Hello-
I am new to R and am trying to use glmpath for LASSO logistic regression. My
data set contains 40 observations and a mix of dichotomous, categorical, and
continuous potential predictor variables.
I am having a problem with the 'weight = rep(1, n)' argument that relates to an
optional ve
I'm fairly new to the silhouette functionality in the cluster package, so
apologize if I'm asking something naive.
If I run the 'agnes(ruspini)' example from the silhouette section of the
cluster package vignette, and assign colours to clusters, two clusters have
what appear to be incorrect co
Dear Deepayan and Dennis,
Both of your proposed fixes worked perfectly. Thank you!
Adam
On 8/24/2011 9:47 PM, Deepayan Sarkar wrote:
On Thu, Aug 25, 2011 at 1:00 AM, Adam Zeilinger wrote:
Hello,
I am using the xyplot() function to create a series of scatterplot panels
with lines of best f
On Aug 25, 2011, at 12:38 AM, Andra Isan wrote:
Hi All,
I have a set of features of size p and I would like to separate my
feature space into two sets so that p = p1 + p2, p1 is a set of
features and p2 is another set of features and I want to fit a glm
model for each sets of features se
Hi:
It's straightforward to write a simple function to do this:
simmat <- function(N, t1) matrix(rep(runif(N, min = -1, max = 1), each
= t1), nrow = N, byrow = TRUE)
v <- simmat(45, 10)
head(v, 3)
HTH,
Dennis
PS: T is a reserved word in R (abbreviation for the logical TRUE) and
should not be us
Hi:
Try this instead:
m <- matrix(rpois(40, 10), nrow = 1)
> dim(m)
[1] 140
r <- m[1:10, 3:6]
dput(r)
structure(c(12, 7, 15, 8, 6, 7, 14, 10, 11, 4, 9, 16, 12, 5,
9, 10, 9, 9, 8, 7, 12, 9, 10, 12, 12, 11, 11, 8, 12, 8, 15, 21,
3, 3, 13, 9, 8, 13, 7, 11), .Dim = c(10L, 4L))
# Alte
Hi:
Since you tried several functions (reasonably so IMO), here is how
they would work in this problem, in addition to the solutions already
supplied.
Some data massaging before starting, taking your data as input, saved
into an object named visits:
visits <- structure(list(unique_id = c(1L, 1L,
On Thu, Aug 25, 2011 at 1:00 AM, Adam Zeilinger wrote:
> Hello,
>
> I am using the xyplot() function to create a series of scatterplot panels
> with lines of best fit. To draw the lines of best fit for each panel, I am
> using a panel function. Here's an example:
>
>> species <- as.character(c(r
Hi All,
I have a set of features of size p and I would like to separate my feature
space into two sets so that p = p1 + p2, p1 is a set of features and p2 is
another set of features and I want to fit a glm model for each sets of features
separately. Then I want to combine the results of two gl
On Wed, Aug 24, 2011 at 6:19 PM, Brian Lunergan wrote:
> Evening all:
>
> Redid my home box using VectorLinux (Slackware variation) and now I'm not sure
> which game trail to follow to reinstall R.
>
> Could somebody with more knowledge in this share their thoughts off-list. Just
> need a pointer
This is for coxph:
The cluster term is used to compute a robust variance for the model. The
term + cluster(id) where each value of id is unique is equivalent to
specifying the robust=T argument, and produces an approximate jackknife
estimate of the variance. If the id variable were not unique, but
Hi:
Here's one way out:
xyplot(y ~ x|species*year, data = ex.data, type = c('p', 'r'))
type = is a very useful argument to know in xyplot(). See p.75 of the
Lattice book.
HTH,
Dennis
On Wed, Aug 24, 2011 at 12:30 PM, Adam Zeilinger wrote:
> Hello,
>
> I am using the xyplot() function to creat
On Aug 24, 2011, at 9:10 PM, Katherine Lizama Allende wrote:
Hi all:
I need to put bold font in the y label, which is an expression at
the same
time
When putting font.lab=2 in plot, it only puts bold font for the x axis
label.. what should I do? Thanks very much
plot(jitter(c(1, 4, 7, 9,
merge() can align the values in the obs columns
so those with the same date can be compared. E.g.,
set up the data with the following copy-and-pastable
code:
A <- read.table(header=TRUE, textConnection(" year mon day obs
2010 03 1212
2010 03 1822
2010 04 1262
2010 0
Hi Kathy,
Try
plot(10, xlab = "Week", ylab = expression(bold("B removal rate
"*"(mg/"*m^3*"-d)")))
HTH,
Jorge
On Wed, Aug 24, 2011 at 9:10 PM, Katherine Lizama Allende <> wrote:
> Hi all:
>
> I need to put bold font in the y label, which is an expression at the same
> time
>
> When putting fo
Hi Alexandra,
Here is an alternative without using a loop:
matrix(sapply(a, rep, T), ncol = 1)
HTH,
Jorge
On Wed, Aug 24, 2011 at 9:10 PM, Soberon Velez, Alexandra Pilar <> wrote:
> Hello,
>
> I want to create a matrix of N random numbers with a uniform distributions.
> Later, I want to repea
thanks.
Merge? I am just looking for a method to comparison on homogeneous sample
between two dataset.
2011/8/25 Jeff Newmiller
> ?merge may be what you are looking for. If not, you should clarify what you
> want to do.
>
Hi all:
I need to put bold font in the y label, which is an expression at the same
time
When putting font.lab=2 in plot, it only puts bold font for the x axis
label.. what should I do? Thanks very much
plot(jitter(c(1, 4, 7, 9, 11, 13), a=0.1), y = Bllim.m, xlab = "Week", ylab
= expression(paste
Evening all:
Redid my home box using VectorLinux (Slackware variation) and now I'm not sure
which game trail to follow to reinstall R.
Could somebody with more knowledge in this share their thoughts off-list. Just
need a pointer to the appropriate trail head. I can take it from there.
Regards.
Hello,
I want to create a matrix of N random numbers with a uniform distributions.
Later, I want to repeat T times each row of this matrix. For this I do the
following loop:
N<-45
T<-10
n<-N*T
a<-matrix(runif(N,min=-1,max=1),nr=N)
mymat<-matrix(rep(NA,n),nr=n,nc=1)
for(i in i:N){
b<-rep(a[i,],
?merge may be what you are looking for. If not, you should clarify what you
want to do.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar
hi ,
Now I have two dataset and want to compare them with same sample.
Dataset A:
year mon day obs
2010 03 1212
2010 03 1822
2010 04 1262
2010 07 24 29
Dataset B:
year mon day obs
2010 03 1215
2010 04 1257
2010 07 2432
2010 08
On Aug 24, 2011, at 8:36 PM, Weidong Gu wrote:
At default, factors (months) are alphabetically leveled. You can
explicitly re-level months
months<-factor(months,levels=c('Jan','Feb','Mar',...,'Dec'))
> ?Constants
> month.abb
[1] "Jan" "Feb" "Mar" "Apr" "May" "Jun" "Jul" "Aug" "Sep" "Oct"
I agree with Ken.. if you can, save it as a CSV file. But if you have a
bunch of these, then it isn't very efficient. I use read.xlsx() from the
package "xlsx".
I notice that you are using the full path.. have you tried changing
directories?... I find it is best to compartmentalize my work and
At default, factors (months) are alphabetically leveled. You can
explicitly re-level months
months<-factor(months,levels=c('Jan','Feb','Mar',...,'Dec'))
Then it should work.
Weidong Gu
On Wed, Aug 24, 2011 at 10:45 AM, Phoebe Jekielek wrote:
> Hi there,
>
> I have length data of an organism o
If I understand you correctly, see ?paste
and the following to extract the values you require:
summary(res)[[4]][1]
summary(res)[[4]][2]
summary(res)[[8]]
HTH
ashz wrote:
>
> Dear all,
>
> How can I covert lm data to text in the form of "y=ax+b, r2" and how do I
> calculate R-squared(r
Hi,
I'm trying to impute a data set consisting of mixed type variables, mostly
logical, but also ordered and non-ordered factors, and numeric variables
with the missForest package. Although the help file of missForest states
that the resulting data matrix 'ximp' has the same type as the original d
Hi all,
Just a small question: After fitting a multivariate mixture using flexmix, I
wish to use refit to get the parameters of covariates and their standard
errors. However using refit I only can see the components for the first
dependent variable. What should I do if I want to see the others?
T
On Aug 24, 2011, at 6:48 PM, Jeffrey Joh wrote:
I have a data frame that is about 40 columns by 1 rows. I want
to get the dput of small portion of that by using
dput(results[1:10,3:6]). The dput is very long and includes all the
values from the original data frame. Why is that?
How are you storing the elements of the data frame? I'm working with a data
frame of doubles with no names and having trouble observing the same
problem. If they are factor levels though, that *might* account for it.
sessionInfo() might also help. Obviously it's not convenient to print this
example
I have a data frame that is about 40 columns by 1 rows. I want to get the
dput of small portion of that by using dput(results[1:10,3:6]). The dput is
very long and includes all the values from the original data frame. Why is
that?
Jeffrey
__
Hi Dan,
You might try
require(gdata)
?read.xls
HTH,
Jorge
On Wed, Aug 24, 2011 at 6:20 PM, Dan Abner <> wrote:
> Hello everyone,
>
> What is the simplest, most RELIABLE way to import data from MS EXCEL (.xls)
> format to R? In the past I have used the read.xls() function from the
> xlsReadWri
Try this:
require(zoo)
lvd <- tapply(df$visit_date, df$unique_id, max)
index <- tapply(df$visit_date, df$unique_id)
df$last_visit_date <- as.Date(lvd[index])
Jean
Kathleen Rollet wrote on 08/24/2011 04:15:45 PM:
>
> Dear R users,
>
> I am encoutering the following problem: I have a dataset wit
-- Forwarded message --
From: Ken Hutchison
Date: Wed, Aug 24, 2011 at 6:27 PM
Subject: Re: [R] Importing data from MS EXCEL (.xls) to R
To: Dan Abner
save as csv.
?read.csv
Ken
On Wed, Aug 24, 2011 at 6:20 PM, Dan Abner wrote:
> Hello everyone,
>
> What is the simpl
-- Forwarded message --
From: Ken Hutchison
Date: Wed, Aug 24, 2011 at 6:06 PM
Subject: Re: [R] help with "by" command
To: amalka
?tapply
or more specifically
?ave
Hope this helps,
Ken
On Wed, Aug 24, 2011 at 2:51 PM, amalka wrote:
> Hello,
>
> I am a new user of R, an
On Aug 24, 2011, at 5:15 PM, Kathleen Rollet wrote:
Dear R users,
I am encoutering the following problem: I have a dataset with a
'unique_id' and different 'visit_date' (formatted as.Date, "%d/%m/
%Y") per unique_id. I would like to create a new variable with the
most recent date of visit
Hello everyone,
What is the simplest, most RELIABLE way to import data from MS EXCEL (.xls)
format to R? In the past I have used the read.xls() function from the
xlsReadWrite package, however, I have been wrestling with it all afternoon
long with no success. I continue to receive the following err
You could try using the numeric representation of date, and split the data
frame using that variable. For example:
src$date.num <- as.numeric(src$date)
Jean
Franc Lucas wrote on 08/24/2011 02:42:58 PM:
>
>Hello everyone,
>I want to split a data.frame by the column date . The data fram
as.POSIXct(518400,origin='2001-01-01')
[1] "2001-01-07 PST"
as.POSIXct(as.numeric(as.POSIXct(518400,origin='2001-01-01')),origin='1970-01-01')
[1] "2001-01-07 08:00:00 PST"
On Wed, Aug 24, 2011 at 9:22 AM, Agustin Lobo wrote:
> Hi!
>
> I'm confused by this:
> > as.numeric(as.POSIXct(518400,**o
Hi Ari,
Try this instead
with(foo, tapply(V2, trust, mean, na.rm = TRUE))
See ?tapply and ?with for more information.
HTH,
Jorge
On Wed, Aug 24, 2011 at 2:51 PM, amalka <> wrote:
> Hello,
>
> I am a new user of R, and I'd be grateful if someone could help me with the
> following:
>
> I would
Hi elh,
You could try using split() and lapply() instead (untested):
mymodels <- lapply(split(usage, ActNo),
function(l) lm(AvgKWh ~ AvgHDD + AvgCDD, data = l)
)
To access the coefficients for all models you can do
lapply(mymodels, coef)
and, to access model number one (first ActNo),
mymodels
Thanks again, you are perfectly right. I checked and saw I was indeed
polluting my machine with unclosed threads.
On 08/23/2011 01:34 AM, Peter Langfelder wrote:
> On Mon, Aug 22, 2011 at 3:12 PM, Immanuel wrote:
>> Hello,
>>
>> thanks for the input. Below is a small example, simpler then expecte
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of shardman
> Sent: Wednesday, August 24, 2011 8:10 AM
> To: r-help@r-project.org
> Subject: [R] unused argument(s) (Header = True) help!
>
> Hi,
>
> I'm really new to R so I aopl
Hi Sam,
It is "header", not "Header". See ?read.table.
HTH,
Jorge
On Wed, Aug 24, 2011 at 11:10 AM, shardman < wrote:
> Hi,
>
> I'm really new to R so I aoplogise if this is a stupid question.
>
> I'm trying to import data from a .txt file into R using the read.table
> command, the headers fo
The proper command is "header = TRUE"
capitalization is important for both *h*eader and T*RUE*
Hope this helps,
Michael Weylandt
On Wed, Aug 24, 2011 at 11:10 AM, shardman wrote:
> Hi,
>
> I'm really new to R so I aoplogise if this is a stupid question.
>
> I'm trying to import data from a .txt
Read the plot documentation by typing ?plot, particularly the optional
argument main.
Hope this helps,
Michael Weylandt
On Wed, Aug 24, 2011 at 1:40 PM, upani1982 wrote:
> Hi All,
>
> I am new to this forum. I have just started learning R. When i use
> plot(decompose(x)), i am getting the titl
Claudio Zanettini wrote on 08/24/2011
04:33:50 PM:
>
> Thank you, this work fine,
> and is not contorted like mine:)
> In this case lastV=LastI but depending on the data that I obtain
> lastV can be = LastA.
>
> Any way it works very good:)
>
> Thank you
> very much :)
>
>
> PS: but I still
Hi,
All the information is contained in your aov() object. Take a look at the
first example at
?aov
HTH,
Jorge
*
*
On Wed, Aug 24, 2011 at 10:33 AM, martinas <> wrote:
> hello
>
> I need to know the dfn and dfd of my Anova. But in the Anova output there
> is
> only "Df".
> Is this the dfn or
I have a data frame (narrow) with 431 rows and 6 columns containing information
on chromosome, position, lod1, lod2, lod3, lod4, looking like this:
> narrow
chr pos lod1 lod2 lod3 lod4
1 1 3.456 -0.025 -0.003 -0.209 -0.057
2 1 5.697 -0.029 -0.005 -0.200 -0.058
3 1
Dear R users,
I am fitting nonlinear mixed effects models with autocorrelated errors
(an AR(1) model on the residuals) using NLME and am comparing a set of
models that contain the same fixed effects structure but with different or
no random effects (nested) . The issue I've come across that
I figured out a solution by myself. In brief, I used different axis commands
to specify the ticks (with labels set to FALSE) and the labels (with tick
set to FALSE). For instance (with width=1 and space=1):
axis(side=1,at=c(2,6),labels=FALSE,tck=-0.1)
axis(side=1,at=c(0,4,8),labels=FALSE,tck=-0.2)
Hello,
I am using the xyplot() function to create a series of scatterplot
panels with lines of best fit. To draw the lines of best fit for each
panel, I am using a panel function. Here's an example:
> species <- as.character(c(rep(list("A", "B", "A"), 10), "B"))
> year <- as.character(c(rep
Dear R users,
I am encoutering the following problem: I have a dataset with a 'unique_id' and
different 'visit_date' (formatted as.Date, "%d/%m/%Y") per unique_id. I would
like to create a new variable with the most recent date of visit per unique_id
as shown below.
unique_id visit_date
Hi folks,
I've a basic question concerning missing data. I'm running mixed effects
analyses using nlme. I've a sizable chunk of missing data on the outcome
being modeled, and am using "na.action=na.omit" when running the models.
After fitting the models, I'm then trying to extract and use the f
Hello everyone,
I want to split a data.frame by the column date . The data frame looks like
this
date time openclose
02.01.201109:00:00 1000 1200
02.01.201109:05:02 1200 1203
...
01.02.2011
Hi All,
I am new to this forum. I have just started learning R. When i use
plot(decompose(x)), i am getting the title " Additive time series
decomposition". How to make this title off and change to some other title.
Any help regarding this is highly appreciated.
With sincerer regards,
Upananda
Hello,
I am a new user of R, and I'd be grateful if someone could help me with the
following:
I would like to compute the mean of variable "trust" in dataframe "foo", but
separately for each level of variable V2. That is, I'd like to compute the
mean of trust at each level of V2.
I have done th
Dear R-users.
I am faced with a problem I dont know how to solve.
I need to calibrate the Heston stochastic volatility model, and have (to my
own belief) created a code for calculating the prices of options by this
model. However, when I calibrate the model using NLMINB I also evaluate my
initial
Hi!
I'm confused by this:
> as.numeric(as.POSIXct(518400,origin="2001-01-01"))
[1] 978822000
I guess the problem is that as.numeric() assumes a different origin, but cannot
find
any default origin.
How can I get back the seconds from the POSIXct format? In other words, which
the inverse funct
This should be easy but it does not work
I have 3 vectors*(activeT,inactT, activeR)*,
the idea is that if the last value in inactT is higher than the last in
activeT
this value has to be append in active T
and the last value in another vector call activeR has to be repeated.
(at the bottom you can
Apologies for the elementary nature of the question (yes, I'm another
newbie)...
I'd like to perform a multiple regression on a single data set containing a
representation of energy consumption and temperatures containing account
number, usage (KWh), heating degree days (HDD) and cooling degree (C
Hello
did you happen to figure this out? I am just learning about using R, i have
a whack of fish data in MSAccess...and i want to take whatever functions
access is limited by with stats, and then call R to do them
i know the package RODBC works great to read data from your mdb, but i want
to hav
hello
I need to know the dfn and dfd of my Anova. But in the Anova output there is
only "Df".
Is this the dfn or the dfd? and how do I get both of it in R?
Thanks for any answers
--
View this message in context:
http://r.789695.n4.nabble.com/df-of-numerator-and-denominator-tp3765526p3765526.htm
answered my own question, just use the call shell function in vb
woohoo
--
View this message in context:
http://r.789695.n4.nabble.com/Controling-R-from-MS-Access-tp2719751p3766037.html
Sent from the R help mailing list archive at Nabble.com.
__
R-h
Many thanks for your response.
unfortunately, it appears that I'm the closest thing in the vicinity to a
local expert (chilling times indeed...), but i will certainly look at the
booklist
in terms of the number of data points, we have:
two shores,
three treatments,
three replicates of each treatm
Hello everyone,
I was asked to repost this again, sorry for any inconvenience.
I'm looking replacement for ddply function from plyr package.
Function allows to apply function by category stored in any column/columns.
Regular loops or lapplys slow down greatly because my unique combination
count
Thank you Dan and Ista!
Both of you are correct, I should have used NA rather than "NA" in my
example. So the correct code should be:
X <-as.data.frame(matrix(c(9, 6, 1, 3, 9, NA, NA,NA,NA,NA,
6, 4, 3,NA, NA, NA, 5, 4, 1, 3), ncol=2))
names(X)<-c("X1","X2")
X$X1[
Hi,
I'm really new to R so I aoplogise if this is a stupid question.
I'm trying to import data from a .txt file into R using the read.table
command, the headers for the data columns are already in the text file so I
add Header = True after the file location. The problem is I keep getting the
erro
Dear all,
How can I covert lm data to text in the form of "y=ax+b, r2" and how do I
calculate R-squared(r2)?
Thanks.
Code:
x=18:29
y=c(7.1,7,7.7,8.2,8.8,9.7,9.9,7.1,7.2,8.8,8.7,8.5)
res=lm(y~x)
--
View this message in context:
http://r.789695.n4.nabble.com/Howto-convert-Linear-Regression-dat
Hi there,
I have length data of an organism over the year and I want to make a
boxplot. I get the boxplot just fine but the months are all out of order. In
the data set they are in order from Jan-Dec...how can I fix this problem?
Thanks so much in advance!!
Phoebe
[[alternative HTML ver
Thank you, this work fine,
and is not contorted like mine:)
In this case lastV=LastI but depending on the data that I obtain
lastV can be = LastA.
Any way it works very good:)
Thank you
very much :)
PS: but I still do not understand what was wrong in the script that I used,
It was not very appr
I'm still a little confused about lastV and lastI. The code you provide
uses lastV, but your description seems to refer to lastI. Test out this
code and see if it is doing what you want it to do.
lastI
lastA
activeT
activeR
if(lastI > lastA) {
activeT <- c(activeT, lastI)
activ
On Aug 24, 2011, at 3:31 PM, Jim Silverton wrote:
Hi all,
I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6)
I want to return a vector with the corresponding probabilities based
on the
amount of times the numbers occurred. For example, I should get the
following vector for xm:
prob.xm = c(1
Claudio Zanettini wrote on 08/24/2011 03:04:39 PM:
> This should be easy but it does not work
> I have 3 vectors*(activeT,inactT, activeR)*,
> the idea is that if the last value in inactT is higher than the last in
> activeT
> this value has to be append in active T
When you say "this value" whi
This should be easy but it does not work
I have 3 vectors*(activeT,inactT, activeR)*,
the idea is that if the last value in inactT is higher than the last in
activeT
this value has to be append in active T
and the last value in another vector call activeR has to be repeated.
(at the bottom you can
Apparently my request to view the help pages is not a popular method among R
users for gaining information. for me these pages are very helpful so I will
follow up to completed this thread for future searchers.
First thanks fo Prof. Brian Ripley. Your idea was spot on what I was looking
for
1. As this is not really appropriate for R, I suggest replies be private.
2. You might try posting on various statistical forums, e.g. on
http://stats.stackexchange.com/
-- Cheers, Bert
On Wed, Aug 24, 2011 at 12:15 PM, Arnaud Mosnier wrote:
> Hi,
>
> In order to find the best models I use AIC
If your numbers are all positive integers, this should work:
(tabulate(xm)[xm])/length(xm)
it can be put into a function for ease of use:
probVec <- function(x) {(tabulate(x)[x])/length(x)}
You'll have some trouble if you have non-positive integers or non-integers.
Let me know if you need to ha
Try this:
prob.xm <- (table(xm)/length(xm))[match(xm, sort(unique(xm)))]
Jean
Jim Silverton wrote on 08/24/2011 02:31:05 PM:
> Hi all,
> I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6)
>
> I want to return a vector with the corresponding probabilities based on
the
> amount of times the nu
Hi all,
I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6)
I want to return a vector with the corresponding probabilities based on the
amount of times the numbers occurred. For example, I should get the
following vector for xm:
prob.xm = c(1/9, 1/9, 1/9, 1/9, 3/9, 3/9, 3/9, 2/9, 2/9)
Any help grea
Hi,
In order to find the best models I use AIC, more specifically I calculate
Akaike weights then Evidence Ratio (ER) and consider that models with a ER <
2 are equally likely.
But the same problem remain each time I do that. I selected the best models
from a set of them, but I don't know if those
On 8/24/2011 12:40 PM, Uwe Ligges wrote:
Actually it is recommended to test for the availability of a valid
package with find.package(), particularly in this case where the name
of the package is already know.
Best,
Uwe
Thanks. So I guess the idiom I'm looking for is
> length((find.package(
Rereading your email, still not sure what the question is -- perhaps you
could give a better code example to illustrate the difference between a[[2]]
and mat1 -- but, since you mentioned briefly lists of lists, have you looked
at unlist(, recursive = F)? If applied to a list of lists, it won't unli
I'm not sure I understand your question: a[[2]] is a matrix.
> a <- list(matrix(1:6,2),matrix(5:10,2))
> is.matrix(a[[2]])
TRUE
x = a[[2]]
> is.matrix(x)
TRUE
> x+2
[,1] [,2] [,3]
[1,] 7 9 11
[2,] 810 12
> a[[2]] + 2
[,1] [,2] [,3]
[1,] 7 9 11
[2,] 810 12
Wha
Hi,
On Wed, Aug 24, 2011 at 2:18 PM, zhenjiang xu wrote:
> Hi R users,
>
> I was using read.table to read a file. The data.fame looked alright, but I
> found not all rows are read by the read.table. What's wrong with it? It
> didn't give me any warning or error messages. Why the data are truncate
Dear All,
As always, I appreciate all your help.
I would like to know the easiest way to convert each of the homogeneous
elements of a numeric list into a matrix. Each element of this list is also a
list such that when displayed, looks like a 2-by-3 matrix , I would like to
convert each of them
Hi R users,
I was using read.table to read a file. The data.fame looked alright, but I
found not all rows are read by the read.table. What's wrong with it? It
didn't give me any warning or error messages. Why the data are truncated?
Thanks.
$ wc -l all/isoform_exp.diff
42847 all/isoform_exp.diff
On Aug 24, 2011, at 1:37 PM, Sebastian Bauer wrote:
Hi!
in R? Basically, what I need is a mixture of order() and rank().
While the former allows to specify multiple vectors, it doesn't
provide the flexibility of rank() such that I can specify what
happens if ties can not be broken.
An exampl
What is your sample size? I've had trouble getting reliable estimates using
simple data splitting when N < 20,000.
Note that the following functions in the rms package facilitates
cross-validation and bootstrapping for validating models: ols, validate,
calibrate.
Frank
Andra Isan wrote:
>
> Hi
Hi!
in R? Basically, what I need is a mixture of order() and rank().
While the former allows to specify multiple vectors, it doesn't
provide the flexibility of rank() such that I can specify what
happens if ties can not be broken.
>>> An example of this "simple problem" would c
Thanks Dennis! I'll check this out.
Just to clarify, I need the total number of switches/changes
regardless of if that state
had occurred in the past. So A-A-B-A, would have 2 changes: A to B and B to A.
Thanks again.
On Wed, Aug 24, 2011 at 1:28 PM, Dennis Murphy wrote:
> Hi Juliet:
>
> Here'
On Aug 24, 2011, at 1:11 PM, Sebastian Bauer wrote:
Hi!
I'd like to rank rows of a data frame similar to what rank() does
for vectors. However, ties should be broken by columns that I
specify. If it is not possible to break a ties (because the row data
is essentially the same), I'd like to ha
Hi Juliet:
Here's a Q & D solution:
# (1) plyr
> f <- function(d) length(unique(d$mygroup)) - 1
> ddply(myData, .(id), f)
id V1
1 1 0
2 2 2
3 3 1
4 4 0
# (2) data.table
myDT <- data.table(myData, key = 'id')
myDT[, list(nswitch = length(unique(mygroup)) - 1), by = 'id']
If one can sw
Hi,
Thanks for the reply. What I meant is that, I would like to partition my dat
data (a data frame) into training and testing data and then evaluate the
performance of the model on test data. So, I thought cross validation is the
natural choice to see how the prediction works on the hold-out d
What you describe is not cross-validation, so I am afraid we do not
know what you mean. And cv.glm does 'prediction for the hold-out
data' for you: you can read the code to see how it does so.
I suspect you mean you want to do validation on a test set, but that
is not what you actually claim.
Hi!
>> I'd like to rank rows of a data frame similar to what rank() does
>> for vectors. However, ties should be broken by columns that I
>> specify. If it is not possible to break a ties (because the row data
>> is essentially the same), I'd like to have the same flexibility that
>> rank() offers
Hi All,
I have a fitted model called glm.fit which I used glm and data dat is my data
frame
pred= predict(glm.fit, data = dat, type="response")
to predict how it predicts on my whole data but obviously I have to do
cross-validation to train the model on one part of my data and predict on the
I have a data set with about 6 million rows and 50 columns. It is a
mixture of dates, factors, and numerics.
What I am trying to accomplish can be seen with the following
simplified data, which is given as dput output below.
> head(myData)
mydate gender mygroup id
1 2012-03-25 F
On Aug 24, 2011, at 11:09 AM, Sebastian Bauer wrote:
Hello,
I'd like to rank rows of a data frame similar to what rank() does
for vectors. However, ties should be broken by columns that I
specify. If it is not possible to break a ties (because the row data
is essentially the same), I'd l
Actually it is recommended to test for the availability of a valid
package with find.package(), particularly in this case where the name of
the package is already know.
Best,
Uwe
On 24.08.2011 18:29, Yihui Xie wrote:
.packages(all = TRUE) will give you a list of all available packages
witho
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