Dear Prof. Ripley,
many thanks for your quick reply.
A character matrix (although clearly not very elegant) would be no problem,
xtable deals
with that.
I tried as.data.frame() before, but if one wants to have the same rows
as in ft, one has to use additional commands (?):
ft # => 16 rows
as.
as.data.frame(ft)
seems straightforward enough.
I don't think you actually want a matrix, as they would have to be a
character matrix and the ftable object is numeric.
On Fri, 27 May 2011, Marius Hofert wrote:
Dear expeRts,
What's the easiest way to convert an ftable object to a matrix suc
On Thu, 26 May 2011, Duncan Murdoch wrote:
On 11-05-26 9:11 PM, sivan aldor wrote:
I ran the exact command this morning on a friend's computer which is
32bit and it worked!
I am now running it from a different computer. As far as I can tell the
only difference is that his OS is VISTA and mine i
Dear expeRts,
What's the easiest way to convert an ftable object to a matrix such that the
row names of the ftable object are shown in the first couple of columns of the
matrix? This is (typically) required, for example, when the final goal is to
print
the matrix via xtable.
Below is a rather
Dear all,
I am using the plm package for both fixed and random effects models on my
country-year panel data. However, for some of the random effects models I
get the following error:
Error in solve.default(OM) :
system is computationally singular: reciprocal condition number =
1.78233e-18
The
Thanks everybody.
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On 11-05-26 9:39 PM, sivan aldor wrote:
The file was copied properly, My instinct is the problem has to do with
the OS but I want to see if there is a way to resolve this.
Sounds like you have to take it up with the OS manufacturer then.
Duncan Murdoch
On Thu, May 26, 2011 at 9:35 PM, Dunc
The file was copied properly, My instinct is the problem has to do with the
OS but I want to see if there is a way to resolve this.
On Thu, May 26, 2011 at 9:35 PM, Duncan Murdoch wrote:
> On 11-05-26 9:11 PM, sivan aldor wrote:
>
>> I ran the exact command this morning on a friend's computer wh
On 11-05-26 9:11 PM, sivan aldor wrote:
I ran the exact command this morning on a friend's computer which is
32bit and it worked!
I am now running it from a different computer. As far as I can tell the
only difference is that his OS is VISTA and mine is Windows 7. Can that
really be the problem?!
I just discovered that some evil spammer has somehow gotten my contacts list
and
used it to send out a bunch of spam. This is just to notify you that if you
get
an email from me on May 26, 2011 (other than this one or one like it - the
problem was more extensive than I first thought) it wasn'
I ran the exact command this morning on a friend's computer which is 32bit
and it worked!
I am now running it from a different computer. As far as I can tell the only
difference is that his OS is VISTA and mine is Windows 7. Can that really be
the problem?!?!
On Thu, May 26, 2011 at 9:05 PM, Dunca
On 11-05-26 8:42 PM, sivan aldor wrote:
The only command I am using is
dyn.load("dbsvls.dll")
Okay, I get the error message you showed if I try to load something that
really isn't a valid Win32 dll, and I don't get it when I load a valid
one. (I'm not sure if I can fix it to replace the %1
> Date: Thu, 26 May 2011 13:50:15 -0700
> From: gunter.ber...@gene.com
> To: ahmed.el-taht...@pfizer.com
> CC: r-help@r-project.org
> Subject: Re: [R] predictive accuracy
>
> 1. This is not about R, and should be taken off list.
Well, depending on wha
The only command I am using is
dyn.load("dbsvls.dll")
On Thu, May 26, 2011 at 8:41 PM, Duncan Murdoch wrote:
> On 11-05-26 4:53 PM, sivan aldor wrote:
>
>> Hi Everyone,
>> I am trying to work with a routine that is from the IMSL called BSVLS. I
>> have the file as a .dll file.
>> I have been tr
On 11-05-26 4:53 PM, sivan aldor wrote:
Hi Everyone,
I am trying to work with a routine that is from the IMSL called BSVLS. I
have the file as a .dll file.
I have been trying to call the routine using the dyn.load function. I am
working on a 32-bit windows 7 OS with 3 GB. For some reason i still
Gabor Grothendieck wrote:
>
> On Thu, May 26, 2011 at 7:35 PM, Pete Brecknock
> wrote:
>> I have a zoo object that contains 2 time series named "A-B" and "V1".
>>
>> When I create a third series "V2", the name of the "A-B" series is
>> changed
>> to "A.B".
>>
>> Although
On 11-05-26 5:09 PM, James McCreight wrote:
I'm still getting used to R's scoping. I've run into the following situation
value=0
thefunction<- function() print( value )
somefunction<- function() { value=99; thefunction() }
somefunction()
now, I understand that somefunction() returns 0 because t
On Thu, May 26, 2011 at 7:35 PM, Pete Brecknock wrote:
> I have a zoo object that contains 2 time series named "A-B" and "V1".
>
> When I create a third series "V2", the name of the "A-B" series is changed
> to "A.B".
>
> Although I could recreate the names for the 3 series I am wondering if there
Jun,
The par() command is to give extra margin on right side to accomodate the y
axis label.
I recognize, like Rolf, that three y axis can be cumbersome, confuse. So, I
would adopt another approach using lattice::xyplot()
require(reshape)
x <- melt(x, id="Time")
str(x)
require(lattice)
xyplot(va
On Thu, May 26, 2011 at 5:09 PM, James McCreight wrote:
> I'm still getting used to R's scoping. I've run into the following situation
>
> value=0
> thefunction <- function() print( value )
> somefunction <- function() { value=99; thefunction() }
> somefunction()
>
> now, I understand that somefun
Thanks to Mark for an off-list suggestion which works.
somefunction <- function() { value=99; environment(thefunction) <-
environment(); thefunction() }
this apparently defines a copy of thefunction within somefunction, exactly
what I wanted. thefunction in the global env is unchanged.
It seems
I have a zoo object that contains 2 time series named "A-B" and "V1".
When I create a third series "V2", the name of the "A-B" series is changed
to "A.B".
Although I could recreate the names for the 3 series I am wondering if there
is a way of preventing the name change from happening ( ... may
On Thu, May 26, 2011 at 6:05 PM, Dennis Fisher wrote:
> Colleagues
>
> Assume that I have a vector containing some text strings, some of which
> contain a particular character. I could like to apply "toupper" to the text
> before the character. For example (in this case, "|" is the particular
> arr<-1:72
> dim(arr)<-c(2,4,3,3)
> apply(arr,1:2,sum)
[,1] [,2] [,3] [,4]
[1,] 297 315 333 351
[2,] 306 324 342 360
> apply(arr,3:4,sum)
[,1] [,2] [,3]
[1,] 36 228 420
[2,] 100 292 484
[3,] 164 356 548
>
apply(arr,1:2,sum)
will sum all elements arr(i1,i2,..)
for each
I think the source of the OP's problem is that
while things like df>30 and is.na(df) return
a logical matrix with the dimensions of the
data.frame df, both is.infinite(df) and is.nan(df)
return a logical vector as long as the number
of columns of df. (`>` and is.na have data.frame
methods but is.i
On Fri, May 27, 2011 at 10:06 AM, marco milella wrote:
> Hi to everybody
> I have an array with dimensions 2,4,3,3.
> Wanting to sum the matrices in the first two dimensions, I'm trying to use
> the "apply" function, but with no results. Have to say I'm quite new with R
> syntax.
Tell apply() whi
Hi to everybody
I have an array with dimensions 2,4,3,3.
Wanting to sum the matrices in the first two dimensions, I'm trying to use
the "apply" function, but with no results. Have to say I'm quite new with R
syntax.
tx in advance
marco
[[alternative HTML version deleted]]
___
Hi
I have 2 time series consisting of sequence of data points measuring flies
activity at time intervals of 20 min over a period of 24 hrs. One time series
is
from control flies while the other is from flies treated with a drug. My
question is to find out if the drug has a significant effect
Dennis -
Here's one way, using a somewhat obscure feature of perl
regular expressions, i.e. the \U and \L escape characters which
modify the case of the strings they appear in:
TEXT<- c("", "bbb|cc", "|ddd")
sub('([a-z]*)(\\|?)([a-z]*)','\\U\\1\\2\\L\\3',TEXT,perl=TRUE)
[1] ""
> If a function uses substitute() or its equivalent to avoid evaluating
> its arguments in the normal way, you are pretty much forced to use
> eval() with the output of substitute() or call() or use do.call()
> to evaluate the arguments it will not evaluate for itself.
Which is why I'd argue all f
On May 27, 2011, at 00:05 , Dennis Fisher wrote:
> Colleagues
>
> Assume that I have a vector containing some text strings, some of which
> contain a particular character. I could like to apply "toupper" to the text
> before the character. For example (in this case, "|" is the particular
>
Hi Vikas,
I've found it nessessary to do
su
R CMD javareconf
After java updates. Then install/update JGR etc. if desired.
Note that for whatever reason 'sudo R CMD javareconf' doesn't seem to
work for me, I have to actually become root with su and then run R CMD
javareconf
Best,
Ista
On Thu, M
> -Original Message-
> From: Brian Diggs [mailto:dig...@ohsu.edu]
> Sent: Thursday, May 26, 2011 2:59 PM
> To: William Dunlap
> Cc: Julian TszKin Chan; r-help@r-project.org
> Subject: Re: Question about ggplot2
>
> On 5/26/2011 2:15 PM, William Dunlap wrote:
> > eval(substitute(
> >
>
Cute!
I don't think your proposed strategy is all that complicated, but see
the gsubfn package for a one-liner. In particular, check out
http://code.google.com/p/gsubfn/
where there is an example for your almost exact task.
-- Bert
On Thu, May 26, 2011 at 3:05 PM, Dennis Fisher wrote:
> Collea
On 05/26/2011 01:32 PM, Petter Brodin wrote:
Dear all
Does anybody know of updated Custom CDFs with binaries for MacOS. I'm
trying to do affy-analysis using R-studio for Mac.
Hi Petter -- ask on the Bioconductor mailing list
http://bioconductor.org/help/mailing-list
Generally, CDF files do n
Colleagues
Assume that I have a vector containing some text strings, some of which contain
a particular character. I could like to apply "toupper" to the text before the
character. For example (in this case, "|" is the particular character):
ORIGINAL:
TEXT<- c("", "bbb|cc", "|
On 27/05/11 03:35, Jun Shen wrote:
Dear list,
We have three time course profiles with very different scales, and we want
to show them in one plot. Is it possible to have three y axis? I guess not,
then what would be other options? something like two 2-y axis plots on a
three dimensional view? Ap
On May 26, 2011, at 5:15 PM, Struckmeier, Nathanael wrote:
Hello,
I've been banging my head against my monitor for some time now
trying to
import data into R. The problem is either in my use of syntax, my data
format, or my memory. I have described what I have tried below. Please
help.
===
On 5/26/2011 2:15 PM, William Dunlap wrote:
eval(substitute(
qplot(x=xVariable,y=yVariable,data=diamonds,geom=c('point','smooth')),
list(xVariable=as.name("carat"), yVariable=as.name("price"))
))
Well, yes, I suppose there is that. I've mentally (perhaps
inappropriately) lumped substitu
Hi Everyone,
I am trying to work with a routine that is from the IMSL called BSVLS. I
have the file as a .dll file.
I have been trying to call the routine using the dyn.load function. I am
working on a 32-bit windows 7 OS with 3 GB. For some reason i still keep
getting the error message
LoadLibra
Dear all
Does anybody know of updated Custom CDFs with binaries for MacOS. I'm
trying to do affy-analysis using R-studio for Mac.
Thanks
Petter
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PLEASE do read the pos
Hey together,
thanks a lot for the quick and helpful answers. I will not use median
on a data.frame from now on and use sapply instead.
Thanks Christoph
On Thu, May 26, 2011 at 11:37 PM, Marc Schwartz wrote:
> On May 26, 2011, at 4:22 PM, Christoph Jäckel wrote:
>
>> Hi together,
>>
>> below is
On May 26, 2011, at 4:22 PM, Christoph Jäckel wrote:
> Hi together,
>
> below is a small example which produces outcome I do not understand,
> namely that the median function works fine on a data.frame without
> negative numbers, but doesn't work on a data.frame with one negative
> number. I'm su
Hi Christoph,
Use:
sapply(df, median)
median() does not have methods for a data frame (read: it is never
meant to be used directly on data frames so do not expect sensical
results).
Cheers,
Josh
On Thu, May 26, 2011 at 2:22 PM, Christoph Jäckel
wrote:
> Hi together,
>
> below is a small exam
Christoph,
After a quick look at the code for median, I'm amazed that it gives a
correct result for any data frame.
median() isn't really intended for use with data frames; there's no
data.frame method. The correct and safe approach is to use
sapply(df, median)
This was recently discussed on the
Summary (df) will also work.
On 5/26/11, Christoph Jäckel wrote:
> Hi together,
>
> below is a small example which produces outcome I do not understand,
> namely that the median function works fine on a data.frame without
> negative numbers, but doesn't work on a data.frame with one negative
> nu
Hi together,
below is a small example which produces outcome I do not understand,
namely that the median function works fine on a data.frame without
negative numbers, but doesn't work on a data.frame with one negative
number. I'm sure there is a reasonable explanation for that or better,
that I'm
Hello,
I've been banging my head against my monitor for some time now trying to
import data into R. The problem is either in my use of syntax, my data
format, or my memory. I have described what I have tried below. Please
help.
=My GOAL==
>import a 4 c
I have used lars before. I could not find a tutorial, so finally asked
a professor at my school. He has a wrapper that nicely prints out all
the variables that were selected and is more stable in cross
validation than the original package. See below for the code and
description.
http://www4.stat.n
On May 26, 2011, at 3:18 PM, Albert-Jan Roskam wrote:
> Hi,
>
> I want to recode all Inf and NaN values to NA, but I;m surprised to see the
> result of the following code. Could anybody enlighten me about this?
>
>> df <- data.frame(a=c(NA, NaN, Inf, 1:3))
>> df[is.infinite(df) | is.nan(df)] <
eval(substitute(
qplot(x=xVariable,y=yVariable,data=diamonds,geom=c('point','smooth')),
list(xVariable=as.name("carat"), yVariable=as.name("price"))
))
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-
I'm still getting used to R's scoping. I've run into the following situation
value=0
thefunction <- function() print( value )
somefunction <- function() { value=99; thefunction() }
somefunction()
now, I understand that somefunction() returns 0 because thefunction() was
defined with value=0 in its
On 5/26/2011 12:29 PM, Julian TszKin Chan wrote:
Hi all,
Is there any way for me to to string in the argument of qplot or ggplot? for
example
qplot(x='carat',y='price',data=diamonds,geom=c('point','smooth'))
instead of
qplot(x=carat,y=price,data=diamonds,geom=c('point','smooth'))
I don't know
You can pass a vector of ticks to the axis() function that expand the actual
range,
par(mar=c(5.1,4.1,4.1,5.1))
plot(x$Time, x$y1, type='l', bty = 'c', col = 'red')
par(new = TRUE)
plot(x$Time, x$y2, type = 'l', axes = FALSE, xlab = '', ylab = '', col=
'green')
##axis(4, col='green')
axis(4, at=se
1. This is not about R, and should be taken off list.
2. You are wading in an alligator infested swamp. Get help from
(other) statisticians at Pfizer (there are many good ones there).
Best,
Bert
P.S. The answer to all your questions is "no" (imho).
On Thu, May 26, 2011 at 1:35 PM, El-Tahtawy,
The strong predictor is the country/region where the study was
conducted. So it is not important/useful for a clinician to use it (as
long he/she is in USA or Europe).
Excluding that predictor will make another 2 insignificant predictors to
become significant!! Can the new model have a reliable
Hello-
I am looking for R function that will give me some proper confidence
intervals on un-transformed mean prediction when performing a linear
regression on log-transformed data. I am referring to the UMVU estimate, the
el-shaarawi and viveros (1997) estimate, or the Wu, wong, and Wei (2005)
est
Thanks for your comments and suggestion. I didn’t show all my own function
here because it has many lines. “x” is the results of another function. I am
calling summary because I want to extract some values from the results.
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Thanks a bunch, Walmes.
One more concern, the new Y axes added do not extend all the way down to
cross with x axis. Is there anyway to make them look like the very first Y
axis on the left?
Jun
On Thu, May 26, 2011 at 1:24 PM, Walmes Zeviani wrote:
> You can use mtext()
>
> par(mar=c(5.1,4.1,4.
> df$a[is.infinite(df$a) | is.nan(df$a) ] <- NA
> df
a
1 NA
2 NA
3 NA
4 1
5 2
6 3
On 5/26/11 3:18 PM, "Albert-Jan Roskam" wrote:
>Hi,
>
>I want to recode all Inf and NaN values to NA, but I;m surprised to see
>the
>result of the following code. Could anybody enlighten me about this?
>
>>
I think the IOTT is more a general testing framework rather than a single test
(like maximum likelihood, least squares, bootstrap, etc.) so a single function
won't capture the whole IOTT. There are already many functions available to do
IOTT for many cases (well the user needs to provide the oc
Hi,
I want to recode all Inf and NaN values to NA, but I;m surprised to see the
result of the following code. Could anybody enlighten me about this?
> df <- data.frame(a=c(NA, NaN, Inf, 1:3))
> df[is.infinite(df) | is.nan(df)] <- NA
> df
a
1 NA
2 NaN
3 Inf
4 1
5 2
6 3
>
Thanks!
Please...
?"["
Online tutorial "An Introduction to R."
I think you'll find everything you need in these.
-- Bert
On Thu, May 26, 2011 at 12:39 PM, Dat Mai wrote:
> When I use the as.matrix, the data.frame does turn into a matrix, but I
> cannot change the dimensions of the matrix. I'd still
When I use the as.matrix, the data.frame does turn into a matrix, but I
cannot change the dimensions of the matrix. I'd still want it to have that
pseudo cartesian format (e.g. [a1,b1], [a2,b2])
On Thu, May 26, 2011 at 6:58 PM, David Winsemius wrote:
>
> On May 26, 2011, at 1:53 PM, Andy Zhu wrot
Wierd... sorry about the double posting...
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Hi all,
Is there any way for me to to string in the argument of qplot or ggplot? for
example
qplot(x='carat',y='price',data=diamonds,geom=c('point','smooth'))
instead of
qplot(x=carat,y=price,data=diamonds,geom=c('point','smooth'))
Thanks!!
Regards,
TszKin Julian
[[alternative HTML ve
HI,
I do it like this :
setwd("C:/Users/mpavlic/Desktop/Temp")
library(tm)
tekst <- Corpus(DirSource("."),readerControl = list(language ="ansi"))
where *.txt files are stored in a folder Temp in my desktop,
HTH, m
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-hel
Well, since we have no idea what x is, that is going to be hard to do.
Are you calling summary because you want the info on the last
iteration of a loop? If so, just put the summary call outside the
loop. Otherwise, why are you calling summary if you don't want a
summary?
Also, the posting guide r
On May 26, 2011, at 1:53 PM, Andy Zhu wrote:
Dat:
1. you can use as.matrix to convert data.frame to matrix;
2. it is likely that the internal representation of your data.frame
may not be numerical value; matrix can only take on numeric.
Not true. Can be any single mode, including "charac
Thank you both. These solutions are far more elegant than anything I could
have come up with, and I appreciate the opportunity to learn new commands
within the context of my own data.
I think I've got it working now. :)
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Duncan,
installing MikTeX solved the problem. thank you.
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Dear all,
I have a question about how to suppress intermediate results in a function
on console. For example, I will use summary() in my own function that looks
like:
myfunction <- function(…)
{
…
Summary(x)
…
}
Then myfunction() will print “x” on console that is intermediate result and
do
Dear all,
I am having a (really) hard time getting pyears to work together with a
ratetable to give me the number of expected events (deaths).
I have the following data:
dos, date of surgery, as.Date
dof, date of last follow-up, as.Date
dos, date of surgery, as.Date
sex, gender, as.factor (femal
Dat:
1. you can use as.matrix to convert data.frame to matrix;
2. it is likely that the internal representation of your data.frame may not be
numerical value; matrix can only take on numeric.
--- On Thu, 5/26/11, Dat Mai wrote:
From: Dat Mai
Subject: [R] matrix not working
To: r-help@r-pro
Federico,
I understand what you are after — you want time-series estimates based on the
Thiessen polygon estimates
taken from the station time-series data. My recommendation is that the process
of doing this would be far
easier using something like GRASS GIS, possibly in conjunction with R (s
On May 26, 2011, at 2:18 PM, Jun Shen wrote:
> David,
>
> I surely tried the labels argument. But it seems for tick marks not
> for a text label. Did you see a different outcome? Thanks.
>
`labels` is for labels, `at` is for tick (and label) locations. They
should be the same length. One lab
You can use mtext()
par(mar=c(5.1,4.1,4.1,5.1))
plot(x$Time, x$y1, type='l', bty = 'c', col = 'red')
par(new = TRUE)
plot(x$Time, x$y2, type = 'l', axes = FALSE, xlab = '', ylab = '', col=
'green')
axis(4, col='green')
mtext(side=4, text="label green", line=2)
par(new = TRUE)
plot(x$Time, x$y3, ty
David,
I surely tried the labels argument. But it seems for tick marks not for a
text label. Did you see a different outcome? Thanks.
Jun
On Thu, May 26, 2011 at 12:38 PM, David Winsemius wrote:
>
> On May 26, 2011, at 1:25 PM, Jun Shen wrote:
>
> Jim,
>>
>> One more question, how do I put a l
Federico,
That's an improvement, but a long way from the reproducible example
requested by the posting guide. I and others who might help are more
interested in the way the data and coordinates are organized and a
detailed explanation of what you expect the results to look like, etc,
than in a ver
Hello list,
I am not sure if the terminology that I am using here is widely used,
however, I provide an example in the hopes that my problem will become
clear. My basic problem is that I am unsure of how to 'constrain' my
model estimates to reproduce the aggregate (by factor levels) observed
depen
That is the strange thing... I DON’T get a warning message. R behaves as if
all data is being imported successfully. I don't receive a memory warning,
or warning of any kind. I've scanned my data for potential problems and I do
not have any ('). However...
The .txt version of my file does use (")
That is the strange thing... I DON'T get a warning message. R behaves as
if all data is being imported successfully. I don't receive a memory
warning, or warning of any kind. I've scanned my data for potential
problems and I do not have any ('). However...
The .txt version of my file does use (")
On May 26, 2011, at 1:59 PM, Peter Langfelder wrote:
On Thu, May 26, 2011 at 10:03 AM, StatBat2
wrote:
Hello!
I am trying import data into R and im running into a snag.
GOAL:
Import a 4 column, 8,000 row table into R including headers.
WHAT I'VE ATTEMPTED:
Original data was in Excel format
On May 26, 2011, at 11:19 AM, 1Rnwb wrote:
Hello R gurus, I have a data set from which i have to extract the
gender and
age matched rows from controls and disease group
You need to define what you mean by "age-matched". Your example
creates a very narrow age range which further adds quest
On Thu, May 26, 2011 at 10:03 AM, StatBat2
wrote:
> Hello!
>
> I am trying import data into R and im running into a snag.
>
> GOAL:
> Import a 4 column, 8,000 row table into R including headers.
>
> WHAT I'VE ATTEMPTED:
> Original data was in Excel format.
> Converted data to both a .txt and .csv
Hello,
I have a x,y,z file.
Z is not corresponding to a simple analytical function of x and y.
I am trying to find the minimum value of z by a spline interpolation or from a
polynomial fit.
I tried the akima package but as the location of the point I am looking for
(the minimum) is outside of t
Dear Sarah,
I have a grid in which 8 raingauges are locted, in my case the dataset is
composed by 8 hourly timeseries, one for each raingauge. I would like to
obtain from these timeseries using the Thiessen method the values of the
precipitation in all the grid. In particular I would like to creat
Hello R gurus, I have a data set from which i have to extract the gender and
age matched rows from controls and disease group
disease<-paste(rep(c('y','n'),11))
gender<-paste(rep(c('m','f'),11))
mcp<-rnorm(700,1400)
age<-rnorm(32,34)
dat<-data.frame(disease=disease,sex=gender,Dr_age=age[1:22],MCP=
Hi,
I'm going to try to do some species distribution modeling in R using
Mahalanobis distance. I have presence point locations for the species
and I have ASCII files for each of the WorldClim variables for North
America. My question is how do I bring these into R?
If this is too complex, I have don
Hello!
I am trying import data into R and im running into a snag.
GOAL:
Import a 4 column, 8,000 row table into R including headers.
WHAT I'VE ATTEMPTED:
Original data was in Excel format.
Converted data to both a .txt and .csv (to see which worked better)
Imported data into R via commands as o
On May 26, 2011, at 1:25 PM, Jun Shen wrote:
Jim,
One more question, how do I put a label on the axes I added? Thanks.
I don't
see any argument in axis() for that? Thanks
You don't see the `labels` argument in the help page for `axis`?
Jun
David Winsemius, MD
West Hartford, CT
Jim,
One more question, how do I put a label on the axes I added? Thanks. I don't
see any argument in axis() for that? Thanks
Jun
On Thu, May 26, 2011 at 11:20 AM, jim holtman wrote:
> Try this:
>
> plot(x$Time, x$y1, type='l', bty = 'c', col = 'red')
> par(new = TRUE)
> plot(x$Time, x$y2, typ
I just found out by setting bty='l' to get rid of the border line on the
top.
On Thu, May 26, 2011 at 11:41 AM, Jun Shen wrote:
> Hi, jim
>
> That's exactly what I wanted. One more trivial thing. How do I get rid the
> border line on the top? Thanks again.
>
> Jun
>
>
> On Thu, May 26, 2011 at 1
On May 26, 2011, at 10:04 AM, Mikkel Grum wrote:
> Thanks Marc,
>
> I had just come up with another, slightly more convoluted solution. Add as.is
> = TRUE to the query and then get the timetoken with
> timetoken <- df$timestamp[df$timestamp == max(as.POSIX(df$timestamp))]
>
> While it looks li
Hi, jim
That's exactly what I wanted. One more trivial thing. How do I get rid the
border line on the top? Thanks again.
Jun
On Thu, May 26, 2011 at 11:20 AM, jim holtman wrote:
> Try this:
>
> plot(x$Time, x$y1, type='l', bty = 'c', col = 'red')
> par(new = TRUE)
> plot(x$Time, x$y2, type = '
Hello All,
I'm trying to create a matrix from a dataframe (let's call it df):
..a..b.c.d
a inputs output
b inputs output
c inputs output
d inputs output
e inputs output
The inputs are represented by columns a and b
The outputs are represented by
Try this:
plot(x$Time, x$y1, type='l', bty = 'c', col = 'red')
par(new = TRUE)
plot(x$Time, x$y2, type = 'l', axes = FALSE, xlab = '', ylab = '', col
= 'green')
axis(4, col='green')
par(new = TRUE)
plot(x$Time, x$y3, type = 'l', axes = FALSE, xlab = '', ylab = '', col = 'blue')
axis(4, col='blue',
Hi, Jim,
Thanks for the information. But I am still not clear how to show the 3
separated Y axis. If I just call par(new=TRUE), the three axes are
overlapped.
attached some test data. Thanks.
Jun
=
structure(list(Time = 1:1
On May 26, 2011, at 10:16 AM, Gabor Grothendieck wrote:
> On Thu, May 26, 2011 at 11:09 AM, Marc Schwartz wrote:
>>
>> On May 26, 2011, at 5:09 AM, vioravis wrote:
>>
>>> I am using read.xls command from the gdata package. I get the following
>>> error
>>> when I try to read a work sheet from
There is nothing to prevent you from putting 3 y-axis on your plot;
might be confusing, but it can be done. What have you tried and why
do you say "guess not"? With the use of par(new=TRUE) or by doing
your own scaling, you can use 'axis' to put as many axises as you want
on your graph.
On Thu,
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