I don't know the specifics of this package, but generally the C code
is called internally by R: it, however, requires compilation before R
can talk to it. You wont need to learn C though.
Look at the link David suggested for a precompiled version, but it may
be older than the development-version
I have a related question...I have a data frame similar with 74 rows that I
created with header=TRUE, but when I try to coerce one of the data frame
columns into a vector, it shows up as having length 1, even though when I
print it, it shows 74 elements:
VAL - c(DailyDiary[1])
VAL
[1] 3 3 3 2
pre-predict(ozonea,groupA,type=terms,terms=NULL,newdata.guaranteed=FALSE,na.action=na.pass)
yeah!
but I don't know how to only show the value of s(ratio,bs=cr)
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Sent from the R help mailing
I'm using the package 'gdata' and 'drop.levels' in the following code as a
simplified example of what I want to do, pinpointing the issue of isolating
variable in a data frame:
/dfNamesAndHeight - data.frame(Name = c('Bill','Bob','Bo'), Height =
c(73,68,83))
name_Bob -
You haven't provided a reproducible example. You haven't even provided a subset
of your data, or the commands you used to read it in.
I might guess that
VAL - DailyDiary[[1]]
might be what you wanted, and the str function might help you understand why.
Also, the c function does not coerce
predict.gam is returning a matrix with a named column for each term.
Select the appropriate column. Example below
library(mgcv)
n-200;sig - 2
dat - gamSim(1,n=n,scale=sig)
b - gam(y~s(x0)+s(I(x1^2))+s(x2)+offset(x3),data=dat)
newd -
hello again,
i used the command
rbind(Mymatrix, t(as.data.frame(Z)))
and it works!
thanks very much for your replys!
marion
2011/10/11 David Winsemius dwinsem...@comcast.net
On Oct 11, 2011, at 2:47 AM, Marion Wenty wrote:
dear r-users,
i have got a problem which i am trying to
Why aren't you using as.character instead of drop.levels?
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer
i'm actually new at R, but to me its going to be big time, i want to do an
analysis for the attached data, like define variables and then analyse get
means, variances, histogram, and other important attributes including cross
tabs
Regards,
Peter
Please send me the r package : ctest to the following e-mail
mak.sta...@gmail.com
Thank you.
Best regards
MAK
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Yup, DailyDiary[[1]] did it, thanks!
On Thu, Oct 13, 2011 at 11:43 PM, Jeff Newmiller [via R]
ml-node+s789695n3903955...@n4.nabble.com wrote:
You haven't provided a reproducible example. You haven't even provided a
subset of your data, or the commands you used to read it in.
I might guess
On Fri, 14 Oct 2011, Khalek wrote:
Please send me the r package : ctest to the following e-mail
mak.sta...@gmail.com
The package is in the CRAN archives, check out
http://CRAN.R-project.org/package=ctest
Do note however the time stamps. You may also want to read FAQ 5.1.1
On 10/14/2011 06:29 AM, Peter Kaiga wrote:
i'm actually new at R, but to me its going to be big time, i want to do an
analysis for the attached data, like define variables and then analyse get
means, variances, histogram, and other important attributes including cross
tabs
Hi,
Thanks for
On 10/14/2011 05:25 AM, Chris Conner wrote:
Dear Help-Rs,
I'm working with a file that contains large numbers and I need to export them
as is. for example take:
x -
c(27104010002005,27104020001805,27104090001810,90050013000140,90050013000120)
y - c(1:5)
df - data.frame(cbind(x,y))
Dear forum users,
It's 3:35am and I am swamped with statistics homework lol
I'm terrible with R and this time I have no idea what the prof wants. Here
is the question:
Consider the (two-‐sample) Wilcoxon rank statistic T = Σrank(Xi). For
n1=106 and n2=192, determine by simulation the α=.05
Hi
I have a related question...I have a data frame similar with 74 rows
that I
created with header=TRUE, but when I try to coerce one of the data
frame
columns into a vector, it shows up as having length 1, even though when
I
print it, it shows 74 elements:
VAL - c(DailyDiary[1])
On 11-10-13 10:16 PM, R. Michael Weylandt wrote:
source(FILE, print.eval = TRUE)
or
source(FILE, echo = TRUE)
which will also echo the inputs - the students will find this helpful.
Duncan Murdoch
Hope this helps good work on getting the next round of R enthusiasts
up and going!
Michael
On 13.10.2011 21:46, Ben Bolker wrote:
lincolnmiseno77at hotmail.com writes:
Hi all,
I have run a (glm) analysis where the dependent variable is the gender
(family=binomial) and the predictors are percentages.
I get a warning saying fitted probabilities numerically 0 or 1 occurred
that
On Oct 14, 2011, at 09:39 , shl2a wrote:
Dear forum users,
It's 3:35am and I am swamped with statistics homework lol
I'm terrible with R and this time I have no idea what the prof wants. Here
is the question:
Consider the (two-‐sample) Wilcoxon rank statistic T = Σrank(Xi). For
n1=106
On 14.10.2011 09:26, Achim Zeileis wrote:
On Fri, 14 Oct 2011, Khalek wrote:
Please send me the r package : ctest to the following e-mail
mak.sta...@gmail.com
The package is in the CRAN archives, check out
http://CRAN.R-project.org/package=ctest
Do note however the time stamps. You may
Hi Carl,
I have no idea what z or f(z) are, but maybe outer will help you:
wireframe(outer(seq(0,5,length.out=50),seq(2,4,length.out=40),function(x,y)sin(x*y)))
cheers.
Am 13.10.2011 23:37, schrieb Carl Witthoft:
Hi all,
I'd like to plot the Real and Imaginary parts of some f(z) as two
As you suggested I had a further look at the profile by changing default
values of stepsize (I tried to modify the others but apparently there was
any change).
Here they go the scripts I have used:
dati-read.table(simone.txt,header=T,sep=\t,as.is=T)
Dear All
I need to generate multivariate NON-NORMAL data in R, which follows a
given mean vector and covariance matrix, say multivariate exponential
data. How can I do that?
Best regards
mra
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On 13.10.2011 18:28, Matt Shotwell wrote:
Would it be worthwhile to update the read.spss implementation using the
more recent discoveries from the PSPP group?
If there are important features, I think so. Getting code into foreign
is not too trivial. Smaller changes are more likely to be
On 14/10/2011 1:00 a.m., ashz wrote:
Dear All,
I have some time series data where X=month and Y=nutrient concentration (I
can have several concentration data for one month). Is there a way to fit
for it an Harmonic Function. Is there a package, script,etc which I can use?
Thx
Possibly
Hi David:
Thank you for your answer.
El vie, 14-10-2011 a las 00:32 -0400, David Winsemius escribió:
Legends are built in columns. You need to find a graphics symbol to
put in the points column or you need to find something that the
lines paramater will turn into a dot (and I'm not sure
Hi,
Thank you David and Michael for your suggestions and comments. I'll try
to get the binary version of the package. I'm just waiting for Gabor to let
me know about the development package which works on Windows.
Regards,
Mohammed
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On Fri, 2011-10-14 at 02:32 -0700, lincoln wrote:
As you suggested I had a further look at the profile by changing default
values of stepsize (I tried to modify the others but apparently there was
any change).
Have you read ?glm, specifically this bit:
Details:
For the background
On 13-Oct-11 20:33, Sarah Goslee wrote:
Hi,
On Thu, Oct 13, 2011 at 2:05 PM, Bailey, Danielbai...@spu.edu wrote:
Thank you Sarah. I tried your suggestion, and if I coerce it into a normal data.frame, that method works. But if you've
already made the data into a SpatialPixelsDataFrame and run
I'm trying to link a hydrological model in FORTRAN with R.
I have a subroutine inside wetall.f90 which calls two contained functions.
When I try
rcmd SHLIB -o wetall.dll wetall.f90
I get a bunch of errors stating undefined reference to `__mingw_vsprintf' from
dos (see below).
When the same is run
On Oct 14, 2011, at 5:11 AM, Petr PIKAL wrote:
Hi
I have a related question...I have a data frame similar with 74 rows
that I
created with header=TRUE, but when I try to coerce one of the data
frame
columns into a vector, it shows up as having length 1, even though
when
I
print it,
Dear all,
I need to run a quantile regression to estimate the coefficients of the
following model: Q_{Y}(Ï|X)=exp(βâ(Ï)+Xâ²Î²â(Ï)).
Since the model is nonlinear, I need to use nlrq(.). However, if I try
nlrq(Y~exp(X), tau=Ï), the software does not accept and also does not
No, the point of my question is how to plot the Re and Im parts as two
separate surfaces in one chart.
On 10/14/11 5:46 AM, Eik Vettorazzi wrote:
Hi Carl,
I have no idea what z or f(z) are, but maybe outer will help you:
I have a list of dataframes i.e. each list element is a dataframe with three
columns and differing number of rows. The third column takes on only two
values. I wish to split the list into two sublists based on the value of the
third column of the list element.
Second issue with lists as well.
It would be nice if you could provide a sample. However, if the data
in the list have the same colnames, you can combine them by
df-do.call('rbind',your_list_data_frame)
Then you can do what you want on the dataframe instead of a list
HTH
Weidong Gu
On Fri, Oct 14, 2011 at 9:06 AM, Juliet
Comments inline:
On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum jpnts...@yahoo.com wrote:
I have a list of dataframes i.e. each list element is a dataframe with three
columns and differing number of rows. The third column takes on only two
values. I wish to split the list into two sublists
On Oct 14, 2011, at 9:26 AM, Weidong Gu wrote:
It would be nice if you could provide a sample.
That is certainly true.
However, if the data
in the list have the same colnames, you can combine them by
df-do.call('rbind',your_list_data_frame)
Then you can do what you want on the dataframe
See Duncan Murdoch's post here for some pointers as well as an
explanation of why this isn't a totally well-formed question:
http://r.789695.n4.nabble.com/generate-two-sets-of-random-numbers-that-are-correlated-td3736161.html
(Specifically, the post at 1:33, but it may be worthwhile to read the
Yes, it is a little bit amazing to me too, so let's cc to the package
maintainer to see if he can do anything.
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Fri, Oct 14,
Hi All,
I installed qcc package and the dependency packages. For the first time I can
use the function : process.capability.sixpack(). But later when ran the code
again I always got the following error: Error: could not find function
process.capability.sixpack. I tried reinstalling the qcc
Did you reload the package for each new session? Sounds like you are
probably missing a line like library(qcc) or require(qcc)
Michael
On Fri, Oct 14, 2011 at 11:10 AM, Li, Yan yan...@ibi.com wrote:
Hi All,
I installed qcc package and the dependency packages. For the first time I can
use
Apologies for my brief reply -- I didn't take the time to examine the
package closely.
On a fresh install, I don't see process.capability.sixpack() as an
available function and process.capability() isn't a generic so it's
not that sort of thing. Looking at the package index, there's no
indication
Hi all
Consider the classic data below from Darwin on the heights of 15 pairs
of zea mays (corn) plants
either cross-fertilized or self-fertilized, where the goal is to see if
it makes a difference.
head(ZeaMays)
pair pot cross self diff
11 1 23.500 17.375 6.125
22 1
A few things in play here:
1) I'm guessing you are new to R, so I'd advise you to take some time
to read some introductory materials at this point. If you type
help.start() into your R session, a good introductory manual will be
available.
2) You don't need any of this textConnection business,
I would like to build a forest of regression trees to see how well some
covariates predict a response variable and to examine the importance of the
covariates. I have a small number of covariates (8) and large number of
records (27368). The response and all of the covariates are continuous
Hello everyone,
I'm trying to change the name variables of a big dataset. Here's more than 300
variables. The point is that I have to match it with another dataset that have
same variables, but in lowercase, the I can use rbind and do my work.
Is there any function for changing uppercase
You'll note that the package version for which you are reading the
manual is not the version that you downloaded from CRAN.
As I suspected, this functionality seems to have existed in an older
version of the package, but seems to have since been deprecated (cf.
the current CRAN files). You can
Hello,
I'm looking for a method to estimate narrow sense heritability of traits in
a RIL population. Papers I've checked either use either SAS or SPSS or do
not give any details at all. I've found some reference to using variance
components in ANOVA, using the kinship or wgaim packages, but I
Hi all,
I was unable to find a solution to my problem in the archives, but this might
be due to a lack knowledge on the correct terminology on my part. Please
forgive me if this has been explained before and please forgive me my probably
clumsy way of explaining things.
This is what I want to
hi all
I have R object look like this:
spl
$SB012XSB044
DPW Cross
1 66.6 SB012XSB044
2 96.5 SB012XSB044
3 78.8 SB012XSB044
4 68.6 SB012XSB044
5 62.0 SB012XSB044
6 72.1 SB044XSB012
7 72.2 SB044XSB012
8 69.6 SB044XSB012
9 87.9 SB044XSB012
10 84.4 SB044XSB012
11 51.9 SB044XSB012
What sort of variables are these? generally toupper() or tolower()
would do the trick, but it can be little trickier if things are stored
as factors.
Michael
On Fri, Oct 14, 2011 at 12:12 PM, Jose Bustos Melo jbustosm...@yahoo.es wrote:
Hello everyone,
I'm trying to change the name
Thanks for your help...actually there is monotonicity in beta so minimizing
the square of the functional constraint works. I verified it with a brute
force search (while loop).
For the sake of knowledge this is what someone else suggested (but didn't
work in my case)
Since x is fixed (given
I am tried
z - read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls, colNames=FALSE,
rowNames=FALSE)
and the following massege is appeared:
Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, :
Incorrect number of arguments (11), expecting 10 for 'ReadXls'
My dear ,
Dear R users,
I got date and time as two separate characters
[1] 2008-04-11
[1] 22:00:00
which correspond to my starting point in time domain.
Now I need to produce series over, 5 sec epochs up to end point, say:
[1] 2008-04-12
[1] 23:00:00
So something like
2008-04-11 22:00:00
2008-04-11
Greetings and gratitude,
I have 19 persons in each group, and each person walks at 3 different speeds.
I can do a nice p value and f value with summary command
Please help me learn how to report the 95% confidence interval for this anova?
This is easier for my fourth, separate condition:
I can't remember the specifics right now, but when you load the
package used to provide read.xls() you probably get a long warning
message saying you need to run
xls.getshlib()
before using the package. If I remember right, this solves your problem.
Michael
On Fri, Oct 14, 2011 at 11:54 AM,
On 14.10.2011 18:12, Jose Bustos Melo wrote:
Hello everyone,
I'm trying to change the name variables of a big dataset. Here's more than 300
variables. The point is that I have to match it with another dataset that have
same variables, but in lowercase, the I can use rbind and do my
?tolower
Weidong Gu
On Fri, Oct 14, 2011 at 12:12 PM, Jose Bustos Melo jbustosm...@yahoo.es wrote:
Hello everyone,
I'm trying to change the name variables of a big dataset. Here's more than
300 variables. The point is that I have to match it with another dataset that
have same
On Oct 14, 2011, at 9:49 AM, Moohbear wrote:
Hello,
I'm looking for a method to estimate narrow sense heritability of
traits in
a RIL population.
I admit to not knowing that TLA.
Papers I've checked either use either SAS or SPSS or do
not give any details at all. I've found some
It's usually standard to provide an example of what code you've tried
and also to put your data in a form that can be more easily
cut-and-pasted into R.
That said, would something like this work if you know you only have
two sorts of cross in each level?
lapply(spl, function(x) {x -
eyildiz
...
if i installed the package (getshlib) appeared to me to selecting a CRAN
mirror ... Any one to be selected ?
Note: i am beginner to used the languge R
/thanks eyildiz ( my prayers to you)/
On Oct 14, 2011, at 11:54 AM, Sarah_R_edu wrote:
I am tried
z - read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls,
colNames=FALSE,
rowNames=FALSE)
The read.xls that I have (from pkg:gdata) does not have arguments
named colNames or rowNames, and those do not look correct to be passed
On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendly frien...@yorku.ca wrote:
Hi all
Consider the classic data below from Darwin on the heights of 15 pairs of
zea mays (corn) plants
either cross-fertilized or self-fertilized, where the goal is to see if it
makes a difference.
head(ZeaMays)
Paul,
Many thanks for your help! I tried the suggestions below and, unfortunately,
they didn't work for me. I actually tried a tweak even:
Here is what I tried (might I be missing something?):
df$x -format(df$x, scientific = FALSE)
write.csv(df ,file=df.csv)
df$x -format(df$x, scientific
Thanks for the sos function, I didn't know about it. Unfortunately, almost
all the entries listed are not relevant to my problem.
qgen seems to be doing what I want, but gives error message when I type
library(qgen): Error: package 'qgen' was built before R 2.10.0: please
re-install it. I've tried
This is somewhat faster on my machine:
t(sapply(seq(var1), function(i) my.list[[var1[i]]] [[var2[i]]] [var3[i],
]))
Jean
Graaf, G de wrote on 10/14/2011 09:23:24 AM:
Hi all,
I was unable to find a solution to my problem in the archives, but
this might be due to a lack knowledge on the
Have you tried setting
options(scipen=500) # big number of digits
? E.g.,
df - data.frame(x=pi*10^seq(-30,30,by=10), d=seq(-30,30,by=10),
s=state.name[31:37])
getOption(scipen)
[1] 0
write.csv(df, stdout())
,x,d,s
1,3.14159265358979e-30,-30,New Mexico
2,3.14159265358979e-20,-20,New York
Here are some tutorials to R:
http://www.cyclismo.org/tutorial/R/
http://www.statmethods.net/
Or just search on the web.
Good luck,
A.
On Fri, 14 Oct 2011 09:29:52 +0300
Peter Kaiga kaigape...@gmail.com wrote:
i'm actually new at R, but to me its going to be big time, i want to do an
On Oct 14, 2011, at 1:44 PM, Moohbear wrote:
Thanks for the sos function, I didn't know about it. Unfortunately,
almost
all the entries listed are not relevant to my problem.
qgen seems to be doing what I want, but gives error message when I
type
library(qgen): Error: package 'qgen' was
Michael Friendly wrote on 10/14/2011 10:38:44 AM:
Hi all
Consider the classic data below from Darwin on the heights of 15 pairs
of zea mays (corn) plants
either cross-fertilized or self-fertilized, where the goal is to see if
it makes a difference.
head(ZeaMays)
pair pot cross
Good afternoon Robert,
Suppose you have your date and time in characters like this:
d.start = 2008-04-11
t.start = 22:00:00
d.end = 2008-04-12
t.end = 15:00:00
then use POSIXct to convert them to a unified time object:
start - as.POSIXct(paste(d.start, t.start))
end - as.POSIXct(paste(d.end,
Hey,
If I understand correctly, library(gplots) plotmeans(). You might also
try TukeyHSD() to see if that gets you where you are trying to go.
Good luck!
Ken Hutchison
On Fri, Oct 14, 2011 at 12:11 PM, Jebb Remelius j...@kin.umass.edu wrote:
Greetings and gratitude,
I have 19
(Sorry if I'm repeating things: working blind b/c of the
nabble-listserv interface)
Since you haven't actually told us what package you are using, I'm
guessing that your problem seems to be the same as the one discussed
here: http://r.789695.n4.nabble.com/ReadWrite-xls-problem-td3078348.html
If
Hi,
That's a tough one, I'll do my best and hope a more knowledgeable person
will correct me.
Since you can measure conditional importance by permuting predictors and
re-evaluating importance, perhaps try the randomForest package and examine
how your results change based on permutation of each
Hi:
Following the lead of others, here's a reproducible example that I
believe achieves what you want.
# Q1:
L - lapply(1:3, function(n)
data.frame(x = rnorm(6), y = rnorm(6), g = rep(1:2, each = 3)))
# Using David's suggestion:
L1 - lapply(L, function(d) subset(d, g == 1L))
L2 - lapply(L,
On Thu, Oct 13, 2011 at 7:47 PM, malhomidi mal...@essex.ac.uk wrote:
Hi again,
I've looked at the links above and I see the development version of
the igraph library. I see the src folder implemented in C. Are these source
codes available in R or I just would have to use the C code?
Thanks a lot for your immediate help and detailed explanation!
About one thing I'm not quite clear:
When the default fit = glm in gefp() is used:
sctest(gefp(Employed ~ Year + GNP.deflator + GNP + Armed.Forces, data =
longley, fit = lm), functional = meanL2BB)
is this then the original Nyblom's
Dear All
Big Thanks! R is wonderful language!
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Andrei*
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PLEASE do read the posting guide
You guys are working too hard.
Rgames y - c(0,1,1,3,3,3,5,5,6)
Rgames rle(sort(y))
Run Length Encoding
lengths: int [1:5] 1 2 3 2 1
values : num [1:5] 0 1 3 5 6
--
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Sent from my Cray XK6
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This has been dogging me for a while. I've started making a lot of tables
via xtable so the way I want to sort things is not always in alphabetical or
numerical order.
As an example, consider I have a dataframe as follows
set.seed(100)
a - data.frame(V1=sample(letters[1:4],100,
On 10/14/2011 1:20 PM, Weidong Gu wrote:
On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendlyfrien...@yorku.ca wrote:
Hi all
Consider the classic data below from Darwin on the heights of 15 pairs of
zea mays (corn) plants
either cross-fertilized or self-fertilized, where the goal is to see if it
Set the levels of the factor a$V1 to the order
in which you want them to be sorted. E.g.,
a - data.frame(V1=letters[rep(4:1,2)], V2=1001:1008)
a[do.call(order,a[c('V1','V2')]),]
V1 V2
4 a 1004
8 a 1008
3 b 1003
7 b 1007
2 c 1002
6 c 1006
1 d 1001
5 d 1005
If I understand what you want to do, it's simple. 2^15 is small (only about
33000), so you can generate all the possible means (sums, actually) and and
find the population quantile for your result. If avals is the vector of 15
absolute values, the complete distribution is:
allsums -
Works great. I did a couple changes so as to not affect the original
data.frame (and had to add levels back b/c I removed them in the original
read.csv).
a - data.frame(V1=letters[rep(4:1,2)], V2=1001:1008)
b - a
levels(b) - unique(a$V1)
b$V1 - factor(b$V1,levels=c('c','d','a','b'))
a.sorted-
On 10/14/2011 4:10 PM, Bert Gunter wrote:
If I understand what you want to do, it's simple. 2^15 is small (only
about 33000), so you can generate all the possible means (sums,
actually) and and find the population quantile for your result. If
avals is the vector of 15 absolute values, the
MATLAB chunk of code
for jj = 1:numOfAA curAAPosInSeqIndex = aaPosInSeqIndex{1,jj};
for kk = 1:K p_c_a_z_given_x_contribution(1,jj,kk) =
sum(sum(p_c_a_z_given_m_x_permuted(:,:,kk) .* curAAPosInSeqIndex, 1),2); % 1 by
numOfAA by K end end
I am having trouble figuring out how to use do.call to call and run a list of
functions.
for example:
make.draw = function(i){i;function()runif(i)}
function.list = list()
for (i in 1:3) function.list[[i]] = make.draw(i)
will result in
function.list[[1]]()
[1] 0.2996515
function.list[[2]]()
Hello
what is the easiest way to generate rpois(m,lambda) but only values greater
than 0 and length = m.
tanks, knut
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/Michael Weylandt/
i used read.xls() and i install the packages : xlsReadWrite and getshlib
before using this command but, did not work.
anyway i waiting for you.
thank you so much. (Flowers)
-
We are all like the bright moon, we still have our darker side
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function.list=c(sin, cos, function(x) tan(x))
for (f in function.list) print(f(pi))
[1] 1.224606e-16
[1] -1
[1] -1.224606e-16
On Fri, Oct 14, 2011 at 5:48 PM, honeyoak honey...@gmail.com wrote:
I am having trouble figuring out how to use do.call to call and run a list
of
functions.
for
package plyr makes it easier,
plyr::each(function.list)(pi)
HTH,
baptiste
On 15 October 2011 11:55, Richard M. Heiberger r...@temple.edu wrote:
function.list=c(sin, cos, function(x) tan(x))
for (f in function.list) print(f(pi))
[1] 1.224606e-16
[1] -1
[1] -1.224606e-16
On Fri, Oct 14,
This is the third time you've said you've done getshlib, but as has
been pointed out to you, the command is:
xls.getshlib()
not just typing getshlib and it must be entered verbatim. The latter
(getshlib) doesn't exist that I'm aware of. Either way, could you do
xls.getshlib() again and provide
On 15/10/11 10:15, knut-o wrote:
Hello
what is the easiest way to generate rpois(m,lambda) but only values greater
than 0 and length = m.
tanks, knut
The rpospois() function from the VGAM package is what you are looking for.
I found this by doing:
RSiteSearch(truncated Poisson)
and
Hello all,
I would like to compute a quantile regression using rq (from the
quantreg package), while keeping one of the coefficients fixed.
Is it possible to set an offset for rq in quantreg? (I wasn't able to
make it to work)
Thanks,
Tal
Contact
Hello,
While exploring if rgl is along the lines of what I need, I checked out
demo(rgl) and didn't find quite what I'm looking for, and am therefore
seeking additional help/suggestions.
The application is geared towards displaying a 3D rendering of a contaminant
plume in the subsurface
*/David/*
i used the following command :
z -
read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls,colNames=FALSE,rowNames=FALSE)
z - read.table(file=C:\\Users\\user\\Desktop\\LTS.xls)
and i have the packages : xlsReadWrite and gdata , my R version is 2.13.2
(2011-09-30)
but all these did not got
Hello everyone,
I am using the alpha version of glmmadmb, and it works for most of the time
except for one of my models. The weird thing is that it has worked before, a
couple of months ago, and for some reason it won't now and nothing has
changed.
The code is:
Thanks for the reference, the each function in the plyr package is exactly
what I wanted.
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On Oct 14, 2011, at 7:37 PM, Sarah_R_edu wrote:
*/David/*
i used the following command :
z -
read.xls(file=C:\\Users\\user\\Desktop\
\LTS.xls,colNames=FALSE,rowNames=FALSE)
As I pointed out earlier this would have produced an error on my
system because the arguemtnts do not exist in
On Oct 14, 2011, at 8:14 PM, honeyoak wrote:
Thanks for the reference, the each function in the plyr package is
exactly
what I wanted.
I doubt it. You have not looked at the `each` code. Its just a wrapper
for a for loop and on StackOverfolw you started out saying that you
were
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