Hello helpers,
I get some curious results, when I apply the MBA algorithm to my test set of
scattered points.
I'm using R v2.11.1 on Win XP with the package MBA v0.0-7 installed.
Here is my code:
rm(list=ls(all=TRUE))
library(MBA)
library(fields)
#regular 3x3 grid with all values zero, except
Dear All,
Can you please suggest me a correct way to load the following R data frame
from the internet and save it to the hard drive?
The following is what I tried:
raceprofiling -
read.table(http://rss.acs.unt.edu/Rdoc/library/twang/data/raceprofiling.RData;)
save(raceprofiling, file =
Dear list,
If have some repeated measurement data which looks something like:
time - rep(1:5 , each=2*4)
groups - rep(c(Case, Control), each=4)
subjects - factor(rep(1:(2*4), 5))
responses - time + rnorm(5*2*4) + as.integer(factor(groups))
data - data.frame(responses, time, groups, subjects)
Hi!
As Ben Bolker told you already, the levels are alphabetically ordered by
default.
When you print rfactor, the last line shows you the different levels, in
the saved order.
rfactor
[1] c c c d b a b d d a a e e b b e c e e a a b b b a b a e a a b d b b
c a b b
[39] d c a e c d e d a a a
Try with load() instead of read.table()
read.table() is for csv, txt... files.
load() is for R files
Ivan
Le 12/6/2010 09:35, zhiji19 a écrit :
Dear All,
Can you please suggest me a correct way to load the following R data frame
from the internet and save it to the hard drive?
The following
I tried the load() already. It does not work. The erro is shown as following:
Error in readChar(con, 5L, useBytes = TRUE) : cannot open the connection
In addition: Warning message:
In readChar(con, 5L, useBytes = TRUE) :
cannot open compressed file
Hi
r-help-boun...@r-project.org napsal dne 05.12.2010 00:19:17:
year1 is the time series data set below
11.64
11.50
11.49
11.16
11.15
11.37
11.37
11.57
11.83
11.87
11.85
11.92
11.77
11.71
11.57
11.24
11.27
11.33
11.17
.
With a total of 1304 rows of data.
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 12/06/2010 10:29 AM, zhiji19 wrote:
I tried the load() already. It does not work. The erro is shown as following:
load() works - I downloaded the data and then used load() - it works.
I don't know if load() accepts an URL as a source, because
On Mon, Dec 6, 2010 at 8:29 PM, zhiji19 zhij...@gmail.com wrote:
I tried the load() already. It does not work. The erro is shown as following:
Error in readChar(con, 5L, useBytes = TRUE) : cannot open the connection
In addition: Warning message:
In readChar(con, 5L, useBytes = TRUE) :
Hi
Im trying to capture the output of HTML but dont really get what I want
As an example:
y-data.frame(a=c(1,2,3),b=c(1,2,3),c=c(1,2,3))
toReturn-capture.output(HTML(y,file=));
This gives me the output of toReturn as an vector:
toReturn
[1]
You can create a connection to the URL and load() the connection.
Like this:
con -
url('http://rss.acs.unt.edu/Rdoc/library/twang/data/raceprofiling.RData')
load(file=con)
close(con)
str(raceprofiling)
But the easiest way is to save the file and then load it.
Ivan
Le 12/6/2010 10:44, Rainer
Why to you always find the answer as soon as you have posted the Q here :P
paste(toReturn, collapse=)
But if anyone have a better solution plz share.
//Joel
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Sent from the R help
Dear R-Helpers:
I am using trying to use *ddply* to extract min and max of a particular
column in a data.frame. I am using two different forms of the function:
## var_name_to_split is a string -- something like var1 which is the name
of a column in data.frame
ddply( df,
Please help! Im trying to run a GAM:
model3=gam(data2$Symptoms~as.factor(data2$txerad)+s(data2$maritalStatus),family=binomial,data=data2)
But keep getting this error:
Error in dl[[i]] : subscript out of bounds
Can someone please tell me what this error is?
Thanks
--
View this message in
Hi,
I am trying to calculate correlation coefficient for gene expression data.
Tab delimited file looks like this
Id v1 v2v3
df 56 9045
gh 87 9878
ty 897867
I used this code
[code]
gse20437 - read.csv(C:/Users//Desktop/data/GSE20437_matrix.txt,header =
TRUE, sep =
Hi
r-help-boun...@r-project.org napsal dne 03.12.2010 16:56:37:
I solved the 1° problem with this command:
matrix2plot2[,sensor_data]-as.numeric(as.character((matrix2plot2[,sensor_data])))
In fact before the previous command I if write:
str(matrix2plot2[,sensor_data])
I get:
Dear Maarten
xyplot has the argument groups which allows you to create nested
groupings.
data$subjectID - paste(data$groups, data$subjects) # create a character
label
xyplot(responses~time|subjectID, groups = groups, data = data,
aspect=xy)
# or change to a factor to define the plot order
Hmm... still not quite what I was hoping for... but thanks anyway.
I would like to have the output in the following form:
f(1)
$result
[1] 0
$warning
[1] NULL # or
$error
[1] NULL # or
f(-1)
$result
[1] NaN
$warning
[1] Warning in log(-1) : NaNs produced # or something similar
$error
This is to announce that we plan to release R version 2.12.1 on Thursday,
December 16, 2010.
Those directly involved should review the generic schedule at
http://developer.r-project.org/release-checklist.html
The source tarballs will be made available daily (barring build
troubles) via
On Mon, Dec 6, 2010 at 11:02 AM, chintan85 chintanpatha...@yahoo.com wrote:
Hi,
I am trying to calculate correlation coefficient for gene expression data.
Tab delimited file looks like this
Id v1 v2 v3
df 56 90 45
gh 87 98 78
ty 89 78 67
I used this code
[code]
Hi
r-help-boun...@r-project.org napsal dne 06.12.2010 11:02:03:
Hi,
I am trying to calculate correlation coefficient for gene expression
data.
Tab delimited file looks like this
Id v1 v2v3
df 56 9045
gh 87 9878
ty 897867
I used this code
[code]
On Dec 6, 2010, at 11:28 , Petr PIKAL wrote:
Hi
r-help-boun...@r-project.org napsal dne 06.12.2010 11:02:03:
Hi,
I am trying to calculate correlation coefficient for gene expression
data.
Tab delimited file looks like this
Id v1 v2v3
df 56 9045
gh 87 9878
On 2010-12-06 02:02, chintan85 wrote:
Hi,
I am trying to calculate correlation coefficient for gene expression data.
Tab delimited file looks like this
Id v1 v2v3
df 56 9045
gh 87 9878
ty 897867
I used this code
[code]
gse20437-
On Monday 06 December 2010 09:33, Stressed1985 wrote:
Please help! Im trying to run a GAM:
model3=gam(data2$Symptoms~as.factor(data2$txerad)+s(data2$maritalStatus),fa
mily=binomial,data=data2)
But keep getting this error:
Error in dl[[i]] : subscript out of bounds
Can someone please tell
Try excluding the first column.
cor(gse20437[, 2:4])
chintan85 wrote:
Tab delimited file looks like this
Id v1 v2v3
df 56 9045
gh 87 9878
ty 897867
I used this code
[code]
gse20437 - read.csv(C:/Users//Desktop/data/GSE20437_matrix.txt,header =
On 2010-12-06 01:58, Sunny Srivastava wrote:
Dear R-Helpers:
I am using trying to use *ddply* to extract min and max of a particular
column in a data.frame. I am using two different forms of the function:
## var_name_to_split is a string -- something like var1 which is the name
of a column in
Here is another approach to try:
require(data.table)
var - g10
df - data.table(df)
str(df)
Classes ‘data.table’ and 'data.frame': 6 obs. of 5 variables:
$ g10: int 1 1 1 10 10 10
$ l1 : num 0.41 0.607 0.64 -1.478 -1.482 ...
$ d1 : num 0.918 0.959 0.773 0.474 0.591 ...
$ l13: num
Hi,
I typically install new versions of R on windows using the downloadable
executable file rather than the full tar.
I need to now document the success of the installation in addition to my
preferred procedure of running an old dataset against the new build.
I found quickly that this is
On Mon, Dec 6, 2010 at 11:30 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
If you have a *tab*-delimited file, then why are your using
read.csv??
Try this:
1. read your data with read.table() or read.delim()
If you're very new to R, try Rcmdr. Data Import Text file.
Liviu
Thank you for your reply!
It says:
This is mgcv 1.5-5 . For overview type `help(mgcv-package)'.
Its version: R 2.9.2
And summary(data2) gives:
id age bmi siblingsgender
Min. : 1.0 Min. :32.19 Min. :16.41 Min. :0.000
I have a question regarding understanding the output from a nested linnear
mixed effect model.
My model looks like the following
lme(Poaceae~Site+Ditch_block+Tree_cut+as.factor(Dist_start)+(as.factor(Dist_start)*Tree_cut)+((as.factor(Dist_ditch))/Ditch_block),random=~1|Mainplot/Transect/Obsplot)
I have a question regarding understanding the output from a nested linnear
mixed effect model.
My model looks like the following
lme(Poaceae~Site+Ditch_block+Tree_cut+as.factor(Dist_start)+(as.factor(Dist_start)*Tree_cut)+((as.factor(Dist_ditch))/Ditch_block),random=~1|Mainplot/Transect/Obsplot)
On 05/12/2010 10:13 PM, pythonomics wrote:
So I am working on an economic model and I need to change the parameters of
the runif statement as time goes on.
Ex.
X-runif(1:50,0,5)
X-runif(51:100,100,150)
X-runif(100:T, 1,2)
Not sure how to go about entering this in to R properly.
It's not
Dear all,
I am using the as.xts function to transfer a data frame to the xts
The following is the code and result:
a-read.csv(price.csv)
a$Date-as.POSIXct(a$Date)
str(a)
'data.frame': 15637 obs. of 2 variables:
$ Date : POSIXct, format: 2010-01-04 09:45:01 2010-01-04 09:45:02
Probably the problem is with trying to smooth maritalStatus, which is a factor
variable. Smooths are generally functions of metric variables (i.e. variables
that take numerical values, where the ordering is meaningful). You can have
random effect smooths and markov random field smooths in more
On Mon, Dec 6, 2010 at 6:52 AM, Ted Zeng (曾振兴) zengzhenx...@gmail.com wrote:
Dear all,
I am using the as.xts function to transfer a data frame to the xts
The following is the code and result:
a-read.csv(price.csv)
a$Date-as.POSIXct(a$Date)
str(a)
'data.frame': 15637 obs. of 2
Ok, i just tried this:
library(mgcv)
model3=gam(data2$Symptoms~as.factor(data2$txerad)+as.factor(data2$maritalStatus),family=binomial,data=data2)
And im still getting this error!
Error in dl[[i]] : subscript out of bounds
--
View this message in context:
OK, try...
model3=gam(Symptoms~as.factor(txerad)+as.factor(maritalStatus),family=binomial,data=data2)
or
model3=gam(data2$Symptoms~as.factor(data2$txerad)+as.factor(data2$maritalSt
atus),family=binomial)
... both types of construction work with the current mgcv:gam (and with glm).
Your
Ye
Thank you so much, finally i have something to look at. Just got to
interpret the bloomin' thing now!
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Sent from the R help mailing list archive at Nabble.com.
Hi
I get this error with Rserve:
eval failedeval failed, request status: R parser: syntax erroreval failed,
request status: R parser: input
incompleteorg.rosuda.REngine.Rserve.RserveException: eval failed
at org.rosuda.REngine.Rserve.RConnection.eval(RConnection.java:233)
at
I am also finding the link between TINN - R (2.3.7.0) and R (2.12.0 2010
- 10 - 15) to be problematic.
source(.trPaths[5], echo=TRUE, max.deparse.length=150)
Error in source(.trPaths[5], echo = TRUE, max.deparse.length = 150) :
object '.trPaths' not found
Steve Friedman Ph. D.
Ecologist
Hi!
I'm a new programmer in R
Last week i wrote many function in C and i used R CMD SHLIB to make dll.
Since friday i have this error:
cygwin warning:
MS-DOS style path detected: C:\PROGRA~1\R\R-211~1.1/bin/SHLIB.sh
Preferred POSIX equivalent is:
Hi,
I usually use optimize function for ML Estimation. Now I´ve got a data frame
with many sets, but I can´t save estimates each time I run the code for each
data set (I´m using a for loop with my loglikelihood function and works ok but
when I apply another for loop to:
Thanks Chris Campbell,
I didn't though about that.
Cheers,
Maarten
On Mon, 2010-12-06 at 10:08 +, Chris Campbell wrote:
data$subjectID - paste(data$groups, data$subjects) # create a
character
label
xyplot(responses~time|subjectID, groups = groups, data = data,
aspect=xy)
--
Correct me if I'm wrong, but isn't the minimal x value in your example the
same regardless of what positive coefficient you apply to x? If that is
the case, you would expect the same min(x) for each iteration.
i.e. in the interval [0,1] the minimum x value of x^2 + x is the same as
x^2 +
On Dec 6, 2010, at 15:15 , Jonathan P Daily wrote:
Correct me if I'm wrong, but isn't the minimal x value in your example the
same regardless of what positive coefficient you apply to x? If that is
the case, you would expect the same min(x) for each iteration.
i.e. in the interval [0,1]
I am using the function apply.weekly() from the xts package and I
realized that for the program the week starts Tuesday.
So for instance, with a series like this:
Close
2006-07-03 3.00
2006-07-04 36738.00
2006-07-05 36207.00
2006-07-06 36534.00
2006-07-07 36434.00
2006-07-10
I have the following problem.
After having created an xts importing data from a csv with:
data - read.csv(mib.csv, header = TRUE, dec = ., sep=\t)
dates - as.POSIXct(strptime(data[,1],format=%m/%d/%Y))
mib - xts(data[,c(2:6)],order.by=dates))
I work out weekly log returns:
p -
I suppose I should have been more clear. I saw that her interval did not
include the actual minimum, but I was asking if (and if, why) she was
expecting the minimum x value to be different for each run. If the y value
were returned the same on each run that would be puzzling.
As for the
Hi,
I use Oracle sql developer to connect to a server. I would like to connect
to the server directly from R by using odbcConnect(DB name) function. I am
trying to make first the connection below Control Pannel-Datasource (ODBC).
Which software should I select here? Which one is suitable for
Dear All,
I am having trouble getting my data into R as i need it! I am used to using
read.delim() to open .txt files to do work on. The function i am using
requires a matrix like the one below.
My data is from excel and then saved as a txt file. I have tried the usual
read.delim() approach
Hi List,
I'm using npRmpi to run some density equality tests and place the
output into a matrix. I've put together some crude functions for the
purpose, but I'm receiving the following error when npdeneqtest()
reached the bootstrap;
FATAL ERROR: Memory allocation failure (type DBL_VECTOR).
I'd like to make a less than full rank design using dummy variables
for factors. Here is some example data:
when - data.frame(time = c(afternoon, night, afternoon,
morning, morning, morning,
morning, afternoon, afternoon),
I suspect that R is reading the data just fine, but is not displaying it as you
want to see it. Read the R Windows FAQ 3.3.
Dan Winetsky dwine...@gmail.com wrote:
To whom it may concern,
I have a database (an SPSS .sav file) with some vectors containing
strings
of words in Cyrillic. My
Hello list reader,
I am trying to form some constraints for an optimization I am working on.
I think I have understand the use of the constraints in matrix form. I use
them like:
constr_mat- -diag(2)
constr_vec- rep(-0.05,2)
constr_mat- rbind(constr_mat, diag(2))
constr_vec- c(constr_vec,
When I try to read data from excel, I save it in csv-format and read the files
via:
fileData -read.csv(fkFileName, header = fkHasHeader, stringsAsFactors=FALSE,
sep=fkSep)
where fkSep is depending on your localization of excel (I think).
For some columns that don’t get the correct data format
On Mon, Dec 6, 2010 at 5:57 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Mon, Dec 6, 2010 at 6:52 AM, Ted Zeng (曾振兴) zengzhenx...@gmail.com wrote:
Dear all,
I am using the as.xts function to transfer a data frame to the xts
The following is the code and result:
Hi,
I have a least square fitting problem with linear inequality
constraints. pcls seems capable of solving it so I tried it,
unfortunately, it is stuck with the following error:
M - list()
M$y = Dmat[,1]
M$X = Cmat
M$Ain = as.matrix(Amat)
M$bin = rep(0, dim(Amat)[1])
Hello,
How can I apply a function on a vector that refers to actual (n) and previous
elements in the vector (e.g. n-1)?
For example:
I would like to calculate the sum of (n-1) + n for each element of a vector and
get a vector as a result.
Besides others I tried this:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Marianne Stephan
Sent: Monday, December 06, 2010 9:13 AM
To: r-help@r-project.org
Subject: [R] How can I refer to actual (n) and previous (n-1) elements
in a vector?
Hello,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Nordlund, Dan (DSHS/RDA)
Sent: Monday, December 06, 2010 9:21 AM
To: r-help@r-project.org
Subject: Re: [R] How can I refer to actual (n) and previous (n-1)
elements in a vector?
Hi Marianne,
You have to be very careful with subsetting when doing something like that -
that's where you went wrong with your original construct. This version works:
v-c(3,6,8,1,1,3,9,5,6,3)
a - numeric(length(v)-1)
for (i in 2:length(v)) {a[i-1] - v[i-1] + v[i]}
a
[1] 9 14 9 2 4 12 14
My favorite book on SVM is Learning with Kernels by Scholkopf and Smola. You
might also want to consider a relevance vector machine, which is a more recent
development. RVM is Bayesian-based and usually produces a sparser
representation than a SVM. Check out Mike Tipping's web site at
Let us look at the objective function:
f(x) = x^2 + Si * x, where Si is the sum of the i-th column. This function
has a stationary point at x = -Si/2, and the second derivative is 2, so it
is a minimum.
Now, the column sums of your data matrix are all positive. So, your minimum
has to be
hi: i don't know that book but there's another pretty new book on svms by
lutz hamel. It's pretty gentle but very nice, especially if you have ZERO
background on svms.
there's also another one titled an introduction to svms by cristianini and
shawne-taylor that's also quite good but much more
I have got a matrix with 100 values.
Is there a easy way to cut the upper 10% (every value which is above 90%)
and the lower 10% of my matrix-values away and create a new matrix with all
the rest.
Thx for your help
Marty
--
View this message in context:
Andreas Jensen bentrevisor at gmail.com writes:
Hello.
I'm trying to solve a quadratic programming problem of the form min
||Hx - y||^2 s.t. x = 0 and x = t using solve.QP in the quadprog
package but I'm having problems with Dmat not being positive definite,
which is kinda okay since I
peter dalgaard pdalgd at gmail.com writes:
On Dec 6, 2010, at 15:15 , Jonathan P Daily wrote:
Correct me if I'm wrong, but isn't the minimal x value in your example
the same regardless of what positive coefficient you apply to x? If that
is the case, you would expect the same min(x) for
Benjamin B. benj.bad.ac at googlemail.com writes:
Hello list reader,
I am trying to form some constraints for an optimization I am working on.
I think I have understand the use of the constraints in matrix form. I use
them like:
[...]
Now I would like to formulate a constraint like
Hi Steven,
thank you again so much for your help and for the transformation of the data.
In fact I knew that there are these empty lines in the data. I couldn't take
them off with the program I have. I tried to solve this problems with using the
levels.
You are right, the statistical tests
You can use quantile and cut:
x - runif(1000)
x.cut - cut(x, c(-Inf, quantile(x, prob=c(.1, .9)), Inf),
labels=c('lo','mid','hi'))
table(x.cut)
x.cut
lo mid hi
100 800 100
On Mon, Dec 6, 2010 at 12:53 PM, Marty_H churc...@alpenjodel.de wrote:
I have got a matrix with 100 values.
Is
Have you tried setting singular.ok=TRUE in the call to lm? This will start
with the full set of contrasts, but only fit those that it is able to.
Otherwise you can set specific contrasts or subsets using the C (note case) or
contrasts functions.
--
Gregory (Greg) L. Snow Ph.D.
Statistical
Ista,
Thanks for this method, it works! Unfortunately 'R' still spits
everything out to std out, but the variable as you describe below grabs only
the final output variable, similar to how functions or scripts work
internally within 'R'. Thanks!
Kjetil,
Thanks also for your input, but
hi,
for some unknown reason i cannot install the package ca in R running in a
ubuntu mint system. i keep getting the following error message:
configure: error: missing required header GL/gl.h
ERROR: configuration failed for package ‘rgl’
* removing
On 06/12/2010 1:45 PM, Manderscheid Katharina wrote:
hi,
for some unknown reason i cannot install the package ca in R running in a
ubuntu mint system. i keep getting the following error message:
You need the development package for OpenGL or MesaGL to install rgl. I
don't know Ubuntu so I
I have a very sparse square matrix which is 20K rows columns and I am
trying to row standardize the matrix for the rows that have non-missing
value as follows:
row_sums - rowSums(M,na.rm=TRUE)
nonzero_idxs - which(row_sums0)
nonzero_M - M[nonzero_idxs,]/row_sums[nonzero_idxs]
M[nonzero_idxs,] -
thanks a lot, duncan for that super-fast reply and help!
i could install rgl via synaptic and afterwards ca in R directly. and it works!
:-)
have a nice evening, kat
Von: Duncan Murdoch [murdoch.dun...@gmail.com]
Gesendet: Montag, 6. Dezember 2010 19:55
Btw, forgot to mention I am using the standard Matrix package and I am
running version 2.10.1 of R.
On Mon, Dec 6, 2010 at 11:04 AM, scott white distributedin...@gmail.comwrote:
I have a very sparse square matrix which is 20K rows columns and I am
trying to row standardize the matrix for the
Hi,
On Fri, Dec 3, 2010 at 10:23 AM, manuel.martin
manuel.mar...@orleans.inra.fr wrote:
Dear all,
I am currently looking for a book about support vector machines for
regression and classification and am a bit lost since they are plenty of
books dealing with this subject. I am not totally new
if(FALSE) {
stuff your don't want executed
}
Switching a block of code off/on with editing a single character
may be done using 0/1 instead of FALSE/TRUE.
Or even F/T
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
|
I was wondering if there is a way to get an exact p-value at times where R
gives me just a range . for example
t.test(x,y)
p-value 2.2e-16
thanks,
--
View this message in context:
http://r.789695.n4.nabble.com/getting-the-exact-p-value-tp3075107p3075107.html
Sent from the R help mailing
Hello-
Does anyone know if R can read in ESRI file geodatabase feature
classes, directly from the file geodatabase? I currently have to
export the data out of Arc, to a csv file, then read the csv file. I'd
like to bypass this step and read directly from the file geodatabase.
Thanks for your
Dear,
I have the data in the following form:
head(matrices_m)
Location Variable Value Week
1 Africa Africa21 4 weeks
2 Asia Africa 0 4 weeks
3 Canada Africa17 4 weeks
4China Africa29 4 weeks
5 Europe AfricaNA 4 weeks
6Japan Africa68 4
In this case I only get three values back. My intention was to get every
value between 10% and 90% i.e. let's say i have a matrix
1 2 3 4
5 6 7 8
9 10 11 12
i only want the values above 10% and under 90%.
Maybe i should try it with a loop.
--
View this message in context:
Hi,
It depends on the function used to calculate it---each has their own
way of extracting particular values. Here's an example for t.test:
x - rnorm(10, 100, .1); y - x - 100
t.test(x, y)$p.value
You could find this yourself by looking at the documentation for t.test()
?t.test
The heading
Dear All,
I'm trying to use waldtest to test poolability (parameter stability) between
two logistic regressions. Because I need to use robust standard errors
(using sandwich), I cannot use anova. anova has no problems running the
test, but waldtest does, indipendently of specifying vcov or
Baoqiang Cao bqcaomail at gmail.com writes:
I have a least square fitting problem with linear inequality
constraints. pcls seems capable of solving it so I tried it,
unfortunately, it is stuck with the following error:
M - list()
M$y = Dmat[,1]
M$X = Cmat
M$Ain = as.matrix(Amat)
Hi Jef,
I would extract the information about the smoothers and plot them using
standard plotting routines. After
Mtest - gam(outcome ~ s(age, by=as.numeric(gender==0)) +
s(age,by=as.numeric(gender==1))+factor(Gender))
run something similar to:
smoothers - predict(s,type='terms')
and you
No there is no open software to read ESRI file geodatabases (the .gdb ones).
If it's not .gdb but .mdb (personal geodatabase, built on Microsoft
Access database) then you can read it with generic tools (such as
RODBC package) or with an appropriate build of GDAL as part of the
rgdal package.
On Mon, 6 Dec 2010, Roberto Patuelli wrote:
Dear All,
I'm trying to use waldtest to test poolability (parameter stability) between
two logistic regressions. Because I need to use robust standard errors (using
sandwich), I cannot use anova. anova has no problems running the test, but
Dear Achim,
Thanks a lot for the superquick reply!
Somehow with your suggestion I can get around the problem, but of course I
run into other problems, such as this:
waldtest(inv.log.leva.base, inv.log.leva, test = Chisq, vcov = sandwich)
Error in bread. %*% meat. : non-conformable arguments
On Mon, 6 Dec 2010, Roberto Patuelli wrote:
Dear Achim,
Thanks a lot for the superquick reply!
Somehow with your suggestion I can get around the problem, but of course I
run into other problems, such as this:
waldtest(inv.log.leva.base, inv.log.leva, test = Chisq, vcov = sandwich)
Error
hello!
I'm trying to make a very simple calculation using R that would called
from bash with two parameters, the names of the input and output files.
the first one is just an 4 by 100 numerical table- the function finds
the index of the maximal elements in each columns and writes it into the
Hi,
I'm trying to fit a logit model to a set of data that was collected under a
split plot scheme. The structure of the data is
Whole plot factors: Watering Frequency (2 levels: Hi/Lo) and Fertilizer type (3
factors A/B/C)
Subplot factor: slope type (2 factors up/down)
Response: Proportion of
Hi
I think the problem is that the viewer you are using for the PS file
cannot find the ComputerModern fonts that are referenced in the PS file.
If you use something like ghostscript to process the file you can see
this happening ...
[pmur...@stat18 Temp]$ gs plot_example.ps
ESP
Hi All,
I have conducted a meta analysis using the metabin function. I want to plot
5 subgroups on the same forest plot. I have managed to do this using the
byvar argument but when i plot the forest plot in R graphics I am unable to
view the very top and very bottom of the image. It is as though
Hi,
Can you give a reproducible example of what you did? My intuition is
that you could do it using par(), but I am not sure what package
metabin() is from (certainly none of the ones that load by default),
and I have even less idea how you created a forest plot (there are
many ways in R). If
With the result of the 'cut' you can get a direct index for the values:
x - runif(20)
x
[1] 0.26550866 0.37212390 0.57285336 0.90820779 0.20168193 0.89838968
0.94467527 0.66079779 0.62911404
[10] 0.06178627 0.20597457 0.17655675 0.68702285 0.38410372 0.76984142
0.49769924 0.71761851 0.99190609
On Mon, Dec 6, 2010 at 3:58 AM, Sunny Srivastava
research.b...@gmail.com wrote:
Dear R-Helpers:
I am using trying to use *ddply* to extract min and max of a particular
column in a data.frame. I am using two different forms of the function:
## var_name_to_split is a string -- something like
Dear all,
I get stuck when i try to export the data into SPSS format/file using
write.foreign()
Do you know how to do it exactly?
What i have done is
1) First i type the following code in R:
df-data.frame(id,year,res1)
names(df)-c(idcode,year,resarrvl)
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