sets have alot NA's. Some columns are all NA's. I need the plots to overlay.
I don't like the
Dear R-users
I need to fit a nonlinear model to a piece of data. The model to be fitted
uses past values of the input and the ouput - something like
y(k) ~ f(y(k-1),y(k-2),u(k),u(k-1) ) (k is time index). As far as I
know I could use earth(MARS), nnet and etc but I am not sure how to deal
Hi:
Here's one approach using the ggplot2 package:
Y=c(15,14,23,18,19,9,19,13)
Xa=c(17,22,21,18,19,25,8,19)
Xb=c(22,22,29,34,19,26,17,22)
dd <- data.frame(Y, Xa, Xb)
ggplot(dd) +
geom_segment(aes(x = Xa, xend = Xb, y = Y, yend = Y,
colour = Xb - Xa), size = 2) +
geom_po
I have been using the mlogit package but can't seem to figure out how to make
constraints on the beta coefficients.
For example, I would like to force that two of my beta's are equal to each
other.
Thanks in advance.
Jonah
__
R-help@r-project.org mai
is there a R function that produces calibration curve on an independetn data
automatically, just like what calibrate() does on the training data itself?
Thanks
John
From: Comcast
Cc: "r-help@r-project.org"
Sent: Monday, August 15, 2011 2:04 PM
Subject: Re:
That works! Thanks.
On Mon, Aug 15, 2011 at 5:09 PM, Bert Gunter wrote:
> Sverre:
>
> As your reproducible example shows, the problem is that the string
> "y1:y2" is not an acceptable term of the formula. One way round this
> is simply convert the non-string part of the formula to a string, pase
Hi:
If you're asking about the p-values, here's a reproducible example
from the gee package:
library('gee')
m <- gee(breaks ~ tension, id=wool, data=warpbreaks, corstr="exchangeable")
# Nosing around the summary method return object:
> names(summary(m))
[1] "call""version"
Why re-invent? Use lrm in rms.
--
David
Sent from my iPhone
On Aug 15, 2011, at 10:07 AM, Mariana Varela
wrote:
> HI there, I have been trying to use a code posted on R help to be able to
> calculate area under the curve for complicated data points and there seems to
> be an issue with th
Alison Waller embl.de> writes:
> I'm having problems installing plotrix. I tried installing it
> through install.packages, and from the
> unix command line, but each time it seems to stall when
> it is installing the help indices.
>
> has anyone had this same problem, is this package still ma
Sverre:
As your reproducible example shows, the problem is that the string
"y1:y2" is not an acceptable term of the formula. One way round this
is simply convert the non-string part of the formula to a string, pase
it with your term, and then reconvert it to a formula:
update(my.lm,formula(paste(
Build a prediction function using 'Function' that gets applied to set2.
Calibrate and validate.
--
David
Sent from my iPhone
On Aug 15, 2011, at 11:31 AM, array chip wrote:
> Hi, the calibrate.cph() function in rms package generate calibration curve
> for Cox model on the same dataset wher
On Mon, Aug 15, 2011 at 3:53 PM, Monsieur Do wrote:
> I have a dataset and a list of labels. I simply want to apply the labels to
> the variables, all at once. The only way I was able to do it was using a loop:
>
> for (i in 1:length(data)) label(data[,i]) <- data.labels[i]
>
> I'd like to f
I have a dataset and a list of labels. I simply want to apply the labels to the
variables, all at once. The only way I was able to do it was using a loop:
for (i in 1:length(data)) label(data[,i]) <- data.labels[i]
I'd like to find the non-loop way to do it, using apply or the like... Any h
Hi all,
I'm having problems installing plotrix. I tried installing it through
install.packages, and from the unix command line, but each time it seems to
stall when it is installing the help indices.
has anyone had this same problem, is this package still maintained ?
any help?
thanks
> inst
HI there, I have been trying to use a code posted on R help to be able to
calculate area under the curve for complicated data points and there seems to
be an issue with the code: no "b" object found. I am not a good R user and
can't find were the problem is. Any help? Thanks!!
This is the code (
Hello All,
I am currently using the vennDiagram function in the limma pkg to
construct venn diagrams. I would like to change the size of the circles
based upon the n of each set (if that makes sense). Does anyone have any
ideas of how to do this?
Thanks,
Phil
[[alternative HTML versi
Hello,
I've a question concerning the display of interval data.
A sample dataset where X is an interval between Xa and Xb
which should be displayed:
Y=c(15,14,23,18,19,9,19,13)
Xa=c(17,22,21,18,19,25,8,19)
Xb=c(22,22,29,34,19,26,17,22)
X = (Xa+Xb)/2
It's easily possible to plot the mean of the
Does anyone know how could I get the significant codes from mixed model
output fitted with a GEE package?
The output I got is the following:
GEE: GENERALIZED LINEAR MODELS FOR DEPENDENT DATA
gee S-function, version 4.13 modified 98/01/27 (1998)
Model:
Link: Logit
Varia
Hi all,
I'm extracting the name of the term in a regression model that
dropterm specifies as the least significant one, and I'm assigning
this name to an object. However, when I use update(), it ignores this
object. Is there a way I can make it not ignore it? A reproducible
example is below:
> lm
Hi:
Try this:
df<-data.frame(x=1:10,y=1:10)
plot.fun.one <- function(dff, x.var, y.var)
print(ggplot(dff, aes_string(x = x.var, y = y.var)) + geom_point() )
plot.fun.two <- function(dff, x.var, y.var) {
x <- names(dff)[x.var]
y <- names(dff)[y.var]
print(ggplot(dff, aes_string(x = x,
R-help -
This code iterates over a function with 2 free parameters to find a list of
values (which are the number of incorrect predictions for a computational
model). I want to find the values of i,e when there is the minimum number
of incorrect predictions. In other words, the value of "i" and
... and ?densityplot in the lattice package.
Please also note that you could have avoided posting just by typing
help.search("density") at the RGui prompt. Please use R's Help
facilities before posting.
-- Bert
On Mon, Aug 15, 2011 at 11:44 AM, Jorge Ivan Velez
wrote:
> Hi Jeff,
> Take a look a
Dear Dr. Gilbert, Dr. Varadhan and all R-help list members,
I'm using the function BBsolve( ) and I have some questions on the stopping
criterion "maxit" and "noimp" specified in the option "control". Here
are two such examples I'm having problem with.In these two examples, the
function BB
On Mon, Aug 15, 2011 at 6:51 AM, Giles wrote:
> Thanks for that Gabor, it works fine from the development version
> you've pointed to.
>
> There is in addition a performance issue: the following benchmark ran
> in under 0.2s in the previous version, now consistently shows elapsed
> time over 14s o
Hi Jeff,
Take a look at ?density
HTH,
Jorge
On Mon, Aug 15, 2011 at 2:38 PM, Jeffrey Joh <> wrote:
>
>
> Is it possible to smooth an ecdf plot and get a probability density plot?
> I have about 8000 points and I was hoping to get a density curve instead of
> a histogram.
>
>
>
> Jeff
> _
Hi:
Thank you for the reproducible example and expected result. Here's one approach:
library('plyr')
x <- data.frame(a=1:10, b=11:20, t=c(1,1,1,2,2,2,3,3,3,3))
x$sdif <- cumsum(with(x, a - b))
subfun <- function(d) tail(d[c('t', 'sdif')], 1)
ddply(x, 't', subfun)
t sdif
1 1 -30
2 2 -60
3 3
Is it possible to smooth an ecdf plot and get a probability density plot? I
have about 8000 points and I was hoping to get a density curve instead of a
histogram.
Jeff
__
R-help@r-project.org mailing list
htt
On Aug 15, 2011, at 6:09 AM, mdvaan wrote:
Hi,
I have a question concerning the selection of data. Let's say that
given
list h created below, I would like to select a section of the 1999
matrix.
For a case (rownames and colnames) I would like to select the cells
that
have a value > 0. S
Hi, the calibrate.cph() function in rms package generate calibration curve for
Cox model on the same dataset where the model was derived using bootstrapping
or cross-validation. If I have the model built on dataset 1, and now I want to
produce a calibration curve for this model on an independent
Hi Albert,
Another option would be the following:
install.packages('gdata')
require(gdata)
?read.xlsx
HTH,
Jorge
On Mon, Aug 15, 2011 at 11:19 AM, albert coster <> wrote:
> Hello,
>
> How can I read a xlsx file using xlsx package?
>
> Thanks
>
> Albert
>
>[[alternative HTML version de
Hi there
Many thanks. I will try to follow what you two said.
Meanwhile I use the dirty solution that I have just come across.
Cheers
Ed
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Monday, August 15, 2011 1:36 PM
To: David Winsemius
Cc: Eduardo
Hi Peter,
Well the problem is that I need to also move the ":" to the front of the
bracketed expression, although I like the simplicity of your statement. I
cannot replace all of the "[" with ":[" because I am actually just using a
small example from a large file, and there are many more "[" that
On Mon, Aug 15, 2011 at 10:34 AM, Rebecca Gray wrote:
> Hello all,
>
> I have what I think is a simple question but I've been unable to solve it. I
> have the following string:
>
> A[&states=1]:[&rate=2]425, B[&states=3]:[&rate=5]500
>
> I would like to combine the two expressions in the [], so th
Dear R-users,
I have an anual info of gross product and I would like to disaggregate
to trimestral data.
Can I import a matlab library of Quilis? (
http://www.mathworks.com/matlabcentral/fileexchange/24438-temporal-disaggregation-library
)
Thanks,
Sebastián.
___
Hello all,
I have what I think is a simple question but I've been unable to solve it. I
have the following string:
A[&states=1]:[&rate=2]425, B[&states=3]:[&rate=5]500
I would like to combine the two expressions in the [], so that only one set
of [] is present after each letter, so that I have t
Tyler,
My apology. It does help to type correctly and eliminate unnecessary spaces in
commands. The following does work correctly:
levels(Data1$Site) <- list(Fw = c("AB"), Est = c("DE"))
Thanks very much,
BJ
> From: deel...@hotmail.com
>
I prefer RODBC and odbcConnectExcel this way I can query subsets with SQL.
Mikhail
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Ivan Calandra
> Sent: Monday, August 15, 2011 10:28 AM
> To: r-help@r-project.org
> Subject: R
Whats going on here?
df<-data.frame(x=1:10,y=1:10)
ggplot()+geom_point(data=df,aes(x=x,y=y)) ## this is the normal usage
right?
ggplot()+geom_point(data=df,aes(x=df[,1],y=df[,2])) ## but I can also feed
it column indices
ggplot()+geom_point(aes(x=df[,'x'],y=df[,'y'])) ## or column names.
##
Hello!
I'd like to have a function to draw correct grid while using log axis with
xyplot from lattice package. Right now I have the following code inside of my
panel function:
lim <- current.panel.limits()
v <- latticeExtra:::logTicks(2^lim$xlim, loc=1)
h <- latticeExtra:::logTicks(2^lim$ylim,
On 11-08-15 12:21 PM, David Winsemius wrote:
Eduardo;
I think you would be more successful if you put your data in a
dataframe, offered it to lm with column names only in the formula,
and then used the newdata argument with predict with the column names
matching the column names in the original
Dear R folks,
I am doing some calculations over an array using sweep and apply.
# Sample Data (from help 'addmargins')
Aye <- sample(c("Yes", "Si", "Oui"), 177, replace = TRUE)
Bee <- sample(c("Hum", "Buzz"), 177, replace = TRUE)
Sea <- sample(c("White", "Black", "Red", "Dead"), 177, replace = T
Eduardo;
I think you would be more successful if you put your data in a
dataframe, offered it to lm with column names only in the formula,
and then used the newdata argument with predict with the column names
matching the column names in the original data.
--
David.
On Aug 15, 2011, at 5
On 15/08/2011 11:47 AM, R Saba wrote:
Reading data with variable column widths.
Here are several lines of a txt data set I would like to read.
The number of variables is fixed at 13 . The problem is how to read the
first variable when it can contain blank space-- for example " Alabama
(Seasonally
Reading data with variable column widths.
Here are several lines of a txt data set I would like to read.
The number of variables is fixed at 13 . The problem is how to read the
first variable when it can contain blank space-- for example " Alabama
(Seasonally Adjusted)" , "St. Clair", etc.
Alabama
Brian,
I believe I had a similar problem. Make sure you have JDK (*sun-java6-jdk)
* installed as well.
Here is a link to a JDK install guide
http://www.cyberciti.biz/faq/howto-ubuntu-linux-install-configure-jdk-jre/
On Mon, Aug 15, 2011 at 10:18 AM, Brian Lunergan wrote:
> Morning All:
>
> Ha
Morning All:
Having problems installing this one into my R setup and wondered if
anyone on the list had the same problem and found a workable solution.
First to the system particulars:
OS is the Hardy Heron edition of Ubuntu Linux
R version is 2.13.1
Java installation is sun-java6
Next up is th
Problem solved!
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-from-function-tp3744428p3744695.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo
It workes using ""
But what is this:
> p <- plot_mi_time(hakat, "hsa-miR-100")
X1h X4h X9h X15h X18h X21h
hsa-miR-100 384 1038 1503 1511 2603 2899
*Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ*
--
View this message in context:
http://r.789695.n4.
?read.xlsx
Le 8/15/2011 17:19, albert coster a écrit :
Hello,
How can I read a xlsx file using xlsx package?
Thanks
Albert
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mailman/listinfo/r-he
> ls(hakat)
[1] "X15h" "X18h" "X1h" "X21h" "X4h" "X9h"
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-from-function-tp3744428p3744592.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing
Hi,
I have a question concerning the selection of data. Let's say that given
list h created below, I would like to select a section of the 1999 matrix.
For a case (rownames and colnames) I would like to select the cells that
have a value > 0. So for case 8025
8025 8026 8027
802511
Two ways to do it, both of which come down to forcing R to recognize "(" as
a string element than an actual parenthesis:
x = data.frame(1:3, 4:6)
colnames(x) = c("(Intercept)","Slope")
x$"(Intercept)"
or x[,"(Intercept)"] # Works with or without the comma in slightly different
ways -- figure out
Hello,
How can I read a xlsx file using xlsx package?
Thanks
Albert
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You might want to compare the brackets in your code and in Tyler's
Ivan
Le 8/15/2011 16:36, B Jessop a écrit :
Tyler,
Tried your solution: levels (Data1$Site<- list(Fw = c("AB"), Est = c("DE")))
but still got a NULL response to str(Data1) and an alternating list of Fw, Est, Fw, Est
unde
Well, it looks like the standard advice here applies to both poster
and responder: try reading the Help (carefully): ?factor
Example:
> z <- factor(letters[1:3])
> z
[1] a b c
Levels: a b c
> factor(z,labels=letters[4:6])
[1] d e f
Levels: d e f
> factor(z,labels=letters[3:1])
[1] c b a
Levels: c
Try recode from Deducer package.
Sent from my BlackBerry® smartphone from !DEA
-Original Message-
From: B Jessop
Sender: r-help-bounces@r-project.orgDate: Mon, 15 Aug 2011 11:36:07
To:
Subject: [R] FW: Renaming levels of a factor in a dataframe
Tyler,
Tried your solution: levels
Tyler,
Tried your solution: levels (Data1$Site <- list(Fw = c("AB"), Est = c("DE")))
but still got a NULL response to str(Data1) and an alternating list of Fw, Est,
Fw, Est under Site when looked at in the Data editor in R console. The use of
the "levels" function would seem to be appro
There is an error in your call to `optim'. The`lower' bound is incorrect:
lower=c(rep(-1,n),1e-5,1e-5,1e5)
This should be:
lower=c(rep(-1,n),1e-5,1e-5,1e-5)
I am not sure if this fixes the problem, but it is worth a try.
I do not understand your scaling. From the lower and upper bounds it se
Hello,
I am using poLCA to run standard latent class analyses. Is there any way that I
can get parameters for each predictors, in terms of their association with the
latent variable ?
Thank you,
David Joubert
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Hello all,
I have some points for x and f(x). I want to fit the best curev through
these points and find the value of x and f(x) for which the gradient is -1.
Anyone knows if this can be done in R?
--
Thanks,
Jim.
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I can't help with the questions on pooling regressions, but for the question
about data-frame usage, I'll note two things.
One,
> x = data.frame(y = 1:3, z = 4:6)
> is.data.frame(x)
TRUE
> is.data.frame(t(x))
FALSE
> is.data.frame(as.data.frame(t(x)))
TRUE
Two, you'll need to check that your r
On 11-08-15 7:55 AM, rmje wrote:
*I have the following function:*
/plot_mi_time = function(mdata, miname) {
mdata2 = mdata[row.names(hakat) == miname, ]
You passed the expression hsa-miR-98 as miname. The simplest fix is to pass
"hsa-miR-98"
instead. A more complicated alternative is t
Two things,
1) You need to use square brackets in this case because model is an object,
not a function.
2) You probably want to use a list object to store a whole bunch of model
objects so you'll want double brackets.
model = list(NULL)
for (i in 1:1000) {
model[[i]] <- lm(Y~X1+X2())
}
This,
I believe it's a problem in your variable name "hsa-miR-98". R wants to
interpret that as "hsa" less "miR" less 98. Since you don't have a variable
called "hsa", R (rightly) complains.
Call ls() and see what R thinks your variables are named -- that will
hopefully make the problem evident.
Hope t
*I have the following function:*
/plot_mi_time = function(mdata, miname) {
mdata2 = mdata[row.names(hakat) == miname, ]
print(mdata2)
xcoords <- c(1,1,2,2,3,3,4,4,5,5,6,6)
plot(c(xcoords), mdata2, xaxt="n", ylab="Expression", xlab="Time(h)", ,
main=miname)
axis(1, at=xcoords
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