You might also want to consider the XLConnect package. I have had
better luck reading/writing Excel files than with xlsReadWrite.
On Fri, Oct 14, 2011 at 5:57 PM, Sarah_R_edu wrote:
>
> /Michael Weylandt/
>
>
> i used read.xls() and i install the packages : xlsReadWrite and getshlib
> before usi
oops *bk not - bk.
Roger Koenker
rkoen...@illinois.edu
On Oct 14, 2011, at 8:40 PM, Roger Koenker wrote:
Ah yes offsets, I've meant to look into this, but never quite
understood
why something like:
rq((y - xk - bk) ~ x1 + x2)
wasn't just as convenient
Roger Koenker
r
Ah yes offsets, I've meant to look into this, but never quite understood
why something like:
rq((y - xk - bk) ~ x1 + x2)
wasn't just as convenient
Roger Koenker
rkoen...@illinois.edu
On Oct 14, 2011, at 6:55 PM, Tal Galili wrote:
Hello all,
I would like to compute a quantile
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 rejectin
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 e
Thanks for the reference, the each function in the plyr package is exactly
what I wanted.
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_
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:
nbin5<-glmmadmb(stainp~beetle.ev+Ca
*/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
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 wi
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
Details:--
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 scr
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
package plyr makes it easier,
plyr::each(function.list)(pi)
HTH,
baptiste
On 15 October 2011 11:55, Richard M. Heiberger 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, 2011
> 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 wrote:
> I am having trouble figuring out how to use do.call to call and run a list
> of
> functions.
>
> for example:
>
>
/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
--
View this message in cont
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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Sent from the R help mailing list archive at Nabble.com.
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]]()
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
aaP
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 compl
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$V1,levels=c('c','d','a','b'))
a.sorted
If I understand what you want to do, it's simple. 2^15 is small (only about
33000), so you can generate all the possible means (sums, actually) and and
find the population quantile for your result. If avals is the vector of 15
absolute values, the complete distribution is:
allsums <- as.matrix(exp
Set the levels of the factor a$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
On 10/14/2011 1:20 PM, Weidong Gu wrote:
On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendly 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 differenc
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, replace=T),V2=1:1
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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Dear All
Big Thanks! R is wonderful language!
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Andrei*
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org
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
On Thu, Oct 13, 2011 at 7:47 PM, malhomidi 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? The reason is
>
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
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 pre
(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
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 wrote:
> Greetings and gratitude,
>
> I have 19 persons in each gro
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(past
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
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 bu
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 wrote:
> i'm actually new at R, but to me its going to be big time, i want to do an
> analysis for the attach
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",3.141592653589
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
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 trie
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, scientif
On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendly 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)
> pair pot
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
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)/
http://r.789695.n4.nabble.com/file/n3905447/47756_429620173219_788668219_5025394_743078_n
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 <- split(x[,1],
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 referen
?tolower
Weidong Gu
On Fri, Oct 14, 2011 at 12:12 PM, 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 lowe
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 work.
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, Sara
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: preferr
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
2008-0
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 ,
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 the
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 wrote:
>
>
> Hello everyone,
>
> I'm trying to change the name variables of a big data
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
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
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 don'
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 ins
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 or
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
variable
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,
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 12.0
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
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 wrote:
> Hi All,
>
> I installed qcc package and the dependency packages. For the first time I can
> use the function
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 pa
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
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Fri, Oct 14, 2011 at 12:06 AM, P
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
w
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
Comments inline:
On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum 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 based on the valu
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 Nd
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. I
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:
wireframe(outer(seq(0,5,length.out=50),seq(2,4,length.out=40),function(x,
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
unders
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,
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 13-Oct-11 20:33, Sarah Goslee wrote:
Hi,
On Thu, Oct 13, 2011 at 2:05 PM, Bailey, Daniel 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 coordinates (b
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 backgroun
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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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 su
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 there
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 accept
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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https://stat.ethz.ch/mai
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)
> glm.sat<-glm(sex~twp+hwp+hcp+hnp,binomi
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
> dif
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 al
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
>
On 13.10.2011 21:46, Ben Bolker wrote:
lincoln 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 is indica
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
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(DailyDia
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 criti
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.fra
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 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
http://CRAN.R-
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 gues
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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P
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
"Countycode","Provincecode","Interviewercode","Locat
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