> On 21 Oct 2015, at 19:57 , Charles C. Berry wrote:
>
> On Wed, 21 Oct 2015, Ravi Varadhan wrote:
>
>> [snippage]
>
> If half the subjects have a value of 5 seconds and the rest are split between
> 4 and 6, your assertion that rounding induces an error of
> dunif(epsilon,-0.5,0.5) is surely
here is one thought:
if you plug in your numbers into any kind of regression you will get
prediction that are real numbers and not necessarily integers, it may be
that you predictions are good enough with this approximate value of Y. you
could test this by randomly shuffling your data by +- 0.5 an
This could be modeled directly using Bayesian techniques. Consider the
Bayesian version of the following model where we only observe y and X. y0
is not observed.
y0 <- X b + error
y <- round(y0)
The following code is based on modifying the code in the README of the CRAN
rcppbugs R package.
Dear R-help,
I am working on what it seems to be a simple problem, but after several
hours trying to come up with a solution, unfortunately I have not been able
to.
I would like to go from "datain" to "dataout", that is, create the NEWREF
variable according with some restrictions, and update the
Got it. During a read.csv I used stringsAsFactors=FALSE
M. Keith Cox, Ph.D.
Principal
MKConsulting
17105 Glacier Hwy
Juneau, AK 99801
U.S. 907.957.4606
On Wed, Oct 21, 2015 at 10:31 AM, Marlin Keith Cox
wrote:
> I do not have a dataset to share as I do not believe it needs it and I am
> not su
Hi Ravi,
Thanks for this interesting question. My thoughts are given below.
If you believe the rounding is indeed uniformly distributed, then the
problem is equivalent with adding a uniform random error between (-0.5,
0.5) for every observation in addition to the standard normal error, which
will
I do not have a dataset to share as I do not believe it needs it and I am
not sure I could reproduce easily.
I have a column of numerical data (Days) and another of a a measurement
(Resistance). After subsetting, I do a linear regression of the two, and
it reorders the day (x axis) into some other
On Wed, 21 Oct 2015, Ravi Varadhan wrote:
Hi, I am dealing with a regression problem where the response variable,
time (second) to walk 15 ft, is rounded to the nearest integer. I do
not care for the regression coefficients per se, but my main interest is
in getting the prediction equation fo
Hi,
I am dealing with a regression problem where the response variable, time
(second) to walk 15 ft, is rounded to the nearest integer. I do not care for
the regression coefficients per se, but my main interest is in getting the
prediction equation for walking speed, given the predictors (age,
To find the bin centers, I've tried gg2plot_build(), but the
center-coordinates it provides are not what's plotted. In the below example
there are several misplaced "o" symbols and some bins w/o a symbol on them.
library(ggplot2)
dat <- data.frame(x = rnorm(10, 6, 2), y = rnorm(10, 6, 2))
Yes, setting the fig.path option will prevent rmarkdown from deleting
the figure files, and the more natural way to preserve these
intermediate files is to set the rmarkdown option keep_tex or keep_md
(depending on your output format) to yes, e.g.
---
output:
pdf_document:
keep_tex: yes
ht
Thanks. replace also looks to work okay.
On Wednesday, October 21, 2015 3:43 PM, PIKAL Petr
wrote:
Hi
I wonder why you are asking that after quite a long use of R.
test[test < (-.5)] <- (-1)
Double loops seems to me the last resort in R if any other approach fails.
Cheers
Petr
Hi
I wonder why you are asking that after quite a long use of R.
test[test < (-.5)] <- (-1)
Double loops seems to me the last resort in R if any other approach fails.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Alaios
> via R-help
Dear all I have a table as that.
test<-matrix(data=rnorm(100),ncol=10)
and I want to find the values that are below my thresholdthreshold<- -0.5and
replace them with a -1 instead.
I can of course write a double nested for loop to check one by one elementif
(test[i,j]<= threshold) test[i,j]<- -
I think the default now is to not save them unless you set the fig.path chunk
option.
http://kbroman.org/knitr_knutshell/pages/Rmarkdown.html
---
Jeff NewmillerThe . . Go Live...
DCN
The calculation appears to be sum(a)/(sum(a)+sum(b)).
library(dplyr)
library(tidyr)
result <- ( this_df
%>% gather( group, truth, -c(a,b) )
%>% group_by( group, truth )
%>% summarise( calc = sum(a)/(sum(a)+sum(b)) )
%>% as.data.frame
)
-
The figures should be saved somewhere. e.g. if you have x.Rmd, you
should have a X_files/ folder with subfolders for the figures (e.g.
X-html or X-latex). At least that's what I have.
Bob
On 20 October 2015 at 18:18, Witold E Wolski wrote:
> I am running r-markdown from r-studio and can't work o
Hello all,
I've been banging my head over what must be a simple solution. I would
like to apply a function across columns of a dataframe for rows grouped
across different columns. These groups are not exclusive. See below
for an example. Happy to use dplyr, data.table, or whatever. Any
g
It may not be elegant but you can just embed a png() command in the knitr code.
Code from RStudio example with png() command added.
```{r, echo=FALSE}
plot(cars)
png("~/Rjunk/pnd.png")
plot(cars)
dev.off()
```
John Kane
Kingston ON Canada
> -Original Message-
> From: wewol...@gmail.c
Greetings R Community,
I am running quantile regressions using quantreg in R. I also plot the
residuals in a QQplot which indicate fat tails. I would like to try using
Student distribution, but I do not know if the R software allows it for my task
in hand.
In my opinion it is very likely that
Thanks!
On 10/21/2015 5:25 AM, Thierry Onkelinx wrote:
Dear Sebastien,
You are looking for geom_polygon().
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assuran
Dear Sebastien,
You are looking for geom_polygon().
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
To call in the
Hi
I would like to use ggplot2 to create a 2d plot showing a series of
shaded areas that are not continuous with respect to the x-axis
variable. The expected result is illustrated below using lattice/grid
functions.
-
pdata <- data.frame(
x=c(1,2,2,1,NA,3,4,4,3,NA,5,6,6,5),
y
thank you very much! Both the two methods work well for my data.
Best wishes,
Chunyu
At 2015-10-20 19:48:18, "William Dunlap" wrote:
>Or use aperm() (array index permuation):
> > array(aperm(x, c(2,1,3)), c(6,3))
> [,1] [,2] [,3]
> [1,]17 13
> [2,]4 10 16
>
Thanks a lot Petr for ur reply and advice. Hope I 'd be able to minmize time as
much as possible.
Regards,
Maram Salem
Sent from my iPhone
> On Oct 21, 2015, at 9:28 AM, PIKAL Petr wrote:
>
> Hi
>
> Several options
>
> Coding in C+ or other similar language (you seem to be more familiar wit
Hi
Several options
Coding in C+ or other similar language (you seem to be more familiar with them)
Using debug and find out how your function behaves in steps
Using function with smaller m, n with Rprof to see where the time is spent.
I believe that coding in R like it was C+ is the way to hel
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