Hi all,
I am sure this is simple, but being a beginner in R,I am finding it
difficult to manage myself.
I am running the following code to get a multiple line plot for two time
series variables, x and y:
plot(as.ts(cbind(x,y)), plot.type = "single", col = 1:2)
I want to change the scales of both
Hi all:
If the explainary variables are ordinal,the result of regression is different
from
"unordered variables".But I can't understand the result of regression from
"ordered
variable".
The data is warpbreaks,which belongs to R.
If I use the "unordered variable"(tension):Levels: L M H
The res
See, e.g.
http://en.wikipedia.org/wiki/Double-precision_floating-point_format
Also in ?print.default please note:
Large number of digits
Note that for large values of digits, currently for digits >= 16, the
calculation of the number of significant digits will depend on the
platform's internal
Dear R-users:
Hi, I read here (
http://stackoverflow.com/questions/2287616/controlling-digits-in-r) that R
is only accurate up to the 15th decimal place, despite the fact that if you
choose to display more decimal places, it will. I wonder if R uses the
information beyond the 15th decimal place in
Hi,
May be this helps.
lines<- readLines(textConnection("NM_019397 // Egfl6 // EGF-like-domain,
multiple 6 // X F5|X 71.5 cM // 54156
NM_019397 // Egfl7 // EGF-like-domain, multiple 6 // X F5|X 71.5 cM // 54158"))
library(stringr)
word(lines,2,sep=" // ")
#[1] "Egfl6" "Egfl7"
lines1<- readLines(
You suggested
> lapply( lines, function(l) strsplit(l, " // ")[[1]][2] )
strsplit is vectorized so the following is equivalent but simpler and quicker:
lapply(strsplit(lines, " // "), function(x)x[2])
The OP probably wants a character vector, not a list so use sapply or vapply
(safer
than s
On May 20, 2013, at 4:45 PM, Hon Kit (Stephen) Wong wrote:
> Dear ALl,
>
> I hope you could help me out on this simple problem. I have many thousand
> lines like this:
> NM_019397 // Egfl6 // EGF-like-domain, multiple 6 // X F5|X 71.5 cM // 54156
>
> I want to extract the string inside the fir
Dear ALl,
I hope you could help me out on this simple problem. I have many thousand lines
like this:
NM_019397 // Egfl6 // EGF-like-domain, multiple 6 // X F5|X 71.5 cM // 54156
I want to extract the string inside the first // //, in this case is Egf16.
How do I apply grep function?
Thanks.
Hello,
You are fitting a vector other than the vector 'x'.
And you are mistaking the parameter scale for rate.
est <- fitdistr(x,"gamma")$estimate
#plot the gamma curve with the found parameters
hist(x, breaks=Size, freq=FALSE, xlab="Drop Size", ylab="No. of Drops")
curve(dgamma(x, rate=est["r
On 05/21/2013 12:54 AM, Xianwen Chen wrote:
Hi Jim,
Thank you for the suggestion. I think overlapped rectangles will well
present the message. I'm now trying ggplot2.
Here is the code:
require(ggplot2)
rect_MNL_Delta <- data.frame(
xmin <- c(1, 3, 5, 7, 9, 11, 13),
xmax <- xmin + 1,
ymin <- c(
I don't know what you plan to do with this list, but lists are quite a bit less
efficient than fixed-mode vectors, so you are likely losing a lot of
computational speed by using this list. I don't hesitate to use simple data
frames (lists of vectors), but processing lists is on par with for loop
On 15.05.2013 17:42, Fabio Berzaghi wrote:
I use to be able to see the layer names of a RasterStack by just typing
its name but now I only get this
class : RasterStack
dimensions : 70, 180, 12600 (nrow, ncol, ncell)
resolution : 0.5, 0.5 (x, y)
extent : -80, 10, 50, 85 (xmin, x
On 20.05.2013 23:24, Andrew Z wrote:
Is there a place to an old version (GBM 1.6 or 2.0) of the Windows
64-bit binary for the GBM package?
In GBM 2.1, CV does not work on any of my data sets, so I reported it
to https://code.google.com/p/gradientboostedmodels/ . However, I would
like to soon c
On 20.05.2013 18:10, Naser Jamil wrote:
Dear R-users,
May I seek your attention please! I have a long R code, which runs most of
the time quite nicely and provides necessary results. However, often it
provides a message like " R for windows GUI front-end has stopped working"
and just stops work
On May 20, 2013, at 2:24 PM, Andrew Z wrote:
> Is there a place to an old version (GBM 1.6 or 2.0) of the Windows
> 64-bit binary for the GBM package?
>
> In GBM 2.1, CV does not work on any of my data sets, so I reported it
> to https://code.google.com/p/gradientboostedmodels/ . However, I woul
Is there a place to an old version (GBM 1.6 or 2.0) of the Windows
64-bit binary for the GBM package?
In GBM 2.1, CV does not work on any of my data sets, so I reported it
to https://code.google.com/p/gradientboostedmodels/ . However, I would
like to soon continue to use GBM for a project, and I p
So it does. I had tried list() but not as.list(). Silly me. :)
Thank you very much, William.
--
Alexandre Sieira
CISA, CISSP, ISO 27001 Lead Auditor
"The truth is rarely pure and never simple."
Oscar Wilde, The Importance of Being Earnest, 1895, Act I
On 20 de maio de 2013 at 18:32:35, William
Try using as.list(x) instead of as.vector(x, mode="list").
The former has a method for POSIXct; the latter does not.
> x <- as.POSIXct(c("2013-05-20 14:28", "2013-11-30 22:10"), tz="US/Pacific")
> x
[1] "2013-05-20 14:28:00 PDT" "2013-11-30 22:10:00 PST"
> as.list(x)
[[1]]
[1] "2013-0
I was trying to convert a vector of POSIXct into a list of POSIXct, However, I
had a problem that I wanted to share with you.
Works fine with, say, numeric:
> v = c(1, 2, 3)
> v
[1] 1 2 3
> str(v)
num [1:3] 1 2 3
> l = as.vector(v, mode="list")
> l
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
[1] 3
> str(
On 13-05-20 2:18 PM, Kimberley Armstrong wrote:
Hello,
I am trying to use R for propensity score matching in SPSS. I have version 21
of SPSS and I downloaded R 2.14.2 as directed as well as the R Essentials
plug-in.
I think you need to talk to SPSS support for this. We have no
connection w
On Mon, May 20, 2013 at 10:01 AM, mary wrote:
>
> Thank you but I want to create a matrix (4,4) or (8,8) not just only a (2,2)
> matrix...
>
provide an m0 matrix of the appropriate dimensions and change 2 to the
new dimension in the relevant lines.
--
Statistics & Software Consulting
GKX Group,
Dear R-users,
May I seek your attention please! I have a long R code, which runs most of
the time quite nicely and provides necessary results. However, often it
provides a message like " R for windows GUI front-end has stopped working"
and just stops working. I'm running the program in Windows 2007
Hello,
I am trying to use R for propensity score matching in SPSS. I have version 21
of SPSS and I downloaded R 2.14.2 as directed as well as the R Essentials
plug-in. I have run a test for R and it appears to be running correctly. I
then downloaded psmatching3 and have tried to use the PS ma
Good afternoon,
I have some rainfall drop size data (frequency count within drop size) that is
already arranged into specific bins (Size). I am looking to fit a gamma curve
onto a histogram of the data.
At the moment I have been able to create estimate the gamma parameters from the
PDF of the
Thank you but I want to create a matrix (4,4) or (8,8) not just only a (2,2)
matrix...
--
View this message in context:
http://r.789695.n4.nabble.com/Generate-positive-definite-matrix-with-constraints-tp4667449p4667509.html
Sent from the R help mailing list archive at Nabble.com.
___
Hi Jim,
Thank you for the suggestion. I think overlapped rectangles will well
present the message. I'm now trying ggplot2.
Here is the code:
require(ggplot2)
rect_MNL_Delta <- data.frame(
xmin <- c(1, 3, 5, 7, 9, 11, 13),
xmax <- xmin + 1,
ymin <- c(16.7026, 14.9968, 1
I am using the iris example came with nnet package to test AMORE. I can see
the outcomes are similar to nnet with adaptative gradient descent. However,
when I changed the method in the newff to the batch gradient descent, even
by setting the epoch numbers very large, I still found all the iris
expe
Hi,
You may need to add "dec=","" in the read.csv.
dat1<- read.table(text="
Time;Mid
31/01/2013 00:00;1,35679
31/01/2013 00:01;1,35678
31/01/2013 00:02;1,356785
31/01/2013 00:03;1,35689
31/01/2013 00:04;1,3569
31/01/2013 00:05;1,3569
31/01/2013 00:06;1,356885
31/01/2013 00:07;1,35691
31/01/2013 00:
arun yahoo.com> writes:
>
> Hi,
>
> Check this link
> https://stat.ethz.ch/pipermail/r-help/2008-May/163274.html
> A.K.
>
> >i have a mer object named "model" :
> >
> >I want to extract t statistics of the coeeficients from "model".
> >Please help me out
> >
> >package used lme4
>
Hi
Over of what Don adviced, do not use HTML posting. Your data could be scrambled
and can not be coppied easily without user intervention.
temp<-structure(list(RGE01 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
Hi,
Check this link
https://stat.ethz.ch/pipermail/r-help/2008-May/163274.html
A.K.
>i have a mer object named "model" :
>
>I want to extract t statistics of the coeeficients from "model".
>Please help me out
>
>package used lme4
__
R-help@r-p
Hi,
Try:
H<-function(x) H1(x)+H2(x)
H1(2)
[#1] 4
H2(2)
#[1] 7
H(2)
#[1] 11
A.K.
- Original Message -
From: Honest Chipoyera
To: "r-help@r-project.org"
Cc:
Sent: Monday, May 20, 2013 8:09 AM
Subject: [R] DEfining and plotting the sum of two functions
I have two functions H1(x) and
On 20-05-2013, at 14:09, Honest Chipoyera wrote:
> I have two functions H1(x) and H2(x) defined separately and I wish to create
> a new function H(x) which is the sum of the two functions. I also need to
> plot the three functions using the command "curve". With the aid of the
> example prog
I have two functions H1(x) and H2(x) defined separately and I wish to create a
new function H(x) which is the sum of the two functions. I also need to plot
the three functions using the command "curve". With the aid of the example
program below, can you please explain how I can do that in R.
H
I have the data frame with the following structure
Operator Score Date Freq
1 A Crap Apr 20130
2 D Crap Apr 20130
3 J Crap Apr 20130
4 L Crap Apr 20130
5 M Crap Apr 20130
6 A Good Apr 20131
7 D Goo
Hi all,
I want to know the moving average order of a time series variable X_t.
I know that ar(x) works for the autoregressive order, but ma(x) is not
doing the same for moving average.I am not sure where to find the correct
function.
Any help is appreciated.
Thanks,
Preetam
--
Preetam Pal
(+91)
Hi,
Try either:
dat2<- structure(list(Operator = structure(c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L,
On 13-05-20 4:34 AM, James Savage wrote:
Hi all,
Just a quick question:
I want to generate a column of counts of a particular variable. The easiest way
seems to be using table(). For reasonably small amounts of data, there seems to
be no problem.
C <- data.frame(A1 = sample(1:1000, 10, repl
On 13-05-20 12:17 AM, Nguyen Hoang wrote:
I'm very new to R, and I'm having trouble figuring out a 3d surface
plot of the data. I typically have something like this:
0.1
0.2
0.3
0.4
0.5
0.001
40960.16
40960.16
40960.16
40960.16
40960.16
0.0025
4096
Dear R Help
Please forgive my lack of knowledge,.I would be very thankful for some help.
Here is my problem:
I was using optim to estimate parameters of a model and I get this error message
"Error in optim(x0, fn = riskll, method = "L-BFGS-B", lower = lbs, upper = ubs,
:
L-BFGS-B needs finite
Dear R Help
Please forgive my lack of knowledge,.I would be very thankful for some help.
Here is my problem:
I was using optim to estimate parameters of a model and I get this error message
"Error in optim(x0, fn = riskll, method = "L-BFGS-B", lower = lbs, upper = ubs,
:
L-BFGS-B needs finite
Hi all,
Just a quick question:
I want to generate a column of counts of a particular variable. The easiest way
seems to be using table(). For reasonably small amounts of data, there seems to
be no problem.
C <- data.frame(A1 = sample(1:1000, 10, replace = TRUE), B1 =
sample(1:1000, 10, r
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