On Wed, 13 Jul 2011, B77S wrote:
> ## Hello.. I have asked a similar question, but this is not fixed as
> before.
>
> ## I am running the following using Ubuntu OS:
>
> R version 2.13.1 (2011-07-08)
> Copyright (C) 2011 The R Foundation for Statistical Computing
> ISBN 3-900051-07-0
> Platform: x
mousy0815 gmail.com> writes:
> I'm trying to make a function that will output the exponent... so f2(2,2) = 4
> and f2(2,3)=8. But I don't want to just use the x^n function, I want to do
> it another way, and without a recursion. I did the follow code but for some
> reason it doesn't work. Help pl
I'm trying to create a formula object to pass on to a function that
applies the function terms.formula() to it.
f <- function(formula, ...)
{
...
mf <- match.call()
term <- terms.formula(mf$formula)
...
}
However, my code below gives an error.
form <- as.formula("y~x")
f(form, ...)
The error
Hi,
I am using the gap R package to do the Hardy Weinberg Case Control test for
many SNP. I am not sure what the values initial1 and initial2 should be for
the test. I tried values but they failed.
I emailed the author but to no avail. There seems to be some documentation
that is deleted at the to
On Jul 12, 2011, at 5:43 PM, Tom La Bone wrote:
Thank you for the reply Dr. Winsemius. Can you take your answer a step
further and, in the context of the simple, reproducible example,
illustrate
how it is done? I would appreciate it.
Tom
The easy way is:
> sfit <- summary(fit)
> sfit$ta
I'm trying to make a function that will output the exponent... so f2(2,2) = 4
and f2(2,3)=8. But I don't want to just use the x^n function, I want to do
it another way, and without a recursion. I did the follow code but for some
reason it doesn't work. Help please?
f2 <- function(x, n) #without r
Dear Susie,
I'm glad the pieces are working. Sorry if I wasn't clear before, when I said
break down into smaller pieces, I meant for testing and figuring out what's
wrong. If you have everything in separate loops, they will be completed
individually, so reading all the datasets will be done a
Take a closer look at your R code. Each loop is returning only the
last value, so you're doing the same thing over and over for the last
value from the previous loop.
Write just the code for *one* set of graphs. Look at what needs to
change to do the next set. Write a function that takes those arg
Hey Josh,
Thank you so much for the help!
Now I have edited my codes--breaking into smaller pieces:) R does not state
the error again:)
However, the problem comes, please see my codes below-- You've suggested
earlier that I should move the "eight" part outside the loop; I changed the
"eight"
On Jul 12, 2011, at 8:40 PM, Eduardo M. A. M. Mendes wrote:
Hello
I need to run an r-file that works on Windows on MAC.
Does that mean you are using a virtual box to run Windoze on Mac
hardware?
The first lines of the file are:
library(R.matlab)
library(Rcompression)
Somehow I cannot
Hello
I need to run an r-file that works on Windows on MAC. The first lines of the
file are:
library(R.matlab)
library(Rcompression)
Somehow I cannot find where Rcompression is. Am I missing something?
Many thanks
Ed
__
R-help@r-project.org mail
On Jul 12, 2011, at 5:43 PM, Tom La Bone wrote:
Thank you for the reply Dr. Winsemius. Can you take your answer a step
further and, in the context of the simple, reproducible example,
illustrate
how it is done? I would appreciate it.
Of course. Would be happy to take a stab at it but
Hello Friends,
I am new to R and stuck with a problem.
i have two columns drug_A and drug_B, i have plotted a scatter plot
using the ggplot2 function.
My problem is with the third column, it is the p-value column.
I want to color and size points differently based on the p_value, the
p_value ran
On Jul 12, 2011, at 7:26 PM, Salifu Wahab wrote:
Please I am new to R. I got the following code from a friend:
gmat <- cbind(gmat,p*(1-p)*wxb)
for (j in seq(1:ncol(gmat))) {
gmat[,j] <- fitted(lm(gmat[,j]~zmat))
}
It is for spatial logit.
After defining all the matrices in it such as P, wx
Thank you for the reply Dr. Winsemius. Can you take your answer a step
further and, in the context of the simple, reproducible example, illustrate
how it is done? I would appreciate it.
Tom
--
View this message in context:
http://r.789695.n4.nabble.com/simple-save-question-tp3429148p3663645.htm
I have a question about running an optimization function on an existing LOESS
function defined in R. I have a very large dataset (1 million observations)
and have run a LOESS regression. Now, I want to run a Newton-Raphson
optimization to determine the point at which the slope change is the
greates
Please I am new to R. I got the following code from a friend:
gmat <- cbind(gmat,p*(1-p)*wxb)
for (j in seq(1:ncol(gmat))) {
gmat[,j] <- fitted(lm(gmat[,j]~zmat))
}
It is for spatial logit.
After defining all the matrices in it such as P, wxb, gmat, and zmat,
I tried to run it in R and go
There is no missing value nor zero values.
- Original Message
From: Uwe Ligges
To: Peter Maclean
Cc: r-help@r-project.org
Sent: Tue, July 12, 2011 2:58:10 AM
Subject: Re: [R] fitdistr() Error
Any NA values or values outside the support region of your distribution?
UWe
On 11.07.2
On 13/07/11 00:57, ruocco wrote:
Dear All,
I have a collections of spatial data. I have to analyze pairs of these
point patterns to test their spatial interaction. I was moving towards
the cross K Ripley's function. The problem, however, are the following:
1) What is the best way to get a singl
On Wed, Jul 13, 2011 at 6:13 AM, Sze Liu wrote:
> Hello all,
> I am using AddHealth data to fit a cure, aka split population model using
> nltm. I am not sure how to account for the complex survey
> design - does anyone have any suggestions? Any help would be greatly
> appreciated!
This will
These sorts of tasks can be automated. See:
?Startup
-- Bert
On Tue, Jul 12, 2011 at 2:48 PM, Steve Taylor wrote:
> My solution to the clutter problem is this: at start-up time, create a list
> of functions, attach the list and then delete the list. I haven't delved
> into making packages ye
My solution to the clutter problem is this: at start-up time, create a list of
functions, attach the list and then delete the list. I haven't delved into
making packages yet.
if (any(search()=="MyFunctions")) detach(MyFunctions)
MyFunctions <- list()
MyFunctions$
"%like%" <- function(x,y) { s
Roger,
I think I know why you are having a problem. Your foreach is using
%dopar% and judging from your output you have active parallel
processing. When running something in paralell, the master is the
processor used by the Rgui, and the slaves are in the background. My
guess is the the functio
On Jul 12, 2011, at 5:28 PM, Bos, Roger wrote:
All,
I am not understanding the scoping used in foreach when it is used
inside a function. I keep getting "could not find function" errors
for
functions that are in memory when I try to use foreach within a
function
call. I have a simple exa
All,
I am not understanding the scoping used in foreach when it is used
inside a function. I keep getting "could not find function" errors for
functions that are in memory when I try to use foreach within a function
call. I have a simple example below. "testFun" is in memory and works
when call
On Tue, Jul 12, 2011 at 5:27 AM, Dimitris.Kapetanakis
wrote:
> Dear all,
>
> I would like to use the apply or a similar function belonging to this
> family, but applying for each column (or row) but let say for each q
> columns. For example I would like to apply a function FUN for the first q
> co
On Tue, Jul 12, 2011 at 2:01 PM, Reith, William [USA]
wrote:
> Do I need to define x in any way before I do the loop?
No, you should not need to define x explicitly. Just pass the data
frame you want to qplot. David was absolutely right though that
inside a loop, you should wrap the call to qpl
Hi Meeta,
yes, there was a bug in the package. Please install the newest version
and try again.
Best,
Peter
On Tue, Jul 12, 2011 at 1:20 PM, mistrm wrote:
> Hi Peter and Raquel
>
> I am following the same tutorial and seem to have the same error appear and
> I am using 30 permutations (code be
Probably not the most elegant, but a workable solution. Assume you have a
matrix x of dimensions 10 x 10. Assume further you want to calculate the
mean for each successive block of two columns. One way to do this is to
create a matrix that indicates the column numbers from/to which to apply the
fun
On Jul 12, 2011, at 4:37 PM, Juliet Hannah wrote:
I have a file in stata format, which I have read in, and I am trying
to create a text file. I have exported the data using various
delimiters, but I'm unable to read it back in. I originally read in
the file with:
library(foreign)
myData <- rea
On Jul 12, 2011, at 3:21 PM, wwreith wrote:
I have 4 columns and 56 rows of made up data that I want to plot as
a series
of bar graphs. The idea is to create one bar graph for each of the 4
columns
using a for loop. I tried the following command in RStudio and when
I type x
in the console
Hi,
You actually did the loop correctly. The problem is that the graphs
were created very quickly so you only see the last one. One way
around this is to make R wait for user input. You can turn this on
and off for a particular device using:
par(ask = TRUE)
See ?par for details on this.
To c
On Jul 12, 2011, at 2:31 PM, Tom La Bone wrote:
Here is a worked example. Can you point out to me where in temp
rmean is
stored? Thanks.
It is not. You need to read the ?print.survfit page:
Value
x, with the invisible flag set to prevent printing. (The default for
all print functions i
Not that I know of, but the paper says that they are easy to compute. If you
did, you could contribute the code.
Best,
Daniel
David Hugh-Jones-3 wrote:
>
> Hi all,
>
> Is there any code to run fixed effects Tobit models in the style of Honore
> (1992) in R?
> (The original Honore article is he
I have a file in stata format, which I have read in, and I am trying
to create a text file. I have exported the data using various
delimiters, but I'm unable to read it back in. I originally read in
the file with:
library(foreign)
myData <- read.dta("mydata.dta")
I then exported it with write.tab
Hi Peter and Raquel
I am following the same tutorial and seem to have the same error appear and
I am using 30 permutations (code below). Is it a bug or something that I can
easily fix? I'm not quite sure how to interpret the error.
multiExpr = list(A1=list(data=t(ctl)),A2=list(data=t(sz)))
multiC
I have 4 columns and 56 rows of made up data that I want to plot as a series
of bar graphs. The idea is to create one bar graph for each of the 4 columns
using a for loop. I tried the following command in RStudio and when I type x
in the console I get just the 4th graph instead of all four graphs.
Hello all,
I am using AddHealth data to fit a cure, aka split population model using nltm.
I am not sure how to account for the complex survey design - does anyone have
any suggestions? Any help would be greatly appreciated!
Sincerely,
Sam
__
R-help
Here is a worked example. Can you point out to me where in temp rmean is
stored? Thanks.
Tom
> library(survival)
> library(ISwR)
>
> dat.s <- Surv(melanom$days,melanom$status==1)
> fit <- survfit(dat.s~1)
> plot(fit)
> summary(fit)
Call: survfit(formula = dat.s ~ 1)
time n.risk n.event survi
On Tue, Jul 12, 2011 at 1:17 PM, Sam Steingold wrote:
> when do I need to use which()?
See ?which
For examples, try:
example(which)
>> a <- c(1,2,3,4,5,6)
>> a
> [1] 1 2 3 4 5 6
>> a[a==4]
> [1] 4
>> a[which(a==4)]
> [1] 4
>> which(a==4)
> [1] 4
>> a[which(a>2)]
> [1] 3 4 5 6
>> a[a>2]
> [1] 3
On Jul 12, 2011, at 4:17 PM, Sam Steingold wrote:
when do I need to use which()?
a <- c(1,2,3,4,5,6)
a
[1] 1 2 3 4 5 6
a[a==4]
[1] 4
a[which(a==4)]
[1] 4
which(a==4)
[1] 4
a[which(a>2)]
[1] 3 4 5 6
a[a>2]
[1] 3 4 5 6
seems unnecessary...
It is unnecessary when `a` is a toy ca
Well ...
which(a==4)^2
??
-- Bert
On Tue, Jul 12, 2011 at 1:17 PM, Sam Steingold wrote:
> when do I need to use which()?
>> a <- c(1,2,3,4,5,6)
>> a
> [1] 1 2 3 4 5 6
>> a[a==4]
> [1] 4
>> a[which(a==4)]
> [1] 4
>> which(a==4)
> [1] 4
>> a[which(a>2)]
> [1] 3 4 5 6
>> a[a>2]
> [1] 3 4 5 6
>>
>
To answer your questions:
Yes, yes, and probably no. You will have to pick up any introductory manual
of R where questions 1 and 2 will be discussed.
For 1: you index x as in x[452,682]. For 2: there are ways to write (and
avoid) loops in R (e.g. for or while loops). Often avoidance is preferable
when do I need to use which()?
> a <- c(1,2,3,4,5,6)
> a
[1] 1 2 3 4 5 6
> a[a==4]
[1] 4
> a[which(a==4)]
[1] 4
> which(a==4)
[1] 4
> a[which(a>2)]
[1] 3 4 5 6
> a[a>2]
[1] 3 4 5 6
>
seems unnecessary...
--
Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X
11.0.60900031
htt
Carson Farmer gmail.com> writes:
>
> Dear list, I'm wondering if anyone can help me calculate the deviance
> of either a zeroinfl or hurdle model from package pscl?
> Even if someone could point me to the correct formula for calculating
> the deviance, I could do the rest on my own.
What a
On Jul 12, 2011, at 3:41 PM, a217 wrote:
Hello,
I have a sample file:
chr22 100 150 125 21 0.145 +
chr22 200 300 212 13 0.05+
chr22 345 365 351 12 0.09+
chr22 500 750 510 15 0.10+
chr22 500 750
Hello:
R has an extensive Help system. Please learn to use it.
?histogram
?help
Also see the online manual tutorial "An Introduction to R"
-- Bert
On Tue, Jul 12, 2011 at 12:41 PM, a217 wrote:
> Hello,
>
> I have a sample file:
>
> chr22 100 150 125 21 0.145 +
> chr22 2
Hello,
I have a sample file:
chr22 100 150 125 21 0.145 +
chr22 200 300 212 13 0.05+
chr22 345 365 351 12 0.09+
chr22 500 750 510 15 0.10+
chr22 500 750 642 9 0.02+
chr22 800 90
Merging two posts (data and questions); see inline below.
On 7/11/2011 7:55 PM, Sigrid wrote:
Thank you, Dennis.
This is my regenerated dput codes. They should be correct as I closed off R
and re-ran them based on the dput output.
NB, this is the test dataset used later
structure(list(year
On 12-Jul-11 17:18:26, mousy0815 wrote:
> Probability <- function(N, f, m, b, x, t) {
> #N is the number of lymph nodes
> #f is the fraction of Dendritic cells (in the correct node) that
have
> the
> antigen
> #m is the number of time steps
> #b is the starting position (som
Dear list, I'm wondering if anyone can help me calculate the deviance
of either a zeroinfl or hurdle model from package pscl?
Even if someone could point me to the correct formula for calculating
the deviance, I could do the rest on my own.
I am trying to calculate a pseudo-R-squared measure based
Dear Susie,
See inline for some suggestions, but generally, I think you would
benefit from breaking this down into smaller pieces. The error you
are getting indicates the problem has to do with the plotting, but
that will be trickier to isolate while also dealing with reading in
data, looping, et
Hello, I'm new to this list. Sorry if my question or parts of it already came
up before.
For my research in geostatistics, I am working with large sets of data in R
(basically large matrices containing discrete x and y coordinates and a
value for a certain parameter). These sets are obtained by kr
It works well. Thanks so much.
--
View this message in context:
http://r.789695.n4.nabble.com/as-numeric-tp3661739p3662671.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/l
Hi,
I'm currently trying to calculate local Getis-Ord Gi* statistics for a
169x315 cell matrix of temperature values, below is the code I currently
have (diffc is the data vector I am removing NaN values from, and I am
moving said values to diffD; -999 represents NaN values; id contains ID
values
Probability <- function(N, f, m, b, x, t) {
#N is the number of lymph nodes
#f is the fraction of Dendritic cells (in the correct node) that have
the
antigen
#m is the number of time steps
#b is the starting position (somewhere in the node or somewhere in the
gap
b
I've written out codes for one particular file, and now I want to generate
the same kind of graphs and files for the rest of similar data files.
For example, a file "8.csv" would look like such:
enc_callee inout o_duration type
A out 342 de
B in 234 de
C
On 7/7/2011 3:23 PM, elephann wrote:
Hi everyone!
I have a data frame with 1112 time series and I am going to randomly
sampling r samples for z times to compose different portfolio size(r
securities portfolio). As for r=2 and z=1,that's:
z=1
A=seq(1:1112)
x1=sample(A,z,replace =TRUE)
x2=s
## Hello.. I have asked a similar question, but this is not fixed as before.
## I am running the following using Ubuntu OS:
R version 2.13.1 (2011-07-08)
Copyright (C) 2011 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-linux-gnu (64-bit)
## when I do this:
Hi Vikas,
Here is one way:
df <- read.table("summary.txt", header = TRUE)
str(df)
df[, "total"] <- rowSums(df[, 3:6])
df[, 3:6] <- apply(df[, 3:6], 2, function(x) x / df[, "total"] * df[,
"new"] * 2)
> head(df)
V1V2 CaseA CaseC CaseG CaseT new total
1 10 135344109 0 024
Dear all,
I have a problem and it is very difficult for me to get a code.
I am reading a file(attached with this mail) using the code-
df=read.table("summary.txt",fill=T,sep="",colClasses = "character",header=T)
and dataframe df is like this-
V1V2 CaseA CaseC CaseG CaseT new
10 13
Hi Harold,
Many (most?) of the statistics function have a similar argument. I
suspect it is sort of to warn the user---you have to be explicit about
it rather than the program just silently removing or ignoring values
that would not work in the function called. I can think of one
example where I
On 12/07/2011 12:26 PM, Doran, Harold wrote:
This is just posed out of curiosity, (not as a criticism per se). But what is
the functional role of the argument na.rm inside the mean() function? If there
are missing values, mean() will always return an NA as in the example below.
But, is there e
In SQL, the default is to ignore NULL (equivalent to NA in R).
However, it can be dangerous to fail to verify how much data was actually used
in an aggregation, so the logic behind the default na.rm setting may be one of
encouraging the user to take responsibility for missing data.
-
This is just posed out of curiosity, (not as a criticism per se). But what is
the functional role of the argument na.rm inside the mean() function? If there
are missing values, mean() will always return an NA as in the example below.
But, is there ever a purpose in computing a mean only to recei
On 07/12/2011 09:53 AM, Heiman, Thomas J. wrote:
## define x and y
x= x<-crs[,9]#predictor variables
y= y<-crs[1:8,] #response variable
This cannot be correct. The response variable is a vector, while the
predictor variables form a matrix. You have the response variable
consisting
Hi Susie,
At a guess, there are no non-missing arguments to min or max.
But no, we can't help you. You haven't provided a minimal reproducible
example, and without knowing anything about your data it is impossible
for the list to offer any constructive suggestions.
The posting guide offers sugge
On 2011-07-12 07:03, Sam Steingold wrote:
[snip]
the totally unnecessary semi-colons)
then why are they accepted?
optional syntax elements suck...
They're accepted because they *can* be useful (multiple
statements on one line).
Is there *any* language that can *not* be abused?
Peter Ehlers
I've written out codes for one particular file, and now I want to generate
the same kind of graphs and files for the rest of similar data files.
When I plugged in these codes, R produced only one plot for the file
"eight", and it states my error(see below) I have edited and checked my
codes so man
Thanks Peter, Ted!
Best, Anirban
On Tue, Jul 12, 2011 at 4:54 AM, Ted Harding wrote:
> On 11-Jul-11 07:55:44, Anirban Mukherjee wrote:
>> Hi all,
>>
>> I wanted to mark the estimation sample: mark what rows (observations)
>> are deleted by lm due to missingness. For eg, from the original
>> exam
Hello,
In my lab we use a four parameter logistic fit model for our ELISA data
(absorbance values). We are currently testing the use of different solvents
and need to find a way to add a correlation value (such as an R squared or
something similar) so we can test different solvents in making this
If you don't need POSIXt types, as Gabor says don't use them. However, there
are good reasons to use them sometimes, and the most workable solution I have
found is to set your default timezone in R to a non-DST timezone before you
convert from character to POSIXct. This is dependent on your OS a
On Jul 11, 2011, at 9:16 PM, Steve Parker wrote:
> Hi there,
> I am using the RODBC library to connect to an Empress database.
> I have installed the ODBC data source with the server DNs number and port,
> and named the source "Trawl".
> It is the odbcDriverConnect that seems to have the problem,
On Tue, Jul 12, 2011 at 8:57 AM, B Laura wrote:
> Dear Gabor
>
> http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows&s=excel
> doesnt describe handling dates with daylight saving time issues.
>
Two references were given and its discussed in the R News article. It
was also mentioned ove
Hi:
Try the cast() function in the reshape package. Using d as the name of
your data frame,
library(reshape)
cast(d, Date ~ Category, value = 'Temperature')
Date A B C
1 2007102 16 17 18
HTH,
Dennis
On Tue, Jul 12, 2011 at 5:42 AM, anglor wrote:
> Hi,
>
> I have a data frame of about
Hi,
(i) As David suggested, please use `dput` to provide examples of data!
(ii) The nut of your problem is that you are giving lars an object
that it is not expecting. It wants a *matrix* for its `x` variable, as
you'll see in the help for ?lars.
So, as long as this expression:
R> is.numeric(x)
On Jul 12, 2011, at 8:42 AM, anglor wrote:
Hi,
I have a data frame of about 700 000 rows which look something like
this:
DateTemperature Category
2007102 16 A
2007102 17 B
2007102 18 C
but need it to be:
Date T
On Jul 12, 2011, at 10:12 AM, Heiman, Thomas J. wrote:
Hi,
Hopefully I got the formatting down.. I am trying to do a lasso
regression using the lars package with the following data (the data
files is in .csv format):
V1 V2 V3 V4
Hi Trying,
It would be helpful if you provided reproducible examples. It would
also be polite to sign a name so that we have something by which to
address you.
On Tue, Jul 12, 2011 at 8:00 AM, Trying To learn again
wrote:
> Hi all,
>
> I have this information on a file ht.txt, imagine it is a da
On Tue, 2011-07-12 at 10:12 -0400, Heiman, Thomas J. wrote:
> Hi,
>
> Hopefully I got the formatting down.. I am trying to do a lasso regression
> using the lars package with the following data (the data files is in .csv
> format):
>
> V1 V2 V3
On Jul 12, 2011, at 10:12 AM, Sam Steingold wrote:
I have two data frames:
str(ysmd)
'data.frame': 8325 obs. of 6 variables:
$ X.stock : Factor w/ 8325 levels
"A","AA","AA-",..: 2702 6547 4118 7664 7587 6350 3341 5640 5107
7589 ...
$ market.cap :
Dear All,
I have a collections of spatial data. I have to analyze pairs of these
point patterns to test their spatial interaction. I was moving towards
the cross K Ripley's function. The problem, however, are the following:
1) What is the best way to get a single real value that represents the
Hi,
I have a data frame of about 700 000 rows which look something like this:
DateTemperature Category
2007102 16 A
2007102 17 B
2007102 18 C
but need it to be:
Date TemperatureA TemperatureB TemperatureC
2007102
Hi,
I am trying to do a lasso regression using the lars package with the following
data (see attached):
FastestTime
WinPercentage
PlacePercentage
ShowPercentage
BreakAverage
FinishAverage
Time7Average
Time3Average
Finish
116.90
0.14
0.14
0.29
4.43
3.29
117.56
117.77
5.00
116.23
Hi all,
I have this information on a file ht.txt, imagine it is a data frame without
labels:
1 1 1 8 1 1 6 4 1 3 1 3 3
And on other table called "pru.txt" I have sequences similar this
4 1 1 8 1 1 6 4 1 3 1 3 3
1 6 1 8 1 1 6 4 1 3 1 3 3
1 1 1 8 1 1 6 4 1 3 1 3 3
6 6 6 8 1 1 6 4 1 3 1 3 3
I want
Many Thanks¡¡¡
I will try this night, I have read this I think could help me.
I´m conscient the question was badly formulated now, I will try to explain
better next time¡¡¡
On a side note: apply always accesses the function you use at least once. If
the input is a dataframe without any rows but
If you switch directly to the multicore package you can use the
mclapply() function. There, check for the parameter mc.preschedule=T /
F. You can use this parameter to improve the load balancing.
I do not know a parameter to tune foreach with this parameter.
Best
Markus
Am Dienstag, den 12.07
How can perform logarithmic binning in the scatterplot? I could only take the
log of the variables and plot them, but I am sure that is not the way. I
have a very huge data, and would want to plot those high density
scatterplots and code then with different colors for the bins/density.
--
View thi
I have two data frames:
> str(ysmd)
'data.frame': 8325 obs. of 6 variables:
$ X.stock : Factor w/ 8325 levels "A","AA","AA-",..: 2702
6547 4118 7664 7587 6350 3341 5640 5107 7589 ...
$ market.cap : num -1.00 2.97e+10 3.54e+08 3.46e+08 -1.00
...
$ X52.
Hi,
Hopefully I got the formatting down.. I am trying to do a lasso regression
using the lars package with the following data (the data files is in .csv
format):
V1 V2 V3 V4
V5 V6 V7
On Jul 12, 2011, at 9:53 AM, Heiman, Thomas J. wrote:
Hi,
I am trying to do a lasso regression using the lars package with the
following data (see attached):
Nothing attached. (And now you have also sent an exact duplicate.)
snipped failed attempt to include data inline that was sabotaged
> * David Winsemius [2011-07-11 18:16:25 -0400]:
>
> What is the point of offering this code?
To illustrate what I was talking about (code is its own specification).
I hoped that there was already a package doing that (and more in that
direction).
> It seems to be doing what you want
yes.
> Ar
Hi,
I am trying to do a lasso regression using the lars package with the following
data:
FastestTime
WinPercentage
PlacePercentage
ShowPercentage
BreakAverage
FinishAverage
Time7Average
Time3Average
Finish
116.90
0.14
0.14
0.29
4.43
3.29
117.56
117.77
5.00
116.23
0.29
0.43
0
I just realised that:
apply(matrix(1:dim(my.data)[2], nrow =3), 2,
function(x){my.function(my.data[,x])})
is the simplest possible method.
bw
F
On 12 Jul 2011, at 14:44, Filipe Leme Botelho wrote:
> Hi Frederico. I would keep the data as it is, create two small vectors
> referring to the r
Hi,
I am trying to do a lasso regression using the lars package with the following
data (see attached):
FastestTime
WinPercentage
PlacePercentage
ShowPercentage
BreakAverage
FinishAverage
Time7Average
Time3Average
Finish
116.90
0.14
0.14
0.29
4.43
3.29
117.56
117.77
5.00
116.23
Eik,
Thanks very much!
Scott
On Tuesday, July 12, 2011 at 8:34 AM, Eik Vettorazzi wrote:
> Hi,
> file.info (http://file.info)()
> does that.
>
> Cheers
>
> Am 12.07.2011 15:29, schrieb Scott Chamberlain:
> > Hello,
> >
> > Are there any built in or user defined functions for printing the d
Hi Frederico. I would keep the data as it is, create two small vectors
referring to the ranges and use a mapply (as a sapply but with multiple
variables) for the function. Hope the example below is helpful, although as
usual someone out there will have a better solution for it.
> dta <- c()
> f
On 12 July 2011 12:27, Mitra, Sumona wrote:
> Dear all,
>
> I am new to programming in R.
>
You sure are ;-)
I deal with microarray data,which is a data frame object type. I need to
> carry out a few statistical procedures on this, one of them being the
> pearson corelation. I need to do this
Hi,
file.info()
does that.
Cheers
Am 12.07.2011 15:29, schrieb Scott Chamberlain:
> Hello,
>
> Are there any built in or user defined functions for printing the date
> created or date updated for a given file? Ideally a function that works
> across operating systems.
>
>
> Thanks!
> Scott
On Jul 12, 2011, at 7:27 AM, Mitra, Sumona wrote:
Dear all,
I am new to programming in R.
You see to think there is a "++" operation in R. That is not so.
I deal with microarray data,which is a data frame object type. I
need to carry out a few statistical procedures on this, one of them
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