Many thanks.
Best wishes,
os
-Original Message-
From: p_conno...@slingshot.co.nz [mailto:p_conno...@slingshot.co.nz]
Sent: 07 July 2011 01:02
To: Ouattara
Cc: r-help@r-project.org
Subject: Re: [R] help with nprmpi
Quoting Ouattara :
> Dear
>
> I have been trying to a program which requi
Hi all,
Is it possible to test mediation effects with a censored outcome
variable using the 'mediation' package?
While on p. 1f. of Imai, Keele, Tingley and Yamamoto (2011) the tobit
model is mentioned for use with 'mediate', Table 1 (p. 8) shows that
tobit via vglm cannot be used to estimat
On Wed, 2011-07-06 at 14:50 +0100, Sarah Leclaire wrote:
> Hi,
>
> I would like to calculate the RELATIVE euclidean distance. Is there a
> function in R which does it ?
>
> (I calculated the abundance of 94 chemical compounds in secretion of
> several individuals, and I would like to have the c
On Jul 6, 2011, at 23:50 , David Winsemius wrote:
>
> On Jul 6, 2011, at 4:43 PM, Data Analytics Corp. wrote:
>
>> Hi,
>>
>> Suppose I have two vectors, not necessarily the same length (in fact, they
>> usually are different lengths): y.1 that has increasing values between 0 and
>> 1; y.2 th
Hello Stacey,
I do not know whether my answer comes late or not, just came across
your post now. I had a similar problem...
First: You might want to think about whether to try to parallelize the
thing or not. Unless coxph takes several minutes, it is probably of no
great help to parallelize it, b
On Jul 7, 2011, at 05:01 , Thomas Lumley wrote:
> On Thu, Jul 7, 2011 at 12:58 PM, Jim Silverton
> wrote:
>> Dear all,
>>
>> I want to simulate from the null distribution of the following 2 x 3 table,
>>
>> 2 5 10
>> 4 8 5
>>
>> I am using a chi-squared test.
>> Anyone has any idea h
I am working on a system to visualize survey responses. Survey responses
typically include factors, numeric, timestamps, textfields and therefore fit
perfectly nice in dataframes, making it easy to visualize using standard R
functions.
However I am currently working on a survey that also include q
Hi,
I did a linear regression with 5 explanatory variables. Now, to see the
contribution of each variable, I use varpart from vegan library. But
varpart don’t accepts my 5 explanatory variables, it accept only 4.
1- How must I do to use my 5 explanatory variables?
2- Is it the sum of variance f
Thanks all!
I will try this later on today...
Omer
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On 07/07/11 08:43, Data Analytics Corp. wrote:
Hi,
Suppose I have two vectors, not necessarily the same length (in fact,
they usually are different lengths): y.1 that has increasing values
between 0 and 1; y.2 that has decreasing values between 1.0 and 0.
You can picture these as being suppl
Hi,
I am reading payment data like so
2010-01-01,100.00
2010-01-04,100.00
...
2011-01-01,200.00
2011-01-07,100.00
and plot it aggregated per month like so
library(zoo)
df <- read.csv("daily.csv", colClasses=c(d="Date",s="numeric"))
z <- zoo(df$s, df$d)
z.mo <- aggregate(z, as.yearmon, sum)
barp
On Wed, Jul 6, 2011 at 9:43 PM, Data Analytics Corp.
wrote:
> close enough. I can't figure out how to find the price value at which the
> two curves intersect. Going back to the economics interpretation, I want
> the price where supply equals demand. Any suggestions as to how I can find
> that
Dear list,
In a function, I don't care if my input has class 'integer' or
'numeric', so I wanted to use 'inherits()' to control for that.
However, this function tells me that an actual object of class 'integer'
does not inherit from class 'numeric'. The class def of 'integer' does
state 'num
Hi
>
> Hi,
>
> I am reading payment data like so
>
> 2010-01-01,100.00
> 2010-01-04,100.00
> ...
> 2011-01-01,200.00
> 2011-01-07,100.00
>
> and plot it aggregated per month like so
>
> library(zoo)
> df <- read.csv("daily.csv", colClasses=c(d="Date",s="numeric"))
> z <- zoo(df$s, df$d)
> z.mo
lapply(rg2, function(x) x$par.ests)
there is no slot residuals! The function gev() does return a S3-object with
class attribute 'gev', see ?gev.
>
> Dr. Pfaff:
>
> After using str; can you give an example on data extration
> (e.g. for $par.ests and @residuals)
>
>
>
> - Original Mess
Petr,
Maybe I did not make it clear, I apologize for that:
I want January to December on the X Axis (as 12 discrete (months))
and then for each month the values for each year as bars in
different colors next to each other, ie Jan-2009, Jan-2011,
Jan-2011...Dec-2009, Dec-2011, Dec-2011 whereas at
Hi,
I am new to R. Does anyone have a perl-style regular expression for use with
the R str_match_all() function for parsing the Combined Log Format [1] that
is commonly used by Apache and other web servers to log data.
Using Google I could find regexs written for many different languages, but I
w
> -Original Message-
> > I would like to calculate the RELATIVE euclidean distance.
> Is there a
> > function in R which does it ?
> >
>
> A simple solution to this is to transform the data and then
> compute the Euclidean distance using dist().
>
> decostand(foo, method = "normalize"
Thanks, Josh!
The index variable (time) was my problem. My R skills are too low! :)
Problem solved!
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Július 7-től 14-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 7 July till 14 July with no access to my
emails.
In urgent cases please contact Ms. Ed
Dear R helpers,
I am not a statistician and right now struggling with Richards curve. Wikipedia
says
(http://en.wikipedia.org/wiki/Generalised_logistic_function)
The "generalized logistic curve or function", also known as Richard's curve is
a widely-used and flexible sigmoid function for growt
OK
> Petr,
>
> Maybe I did not make it clear, I apologize for that:
>
> I want January to December on the X Axis (as 12 discrete (months))
> and then for each month the values for each year as bars in
> different colors next to each other, ie Jan-2009, Jan-2011,
> Jan-2011...Dec-2009, Dec-2011,
I am looking for the implementation of sparse kernel regression
approach e.g. as in this paper:
The Generalized LASSO. Volker Roth
IEEE Transactions on Neural Networks, Vol. 15, NO. 1, January 2004.
I would appreciate any pointers.
Best regards,
Ryszard
---
2 comments below.
On 07/07/2011 06:00 AM, r-help-requ...@r-project.org wrote:
> Date: Wed, 6 Jul 2011 20:39:19 -0700 (PDT)
> From: EdBo
> To: r-help@r-project.org
> Subject: Re: [R] loop in optim
> Message-ID: <1310009959045-3650592.p...@n4.nabble.com>
> Content-Type: text/plain; charset=us-ascii
On Thu, Jul 7, 2011 at 5:41 AM, Dr Eberhard Lisse wrote:
> Hi,
>
> I am reading payment data like so
>
> 2010-01-01,100.00
> 2010-01-04,100.00
> ...
> 2011-01-01,200.00
> 2011-01-07,100.00
>
> and plot it aggregated per month like so
>
> library(zoo)
> df <- read.csv("daily.csv", colClasses=c(d="D
I had a similar survey, and ended up stuffing everything into one field
using the bitops library.
On Thu, Jul 7, 2011 at 4:18 AM, jeroen00ms wrote:
> I am working on a system to visualize survey responses. Survey responses
> typically include factors, numeric, timestamps, textfields and therefore
I have to construct a vector of date with a cycle "for". I use the function
"seq", but when I allocate in a vector, this becomes a number!!!
How do I have? thank you
Example:
dataval=as.Date("2011/07/01")
date_val=seq(dataval,length=260,by="-7 day")
date_inizio=c()
date_con
Vincy Pyne yahoo.ca> writes:
> Dear R helpers, I am not a statistician and right now struggling
> with Richards curve. Wikipedia says
> (http://en.wikipedia.org/wiki/Generalised_logistic_function) The
> "generalized logistic curve or function", also known as Richard's
> curve is a widely-used an
You are storing the results in to a vector that is converted to
numeric; that is why you see the numbers. Try this:
> dataval=as.Date("2011/07/01")
> date_val=seq(dataval,length=260,by="-7 day")
> date_inizio=c()
> date_condizione=c()
> for (k in 1:length(date_val)){
+ date_inizio[
Dear Josh,
thanks for pointing this out - the idea behind writing this function is
plotting gradients on branches of phylogenetic trees - 'tree' refers to
a phylogenetic tree. It's easy to create a random phylogenetic tree in R:
library(ape)
library(plotrix)
rtree(15) -> tree
This gives you
Hi,
I am currently a new user in R and was working on the randomForest package. I
am trying to predict price points using this statistical package. The issue is
that I need to setup a tool so that I can give it to Sales Executive who can
plug in the necessary variables and get the output. Is th
Hi,
I've a data frame like this:
> as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
V1 V2
1 -1.30849402 -0.52094136
2 0.96157302 0.76217871
3 -0.44223351 -1.72630871
4 -0.10432438 -1.04732942
5 -1.38748914 0.95877311
6 -0.63965975 0.65494811
7 -0.24058318 0.19496830
On Jul 7, 2011, at 6:01 AM, Janko Thyson wrote:
Dear list,
In a function, I don't care if my input has class 'integer' or
'numeric', so I wanted to use 'inherits()' to control for that.
However, this function tells me that an actual object of class
'integer' does not inherit from class '
Give them an Excel spreadsheet that they can fill in the values. They
can then send the spreadsheet to you can you can have your R script
read the information from it and send it back. You did not mention
how they are supposed to"get the output". Do you want to setup a
central server that can re
On 07.07.2011 16:09, David Winsemius wrote:
On Jul 7, 2011, at 6:01 AM, Janko Thyson wrote:
Dear list,
In a function, I don't care if my input has class 'integer' or
'numeric', so I wanted to use 'inherits()' to control for that.
However, this function tells me that an actual object of cla
Hi
divide and conquer
Split your data frame to required portions
dat<-as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
dat.s<-split(dat, rep(1:4, each=3))
apply lm to each portion
sapply(dat.s, function(x) coef(lm(V2~V1, x)))
1 2 3 4
(Intercept) 1.2
justin bem yahoo.fr> writes:
>
> When try to use install.package I have this error message :
>
> Error in m[, 1L] : incorrect dimensions number
>
> What the matter with my installation ? I run R2.13 on Window 7 32bits.
>
We have absolutely no idea. This is not enough information
for us t
Dear R People:
Here is some output from AR and ARIMA functions:
> xb <- arima.sim(n=120,model=list(ar=0.85))
> xb.ar <- ar(xb)
> xb.ar
Call:
ar(x = xb)
Coefficients:
1
0.6642
Order selected 1 sigma^2 estimated as 1.094
> xb.arima <- arima(xb,order=c(1,0,0),include.mean=FALSE)
> xb.arima
Dear all,
I have a input file like following :
T
TTTAG
TTAAC
GGATT
ACGTA
How can I make a single vector with this like
following: AGTTAACGGATTACGTA
Best regards
Albert
[[alternative HTML version deleted]]
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R-help@r-project.org
On Thu, Jul 7, 2011 at 4:37 PM, albert coster wrote:
> Dear all,
>
> I have a input file like following :
>
> T
> TTTAG
> TTAAC
> GGATT
> ACGTA
>
> How can I make a single vector with this like
> following: AGTTAACGGATTACGTA
>
?paste, specifically the collapse argument
Rainer
>
> B
Dear R users,
I use "fitted" function in JM package to get fitted marginal survival
function for each subject and then plot marginal survival function for the
event process (x-axis: time, y-axis: survival). Is it possible that I could
get pointwise 95%CI for this survial function?
Thanks,
Kate
WARNING: The following might be **complete baloney** (and my apologies if so).
Erin:
I hope you get a definitive reply on this from a real expert, but if
memory serves, they might be using two different estimation
algorithms. ar() is just doing Yule-Walker recursive calculation as
described in Box
Dear Annemarie,
Look at what you are passing to your function:
lapply(tree$edge, print)
I am guessing you want to be passing:
lapply(1:nrow(tree$edge), print)
so that you are using each row of tree$edge. Also, take a look at the
code below for some examples of ways you can simplify (and vastl
> x <- readLines(textConnection("T
+ TTTAG
+ TTAAC
+ GGATT
+ ACGTA"))
> closeAllConnections()
> paste(x, collapse = '')
[1] "AGTTAACGGATTACGTA"
>
On Thu, Jul 7, 2011 at 10:37 AM, albert coster
wrote:
> Dear all,
>
> I have a input file like following :
>
> T
> TTTAG
> TTAAC
> GGA
This indeed seems to be the case.
Running
ar(xb, order=1, method="mle")
and
arima(xb,order=c(1,0,0),include.mean=FALSE)
give essentially the same results.
It looks to me that ar with method="mle" turns around and calls arima
function, so there is no big surprize there.
Cheers,
Hi there, I have to extract some relevant portion from a defined string,
which is a mix of numeric and character. However this has following
sequence:
Some String - Some numerical - "c/C" (or "p/P") - then again some set of
numbers.
Examples of such string is "fdahsdfcha163517253c463278643"
On Jul 7, 2011, at 11:21 AM, Bogaso Christofer wrote:
> Hi there, I have to extract some relevant portion from a defined string,
> which is a mix of numeric and character. However this has following
> sequence:
>
>
>
> Some String - Some numerical - "c/C" (or "p/P") - then again some set of
> n
On Jul 7, 2011, at 10:25 AM, Janko Thyson wrote:
On 07.07.2011 16:09, David Winsemius wrote:
On Jul 7, 2011, at 6:01 AM, Janko Thyson wrote:
Dear list,
In a function, I don't care if my input has class 'integer' or
'numeric', so I wanted to use 'inherits()' to control for that.
However
Thanks Marc for your reply and detailed explanation. As you said, I also
agree that, using stringr package I wont get anything really important,
however I already have created a long code-book and now I do not want to
change anything. However function names are here better meaningful.
I have one m
Happy to help.
Your interpretation is correct on the use of "\\1". This returns the value
contained in the first back reference in the regex. If you wanted to return
multiple back references, these would be "\\2", "\\3" and so on, each referring
to successive paren pairs in the regex. Note the
Dear forum members,
hope you can understand my unprofessionell English. I have never been
working with RStudio before (even not with R), so I immediately need some
help, because I want to carry out an automated multiple stepwise linear
regression between temperatures and different surface paramete
Hi,
I have a dataset which has var1 from 1 sourse and var2 from 2 different
methods. I need a new variable such that var2 values from both methods can
beused as 1 variable.
I believe deming regression can be used to do this. I just don't know how to
do it.
My data looks like:
idvar1var2method1var
Hello,
I want to use lme to fit two (or more) models, and then compare the fits
on each individual. I know how to write my own code to do this (for each
individual, plot the raw data, followed by lines() to plot each fitted
curve) but I would like to use plot(augPred(... as it produces a nice
Hi there
I'm working with ncdf data. I have different dataset for 4 runs, 4 seasons
and 3 timeslices (48 datasets in total). The datasets have the following
dimensions: 96 longitudes, 48 latitudes and 30 time steps. To read all of
them in, I wrote the following loop:
runs <- c("03","04","05","06"
HI everyone, I'm just starting to get into graphing with R and I need to
generate one graph that illustrates the pattern of gene expression for
various patients. My data is in .csv format and is as follows and i'm
showing below a portion of the data.
Pt Coordinate Log2Ratio
1 34046
On Jul 7, 2011, at 11:28 AM, klimator wrote:
Dear forum members,
hope you can understand my unprofessionell English. I have never been
working with RStudio before (even not with R), so I immediately need
some
help, because I want to carry out an automated multiple stepwise
linear
regressi
--- Begin Message ---
For sure some smart cookie has a much better solution but this can do
the trick
dt <- as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
regs <- sapply(seq(1,9,by=4), function(x) coef(lm(dt[(x:(x+3)),1] ~
dt[(x:(x+3)),2])))
tmp <- c()
for (i in 1:ncol(regs)) {
tmp <-
Yes, ar and arima are using different estimation methods: arima is mle
whereas the default is method-of-moments.
With such a large ar coefficient the end effects will matter, and the
mle (done by arima or ar.mle or ar(method="mle")) is the more accurate
method since it makes maximal use of the
Dear R Users,
I'm having trouble reproducing the results in Section 5.1 of
Culp, M., Johnson, K., Michailidis, G. (2006). ada: an R Package for
Stochastic Boosting Journal of Statistical Software, 16
They build and display a boosting model with the code:
library("ada")
n <- 12000
p <- 10
set.se
On Thu, Jul 7, 2011 at 9:10 AM, confused wrote:
> Hi there
>
> I'm working with ncdf data. I have different dataset for 4 runs, 4 seasons
> and 3 timeslices (48 datasets in total). The datasets have the following
> dimensions: 96 longitudes, 48 latitudes and 30 time steps. To read all of
> them i
Dear All,
I am trying to analysis traffic data with one timestamp column and speed
column. This data set contains six years data; for each year, I want to make
a matrix in (day of the year) * (hour)
0 1 2 . . . 23
1
2
.
.
365
However random day's record is missing(e.g. there are
Hi, I'm new to R. I'm trying to do some extreme value theory analysis,
looking at the Mean Excess Plot of a series. These are the commands I type
to get the plot:
x=read.table("data.txt",header=T)
goa=(x[,3])
meplot(goa)
I can see the plot, but I would like to see the values of the x and y axis.
Hi! I fairly new to R, have only done pretty basic things so far, so this may
be a very basic question. But I did search the forums and didn't see a
solution
I'm trying to get going with kripp.alpha(). I'm loading data from a file,
like this:
> library(irr)
Loading required package: lpSol
How does the R module ARIMA account for unspecified deterministic structure
such as seasonal pulses, level shifts, local time trends and regular pulses
without needing to ask the user to intervene to specify this?
I have attached a Makradakis paper which hammers Box-Jenkins approach to
this probl
Dear all,
I am trying to use apply or similar functions in order to improve the
efficiency of my code but I guess I am misunderstanding the function of
these commands. What I want to do is possible to see with the following code
that I use a for loop command. In words what I want to do are two thi
On Jul 7, 2011, at 3:21 PM, xin123620 wrote:
Dear All,
I am trying to analysis traffic data with one timestamp column and
speed
column. This data set contains six years data; for each year, I want
to make
a matrix in (day of the year) * (hour)
0 1 2 . . . 23
1
2
.
.
365
Howev
Are there any packages with functions that can fit quasi-symmetry and
quasi-indepedence models to square contingency tables?
M. Laviolette
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On Jul 7, 2011, at 3:00 PM, francisco.ahued wrote:
Hi, I'm new to R. I'm trying to do some extreme value theory analysis,
looking at the Mean Excess Plot of a series. These are the commands
I type
to get the plot:
x=read.table("data.txt",header=T)
goa=(x[,3])
meplot(goa)
I can see the plot
Hi,
After performing a multiple linear regression,
I am looking for an R package that can calculate the fraction [a] a
partitioning of variation. This fraction measures the proportion of variance
of y explained by the explanatory variable x1 (for example) when other
variables (x2, x3 ...) are hel
On Jul 7, 2011, at 2:54 PM, sderose wrote:
Hi! I fairly new to R, have only done pretty basic things so far, so
this may
be a very basic question. But I did search the forums and didn't
see a
solution
I'm trying to get going with kripp.alpha(). I'm loading data from a
file,
like
Make factors out of the dayOfYear and HourOfDay
columns and specify the levels you want. Then
pass the factors to tapply.
E.g., you may have something like
> d <- data.frame(Season=c(1,1,3,4),
Type=c(20,15,15,20),
Speed=11:14)
> tapply(d$Speed, d[c("Season","Ty
On Jul 7, 2011, at 19:52 , Prof Brian Ripley wrote:
> Yes, ar and arima are using different estimation methods: arima is mle
> whereas the default is method-of-moments.
>
> With such a large ar coefficient the end effects will matter, and the mle
> (done by arima or ar.mle or ar(method="mle"))
On Thu, 7 Jul 2011, peter dalgaard wrote:
On Jul 7, 2011, at 19:52 , Prof Brian Ripley wrote:
Yes, ar and arima are using different estimation methods: arima is
mle whereas the default is method-of-moments.
With such a large ar coefficient the end effects will matter, and
the mle (done by
Thank you William Your method does work I highly appreciate that. :D
Also thank you for replying, David. You are absolutely correct. I re-edited
my posting to make question clear according to the guidelines in case other
person meet the similar problem as I have.
Thank you,
Jessica
Thanks! How do I do that?
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Dear All,
When I use read.xls() in gdata package to read xls files, I noticed an issue
and couldn't find any solutions after I serched all previous posts.
In the excel file, the number value, for example, is actually 2.3456789 but
formatted as 2.3 (Format Cells ---> Decimal places:1). After I use
On Jul 7, 2011, at 3:52 PM, xin123620 wrote:
Thank you William Your method does work I highly appreciate
that. :D
Also thank you for replying, David. You are absolutely correct. I
re-edited
my posting to make question clear according to the guidelines in
case other
person meet
On Jul 7, 2011, at 3:41 PM, savage.arrow wrote:
Adding back context from earlier posting (WHICH IS YOUR DUTY
savage.arrow)
--
x=read.table("data.txt",header=T)
goa=(x[,3])
meplot(goa)
I can see the plot, but I would like to see the values of the x and y
axis.
According to this p
Hi lt2,
I would use the ggplot2 or lattice package. It strikes me as more
effort to do in traditional graphics. Anyway, here are some examples.
Lattice is a very nice package, but I am not quite as familiar with
it, so my examples for it are not representative of its full power.
Cheers,
Josh
Hello R users,
I have two data sets like the following. Form of dataset:
data:
X1 X2X3X4 X5
1902 RE3 5949
1903 RE3 13407
1904 AA3 760 14
1908 RE4 1
Thank you!
LT
On 7/7/11 3:46 PM, "Joshua Wiley" wrote:
Hi lt2,
I would use the ggplot2 or lattice package. It strikes me as more
effort to do in traditional graphics. Anyway, here are some examples.
Lattice is a very nice package, but I am not quite as familiar with
it, so my examples
Hi,
I am working on a ROC plot problem. Thanks in advance for any help.
We run 20 cross validation (CV) on our data for each customized threshold,
i.e. from 0.1 to 0.9.
Then for these 9 threshold values, we get the overall performance, that is
we are able to get overall sensitivity and specifi
On Jul 7, 2011, at 4:40 PM, Shant Ch wrote:
Hello R users,
I have two data sets like the following. Form of dataset:
data:
X1 X2X3X4 X5
1902 RE3 5949
1903 RE3 13407
1904 AA3 760
Hi:
Here's one way, using the latticeExtra package:
library(lattice)
library(latticeExtra)
p1 <- plot(augPred( fit.lme ), ylim=c(20,65))
p2 <- plot(augPred(update(fit.lme, y ~ log(age))), ylim=c(20,65),
col.line = 'red')
p1 + p2
HTH,
Dennis
On Thu, Jul 7, 2011 at 10:09 AM, R
On Jul 7, 2011, at 22:21 , Prof Brian Ripley wrote:
> On Thu, 7 Jul 2011, peter dalgaard wrote:
>
>>
>> On Jul 7, 2011, at 19:52 , Prof Brian Ripley wrote:
>>
>>> Yes, ar and arima are using different estimation methods: arima is mle
>>> whereas the default is method-of-moments.
>>>
>>> With
Hi,
I have the following data:
> est
sch190 sch107 sch290 sch256 sch287 sch130
sch139
4.16656026 2.64306071 4.22579866 6.12024789 4.49624748 11.12799127
1.17353917
sch140 sch282 sch161 sch193 sch156 sch288
sch352
3.
On Thu, Jul 7, 2011 at 4:23 PM, Victor11 wrote:
> Dear All,
>
> When I use read.xls() in gdata package to read xls files, I noticed an issue
> and couldn't find any solutions after I serched all previous posts.
>
> In the excel file, the number value, for example, is actually 2.3456789 but
> forma
You can easily do this by:
qplot(x=as.factor(sch),y=est, geom='point', colour='red') +
geom_pointrange(aes(x=as.factor(sch), y=est, ymin=lower.95ci, ymax=upper.95ci))+
xlab('School') + ylab("Value-added")+theme_bw()
On 07/07/2011 05:55 PM, Christopher Desjardins wrote:
Hi,
I have the followi
Thanks that worked perfectly. One thing if I may. Is it possible to make the
center dot red and the lines connecting the dots black?
Thanks,
Chris
On Jul 7, 2011, at 5:10 PM, Abhijit Dasgupta, PhD wrote:
> You can easily do this by:
>
> qplot(x=as.factor(sch),y=est, geom='point', colour='red'
Hi:
Here's another approach using the plyr package:
library(plyr)
df <- data.frame(gp = factor(rep(1:3, each = 4)), x = rnorm(12), y = rnorm(12))
mylst <- split(df, df$gp)
mycoefs <- ldply(mylst, function(d) coef(lm(y ~ x, data = d)))
names(mycoefs) <- c('gp', 'intercept', 'slope')
merge(df, myc
On Thu, Jul 7, 2011 at 5:24 PM, Dennis Murphy wrote:
> Hi:
>
> Here's another approach using the plyr package:
>
> library(plyr)
> df <- data.frame(gp = factor(rep(1:3, each = 4)), x = rnorm(12), y =
> rnorm(12))
> mylst <- split(df, df$gp)
> mycoefs <- ldply(mylst, function(d) coef(lm(y ~ x, da
tomreilly autobox.com> writes:
>
> How does the R module ARIMA account for unspecified deterministic structure
> such as seasonal pulses, level shifts, local time trends and regular pulses
> without needing to ask the user to intervene to specify this?
It doesn't.
> I have attached a Makrad
I am trying to use the family of apply functions on mpfr1-class
variables, but run into problems with the output. Consider a simple
example:
foo <- sapply(1:15,FUN=function(x) sin(mpfr(x,20)))
foo is a list, each element of which contains a mpfr1-class value.
What I don't understand is why u
On Jul 7, 2011, at 8:47 PM, Gang Chen wrote:
I define the following function to convert a t-value with degrees of
freedom
DF to another t-value with different degrees of freedom fullDF:
tConvert <- function(tval, DF, fullDF) ifelse(DF>=1, qt(pt(tval, DF),
fullDF), 0)
It works as expected wi
I define the following function to convert a t-value with degrees of freedom
DF to another t-value with different degrees of freedom fullDF:
tConvert <- function(tval, DF, fullDF) ifelse(DF>=1, qt(pt(tval, DF),
fullDF), 0)
It works as expected with the following case:
> tConvert(c(2,3), c(10,12)
On Jul 7, 2011, at 8:52 PM, David Winsemius wrote:
On Jul 7, 2011, at 8:47 PM, Gang Chen wrote:
I define the following function to convert a t-value with degrees
of freedom
DF to another t-value with different degrees of freedom fullDF:
tConvert <- function(tval, DF, fullDF) ifelse(DF>=1,
Thanks for the help! Are you sure R version plays a role in this case? My R
version is 2.13.0
Your suggestion prompted me to look into the help content of ifelse, and a
similar example exists there:
x <- c(6:-4)
sqrt(x) #- gives warning
sqrt(ifelse(x >= 0, x, NA)) # no warning
On Jul 7, 2011, at 10:17 PM, Gang Chen wrote:
Thanks for the help! Are you sure R version plays a role in this
case? My R version is 2.13.0
I'm not sure, but my version is 2.13.1
Your suggestion prompted me to look into the help content of ifelse,
and a similar example exists there:
It's basically a question of layering, and the order in which the layers are
drawn. Draw the pointranges first and then the points:
qplot(x=as.factor(sch), y=est, ymin=lower.95ci, ymax=upper.95ci,
geom='pointrange')+
geom_point(aes(x=as.factor(sch), y=est), color='red')+...
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