First, a very easy question: What is the difference between using
what="character" and what=character() in scan()? What is the reason for
the character() syntax?
I am working with some character vectors that are up to about 27.5 million
elements long. The elements are always unique. Specif
On 16/01/2015 1:18 PM, Nafiseh Mehdipour wrote:
> Hi,
>
> I am writing regarding to my problem about using "plot" function. I am using
> "R x64 3.1.1" (installed on Windows 7 SP1 (laptop Sony VGN-SR) ), and when I
> use type this code:
>
>
>> x <- c(-30,30)
>> y <- c(-5,5)
>> plot.window(x , y
plot.window() is an internal support function, not a user function.
use
plot()
x <- c(-30,30)
y <- c(-5,5)
plot(x , y)
On Fri, Jan 16, 2015 at 4:18 PM, Nafise Mehdipour
wrote:
> Hi,
>
> I am writing regarding to my problem about using "plot" function. I am
> using "R x64 3.1.1" (installed on Win
Hi,
I am writing regarding to my problem about using "plot" function. I am
using "R x64 3.1.1" (installed on Windows 7 SP1 (laptop Sony VGN-SR) ), and
when I use type this code:
> x <- c(-30,30)
> y <- c(-5,5)
> plot.window(x , y)
or any other form of "plot" function, it doesn't show anything.
Hi all
Can someone assist me in creating a table in SQL Server using a Rdataframe
as well as updating it with new records
Here's my sample script in R.
library(RODBC)
#Connecting to Datum
myconn <-odbcConnect("Datum", uid="bihy", pwd="tarta") ### Server name is
Datum
Now I would like to save
Hi,
I am writing regarding to my problem about using "plot" function. I am using "R
x64 3.1.1" (installed on Windows 7 SP1 (laptop Sony VGN-SR) ), and when I use
type this code:
> x <- c(-30,30)
> y <- c(-5,5)
> plot.window(x , y)
or any other form of "plot" function, it doesn't show anything
Hello,
I have problems to label countries in my map.
I use the following code to draw the map:
epimap=joinCountryData2Map(epidata,joinCode="NAME",nameCountryColumn="Country",nameJoinColumn="Country")
mapCountryData(epimap, nameColumnToPlot="EPI.2.0" ,mapRegion="europe",
oceanCol="slateblue1",
Hi
I just started to study the "deepnet" package:
http://cran.r-project.org/web/packages/deepnet/index.html
It is about "deep leaning", so about the usage of multi-layer neural networks.
I've started to use the train() functions available in the package,
but I really cannot understand how to add m
On 1/16/15 9:34 AM, Bert Gunter wrote:
Chee Hee's approach is both simpler and almost surely more efficient,
but I wanted to show another that walks the tree (i.e. the list)
directly using recursion at the R level to pull out the desired
components. This is in keeping with R's "functional" progra
Hi R Users,
I would like to specify strip names in xyplot. I have 5 sources (S) and 3
locations (Loc). How I can force R to specifiy the strip names for each
source and location.
Below is my code;
S*Loc represents source by location. I do not know how to specify which
panel or factors.levels. I
Dear R Geniuses:
I'm a C++ and Perl, not an R System consultant, but a client wants me
to see if R can help him predict whether daily sales for some auto
parts stores will be less than, greater, or equal to the median daily
sales value.
(equal to is defined as within 2%, otherwise there would nev
Chee Hee's approach is both simpler and almost surely more efficient,
but I wanted to show another that walks the tree (i.e. the list)
directly using recursion at the R level to pull out the desired
components. This is in keeping with R's "functional" programming
paradigm and avoids the use of regu
You may use 'eval()' with 'parse()' if you wish to use 'formula()' as below:
> a <- params[[1]]
> b <- params[[2]]
> eval(parse(text=func1))
[1] 5.00 5.652893 6.338843 7.057851 7.809917 8.595041
[7] 9.413223 10.264463 11.148760 12.066116 13.016529 14.00
[13] 15.01
This approach may not be fancy as what you are looking for.
> xl <- unlist(x)
> xl[grep("A", names(xl))]
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
>
I hope this helps.
Chel Hee Lee
On 01/16/2015 04:40 AM, Rainer M Krug wrote:
Hi
Consider the following variable:
--8<---
Dear Renato,
There's not enough information here to diagnose the problem, but perhaps the
following will help:
First, I have no trouble loading the RcmdrPlugin.temis package, either directly
via library(RcmdrPlugin.temis), in which case it brings up the Rcmdr interface,
or indirectly via the R
Sorry, I see that the formatting of the e-mail went all wrong and was
completely unreadable. you can find a readable version in the attachment and
down below (if it will work this time).
--
Hi all,
I have a question about the arules packag
Hi to all,
i'm using R 3.1.2 under Mac, OSX version 10.10.1
i'm trying to launch Rcmdr but i receive an Error and i'm not able to fix,
could you help me?
# Error screen
> library(Rcmdr)
Carico il pacchetto richiesto: splines
Carico il pacchetto richiesto: RcmdrMisc
Carico il pacchetto richiesto:
Dear all,
Is there a possibility to perform in R wave length dependent calibration
analysis? I want to decompose a time series into high, low and band pass
components using a n-years smoothing spline.
Thank you very much!
Best regards!
--
---
Catalin-Constantin ROIBU
Lecturer PhD, Forestry eng
On Fri, Jan 16, 2015 at 3:16 AM, philippe massicotte
wrote:
> Hi all.
>
> How we evaluate a formula in R?
>
> Ex.:
>
> params <- list(a = 2, b = 3)
> x <- seq(1,10, length.out = 100)
>
> func1 <- as.formula("y ~ a*x^2 + b*x")
>
> ##How to evaluate func1 using x and the params list
> ???
>
>
> Than
>> Funnily, this problem disappears when I use RTools31.exe. And, I am not the
>> only one facing this issue. A lot of people in my group (in which we all are
>> learning R) are facing the same problem. I am really puzzled as to why
>> RTools32.exe isn't compatible with R 3.1.2 !!
>> Thanks agai
On 16/01/2015 3:55 AM, PRAMEET LAHIRI wrote:
> find_rtools() is a function of the 'devtools' package. Maybe it's an
> issue with that package and not R, and I'm pretty sure Duncan Murdoch
> put great efforts in asserting Rtools is working well with R (that's
> been my experience for the last 5-10
find_rtools() is a function of the 'devtools' package. Maybe it's an
issue with that package and not R, and I'm pretty sure Duncan Murdoch
put great efforts in asserting Rtools is working well with R (that's
been my experience for the last 5-10 years).
I also know that devtools 1.7.0 have been su
Hi
Consider the following variable:
--8<---cut here---start->8---
x1 <- list(
A = 11,
B = 21,
C = 31
)
x2 <- list(
A = 12,
B = 22,
C = 32
)
x3 <- list(
A = 13,
B = 23,
C = 33
)
x4 <- list(
A = 14,
B = 24,
C = 34
)
y1 <- list(
x1 =
Jeff Newmiller writes:
> It x is a list...
>
> do.call(fun,x)
Thanks - works perfectly. Only one further question:
In my original usage, fun() is an S3 Generic Function and takes as the
first argument an object which obviously *must not be* expanded, while
the ... need to be extended.
--8<
Hi all.
How we evaluate a formula in R?
Ex.:
params <- list(a = 2, b = 3)
x <- seq(1,10, length.out = 100)
func1 <- as.formula("y ~ a*x^2 + b*x")
##How to evaluate func1 using x and the params list
???
Thank you in advance,
Phil
[[alternati
25 matches
Mail list logo
