Hello all,
I'm a long time R user, but recently also using Python. I noticed that
RStudio rolled out Python through reticulate. It's great so far!
My question is, how do you debug in Python?
In R, I simply step through the code script in my console with cmd+enter.
But you can't do that with Pyth
Dear R list,
I am an old-school R user. I use apply(), with(), and which() in base
package instead of filter(), select(), separate() in Tidyverse. The idea of
pipeline (i.e. %>%) my code was foreign to me for a while. It makes the
code shorter, but sometimes less readable?
With ggplot2, I just do
h Date and POSIXt types
> in the same analysis... I recommend sticking with one or the other.
>
> The trunc.POSIXt function is more appropriate for getting POSIXt dates
> than converting to character and back.
>
> On February 2, 2019 9:03:31 AM PST, C W wrote:
> >Also, I
uot;2015-07-13"
[36] "2015-07-13" "2015-07-13" "2015-07-13" "2015-07-13" "2015-07-13"
> class(as.Date(dat, "%m/%d/%y"))
[1] "Date"
> class(strftime(strptime(dat, format="%m/%d/%y %H:%M"), format =
"%
this can be as simple as
>
> Sys.setenv(TZ="GMT")
>
> but it can be specific to your data set also.
>
> On February 2, 2019 7:09:46 AM PST, Duncan Murdoch <
> murdoch.dun...@gmail.com> wrote:
> >On 01/02/2019 10:45 p.m., C W wrote:
> >> Dea
Dear R community,
I am working with dates. And I get the following error:
> strftime(dat[20], format="%H:%M")
Error in as.POSIXlt.character(as.character(x), ...) :
character string is not in a standard unambiguous format
Here's the original data:
dat <- structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8
.
[2,] . .
[3,] . .
[4,] . .
[5,] . .
On Fri, Oct 20, 2017 at 3:51 PM, C W wrote:
> Thank you for your responses.
>
> I guess I don't feel alone. I don't find the docume
though it *looks* like the
opposite.
Cheers!
On Fri, Oct 20, 2017 at 3:22 PM, David Winsemius
wrote:
>
> > On Oct 20, 2017, at 11:11 AM, C W wrote:
> >
> > Dear R list,
> >
> > I came across dgCMatrix. I believe this class is associated with sparse
> >
Dear R list,
I came across dgCMatrix. I believe this class is associated with sparse
matrix.
I see there are 8 attributes to train$data, I am confused why are there so
many, some are vectors, what do they do?
Here's the R code:
library(xgboost)
data(agaricus.train, package='xgboost')
data(agari
the same way:
> >
> > names(dat)[3:5] <- names(dat)[c(5,3,4)]
> >
> > Better(?):
> >
> > perm <- c(1,2,5,3,4)
> > dat <- dat[perm]
> >
> > dat is a list.
> >
> > Göran
> >
> >
> >>
> >> ___
uffle the names in the same way:
>
> names(dat)[3:5] <- names(dat)[c(5,3,4)]
>
> Better(?):
>
> perm <- c(1,2,5,3,4)
> dat <- dat[perm]
>
> dat is a list.
>
> Göran
>
>
>> From: Rolf Turner
>> Sent: Frida
275 0.2213842
6 0.90691210 0.7247171 0.8244184 0.73328097 -1.05284737 0.822489552
0.52521494 0.6796657
But, you would NOT ONLY get undesired variable names, BUT ALSO duplicated
names. I suppose I can use paste() to solve that?
Any better ideas?
On Fri, Apr 28, 2017 at 8:57 PM, C W wrote:
>
Dear R list,
I am am a little unsure what is the best way to approach this. I suppose I
have
> dat <- matrix(rnorm(30), ncol = 5)
> dat <- data.frame(dat)
> dat
X1 X2 X3 X4 X5
1 -1.1317 -0.87868106 -0.33000492 1.5241765 -0.92483388
2 -0.56168006
I figured it out. You can first do,
saveGIF({
for (j in 1:20){
plot(-4:4, -4:4, type = "n")
for(i in 1:10){
x[i] <- rnorm(1)
y[i] <- rnorm(1)
points(x[i], y[i])
}
}
})
It looks like you have to re-plot every time.
On Wed, Mar 15, 2017 at 1:4
p coming along
> > and sticking things into it."
> > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
> >
> >
> > On Tue, Mar 14, 2017 at 10:21 PM, C W wrote:
> >> Dear list,
> >>
> >> I am trying to use saveG
Dear list,
I am trying to use saveGIF() from library animation.
https://cran.r-project.org/web/packages/animation/animation.pdf
saveGIF() has dependency on package ImageMagick.
I got the following,
> install.packages("ImageMagick")
Warning in install.packages :
package ‘ImageMagick’ is not ava
ent from my phone. Please excuse my brevity.
>
> On February 24, 2017 5:55:15 PM PST, C W wrote:
> >I suppose for loop will suffice.
> >
> >I simply copy & paste the code from R editor. From my email, it looks
> >plain. Is there a way to tell?
> >
> >
to expand your example to illustrate how the data comes to you
> in order to suggest alternatives.
>
> Also post using plain text to prevent your code from being mangled on its
> way to us.
> --
> Sent from my phone. Please excuse my brevity.
>
> On February 24, 2017 5:27:07 PM PS
gt; And what is this "no loop drama" you refer to? I use loops frequently to
> loop around large memory gobbling chunks of code.
>
> --
> Sent from my phone. Please excuse my brevity.
>
> On February 24, 2017 5:02:46 PM PST, C W wrote:
> >Dear R,
> >
>
Dear R,
I wanted to simulate a 5 by 3 matrix which fills up by either rows or
columns?
I started with the following filling the matrix by rows,
dat <- matrix(NA, nrow=5, ncol = 3)
for(i in 1:5){
dat[i, ] <- rnorm(3)
}
But, R is known for no loop drama. Any suggestions?
Thanks!
Hi Carl,
I have not fully learned dplyr, but it seems harder than tapply() and the
?apply() family in general.
Almost every ggplot2 data I have seen is manipulated using dplyr. Something
must be good about dplyr.
aggregate(), tapply(), do.call(), rbind() will be sorely missed! :(
Thanks!
On Tu
Dear R list,
I am having a little trouble understanding the R code. I want to compute
expectation of normal pdf.
I did the following:
integrate(x*dnorm(x, rate=1), -Inf, Inf)
Error in match.fun(f) : object 'x' not found
If I did this, I get,
integrate(dexp(x, rate=1), -Inf, Inf)
Error in dexp(x
to have some kind of graph conditioned on species, by grade . What's
a good lead to learn about plotting this?
Thank you!
On Mon, Feb 20, 2017 at 11:12 AM, Hadley Wickham
wrote:
> On Sun, Feb 19, 2017 at 3:01 PM, David Winsemius
> wrote:
> >
> >> On Feb 19, 2017, at 11
Hi R,
I am a little confused by the data.table package.
library(data.table)
df <- data.frame(w=rnorm(20, -10, 1), x= rnorm(20, 0, 1), y=rnorm(20, 10, 1),
z=rnorm(20, 20, 1))
df <- data.table(df)
#drop column w
df_1 <- df[, w := NULL] # I thought you are supposed to do: df_1 <- df[, -w]
df_2
20 PM, C W wrote:
> Hi R list,
>
> Is it possible to use R markdown with beamer palo alto theme? Are they
> compatible?
>
> I copy pasted my LaTex code over to R, I get the following error message:
>
> ! LaTeX Error: Can be used only in preamble.
>
> See the
Hi R list,
Is it possible to use R markdown with beamer palo alto theme? Are they
compatible?
I copy pasted my LaTex code over to R, I get the following error message:
! LaTeX Error: Can be used only in preamble.
See the LaTeX manual or LaTeX Companion for explanation.
Type H for immediate
es utils datasets methods
[7] base
On Fri, Nov 4, 2016 at 4:50 PM, C W wrote:
> I just got the following error instead of the previous one,
>
> 2016-11-04 16:50:08.446 R[23372:326103] -[NSDrawerWindow
> setRuleThickness:]: unrecognized selector sent to instance 0x7fabc94a74b0
> 20
) R.app R 3.3.2 GUI 1.68 Mavericks build
Consider saving your work soon in case this develops into a problem.
On Fri, Nov 4, 2016 at 4:39 PM, C W wrote:
> Dear R list,
>
> Every time I start R, the following error message come up:
>
> *** caught segfault ***
> address 0
Dear R list,
Every time I start R, the following error message come up:
*** caught segfault ***
address 0x0, cause 'unknown'
Possible actions:
1: abort (with core dump, if enabled)
2: normal R exit
3: exit R without saving workspace
4: exit R saving workspace
>
Selection:
I am unable to do a s
On Wed, May 25, 2016 at 1:13 PM, S Ellison wrote:
> > -Original Message-
> > My data come from statistical model N(5, 2), with n=100, call this
> model_1
> > Then, I add bias to that data with N(3, 1), with n=100, call this model_2
> Do you mean you have data from N(5,2) that has had data
Hi everyone,
I am searching for some toy models in R. My goal is do to model checking.
For example,
My data come from statistical model N(5, 2), with n=100, call this model_1
Then, I add bias to that data with N(3, 1), with n=100, call this model_2
Ultimately, I want to see model_1+ model_2 gi
On a side note, is it ok to do?
> which(max(p_x))
and use that instead of numerical integration to get E[X]?
I will try both and report back! Thank you expeRts
On Fri, Feb 12, 2016 at 11:29 AM, C W wrote:
> Hi Peter,
>
> Great, let me try that and get back to you on my findi
't
> work.
>
> -pd
>
> On 12 Feb 2016, at 16:57 , C W wrote:
>
> > Hi Bert,
> >
> > Yay fantasyland!
> >
> > In all seriousness, You are referring to this?
> >
> https://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R
sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Fri, Feb 12, 2016 at 7:36 AM, C W wrote:
> > Hi David,
> >
> > This is the Gaussian looking distribution I am trying to integrate.
> >
> h
but that doesn't seem to be good enough, as we know, it should integrate to
1.
On Thu, Feb 11, 2016 at 3:32 PM, David Winsemius
wrote:
>
> > On Feb 11, 2016, at 11:30 AM, C W wrote:
> >
> > Hi David,
> >
> > My real function is actually a multivariate norma
06 PM, David Winsemius
wrote:
>
> > On Feb 11, 2016, at 9:20 AM, C W wrote:
> >
> > I want to do numerical integration w.r.t. mu: P(mu) × N(mu, 0.1)
> >
> > Because the variance is small, it results in density like: 7.978846e+94
> >
> > Is the
I want to do numerical integration w.r.t. mu: P(mu) × N(mu, 0.1)
Because the variance is small, it results in density like: 7.978846e+94
Is there any good suggestion for this?
Thanks so much!
On Thu, Feb 11, 2016 at 9:14 AM, C W wrote:
> Wow, thank you, that was very clear. Let me g
e(..., -2, 4, 100) will not show the bump.
> The same principal holds for numerical integration.
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Wed, Feb 10, 2016 at 6:37 PM, C W wrote:
>
>> Dear R,
>>
>> I am graphing the following normal d
Dear R,
I am graphing the following normal density curve. Why does it look so
different?
# the curves
x <- seq(-2, 4, by=0.1)
curve(dnorm(x, 2, 10^(-100)), -4, 4) #right answer
curve(dnorm(x, 2, 10^(-100)), -3, 4) #changed -4 to -3, I get wrong answer
Why the second curve is flat? I just
Hi R list,
I am using optim() to optimize a function with 3 parameters.
#My 1-d toy example: loglikelihood of normal with x=c(2,5,3,7,-3,-2,0),
find MLE of mean.
p1 <- function(theta){
sum(log(dnorm(c(2,5,3,7,-3,-2,0), mean = theta, sd = 1))) +log(dnorm(
theta, mean = 0.8, sd = 2))
}
optimiz
;
On Thu, Nov 19, 2015 at 10:29 AM, peter dalgaard wrote:
>
> On 19 Nov 2015, at 16:17 , C W wrote:
>
> > Hi Rolf,
> >
> > I think the MLE should be 1.71, no? And yes, I am aware of the
> > maximum=TRUE argument. I still feel something is wrong here.
> >
Hi Rolf,
I think the MLE should be 1.71, no? And yes, I am aware of the
maximum=TRUE argument. I still feel something is wrong here.
Thanks!
On Wed, Nov 18, 2015 at 6:23 PM, Rolf Turner
wrote:
> On 19/11/15 11:31, C W wrote:
>
>> Dear R list,
>>
>> I am tryin
Dear R list,
I am trying to find the MLE of the likelihood function. I will plot the
log-likelihood to check my answer.
Here's my R code:
xvec <- c(2,5,3,7,-3,-2,0)
fn <- function(theta){
sum(0.5 * (xvec - rep(theta, 7)) ^ 2 / 1 + 0.5 * log(1))
}
gn <- Vectorize(fn)
curve(gn, -5, 20)
opti
Hi,
I am trying to do add a legend to an overplot, something like this:
ggplot() +
geom_density(data = df1, aes(x = x), fill = "green", show_guide =
FALSE) +
geom_area(data = df2, aes(x = x), fill = "yellow", show_guide = FALSE) +
scale_color_manual(values = c("green", "yellow"), labe
Never mind, I figured it out.
You need to use sapply(), for instance, curve(sapply(x, p), from = 0, to
=10)
Thanks all!
On Tue, Oct 27, 2015 at 11:14 AM, C W wrote:
> Dear R list,
>
> I am trying to plot the curve of a function.
>
> Here's the R code:
>
> library(mv
Dear R list,
I am trying to plot the curve of a function.
Here's the R code:
library(mvtnorm)
p <- function(x, mu){
mu <- c(mu, 0)
dmvnorm(c(x, 1), mu, diag(2))
}
> curve(p(x, 2), from = 0, to =1)
Error in dmvnorm(c(x, 1), mu, diag(2)) :
mean and sigma have non-conforming size
I think
on, Jun 22, 2015 at 2:19 PM, Daniel Nordlund
wrote:
> On 6/22/2015 9:42 AM, C W wrote:
>
>> Hello R list,
>>
>> I am have question about sampling unique coordinate values.
>>
>> Here's how my data looks like
>>
>> dat <- cbind(x1 = rep(1:5
Hello R list,
I am have question about sampling unique coordinate values.
Here's how my data looks like
> dat <- cbind(x1 = rep(1:5, 3), x2 = rep(c(3.7, 2.9, 5.2), each=5))
> dat
x1 x2
[1,] 1 3.7
[2,] 2 3.7
[3,] 3 3.7
[4,] 4 3.7
[5,] 5 3.7
[6,] 1 2.9
[7,] 2 2.9
[8,] 3 2.9
viewdoc/download?doi=10.1.1.208.9970&rep=rep1&type=pdf
>
>
> On 15-05-29 05:11 PM, C W wrote:
> > Wow, thanks Ben. That worked very well.
> >
> > I guess I didn't have R.methodS3? But that doesn't make sense,
> > because I was using R.matlab
n Bolker wrote:
> C W gmail.com> writes:
>
> >
> > Hi Henrik,
> >
> > I don't quite get what I should do here. I am not familiar with
> > R.methodS3. Can you tell me what command exactly do I need to do?
> >
> > Thanks,
> >
> >
h the other
> dependencies (R.oo and R.utils), do the same.
>
> /Henrik
>
>
>
> On Thu, May 28, 2015 at 11:47 AM, C W wrote:
> > Dear R list,
> >
> > I am trying to do use the R.matlab library, I did the following, but it
> > does not work.
> &g
Dear R list,
I am trying to do use the R.matlab library, I did the following, but it
does not work.
> library(R.matlab)
Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()), versionCheck
= vI[[j]]) :
there is no package called ‘R.methodsS3’
Error: package or namespace load failed for ‘R
Dear R list,
I am trying install the RcppOctave package to run Matlab packages in R. I
did the following,
> install.packages("RcppOctave")
After some time of installation, I get the following error. I have
homebrew installed, I also have Octave and Matlab install.
installing the source packag
Hi David,
Thanks for the detail information. I think I will probably work in Octave
instead, and fix the problem when I have a little more time.
Mike
On Sat, Mar 14, 2015 at 2:13 AM, David Winsemius
wrote:
>
> > On Mar 13, 2015, at 11:50 PM, C W wrote:
> >
> > Hi eve
Hi everyone,
When I tried to install RcppOctave package I got the following.
> install.packages("RcppOctave")
--- Please select a CRAN mirror for use in this session ---
package ‘RcppOctave’ is available as a source package but not as a binary
Warning message:
package ‘RcppOctave’ is not avai
.00154643121052443,
0.0248938938021518, 0.187306626057348, 0.64822387977866, 1), .Dim = c(5L,
5L))
Now if I do,
> GP_fit(xx, yy, corr=cor.mat)
Error in corr$type : $ operator is invalid for atomic vectors
How could I fix it?
Thanks!
-M
On Mon, Mar 9, 2015 at 2:14 PM, C W wrote:
> Hi R list,
Hi R list,
I am using the GPfit package to fit Gaussian Process model.
The kernel in the package is,
K(x, x') = sigma^2 * exp(x-x')^2
My kernel have an extra term,
K((x, z), (x', z')) = sigma^2 * exp(x-x')^2 * exp(z-z')^2
The function corr_matrix() is,
corr_matrix(X, beta, corr=list(type="expon
e answer can be extracted from a given body of data.
> ~ John Tukey
>
> 2015-02-18 17:44 GMT+01:00 C W :
>
>> Hi Ben and JS,
>>
>> Thanks for the reply.
>>
>> I tried using: hessian(func = h_x, x, method = "complex"), it gives zero,
>> that'
Hi Ben and JS,
Thanks for the reply.
I tried using: hessian(func = h_x, x, method = "complex"), it gives zero,
that's good.
# R code
> hess.h <- hessian(func = h_x, x, method = "complex")
> mat <- h_x(x)*hess.h - grad(h_x, x) %o% grad(h_x, x)
> mat
[,1][,2] [,3][,4]
[1,]
Hi list,
I am running the following R code, the answer should be zero. But R gives
a very small negative number, what should I do?
##R code
library(numDeriv)
h_x <- function(x){
a = x[1]
b = x[2]
c = x[3]
d = x[4]
(a^2 + c^2 + d^2) * (b^2 + c^2 + d^2)
}
x1 = 10
x2 = 1
x3 = 0
x4
And I want zero
everywhere else.
Thanks,
Mike
On Thu, Jan 29, 2015 at 10:34 PM, C W wrote:
> Hi Bill,
>
> You solved by problem. For some reason, I thought xname was only
> referring to name of the x-axis.
>
> I remember last time I fixed it, it was something about xname, co
ent variable
> and func(rnorm(10), rnorm(10), mu=seq(0,5,len=501)) won't
> work right.
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Thu, Jan 29, 2015 at 5:43 PM, C W wrote:
>
>> Hi Rui,
>>
>> Thank you for your help. Th
rk, but I don't know if it's what you want. func2
> will get x and y from the global environment.
>
> func2 <- function(mu){
>x + y + mu ^ 2
> }
>
> curve(func2, from = 0, to = 10)
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 29-01-2015 21
Hi all,
I want to graph a curve as a function of mu, not x.
Here's the R code:
x <- rnorm(10)
y <- rnorm(10)
func <- function(x, y, mu){
x + y + mu ^ 2
}
curve(f = func(x = x, y = y, mu), from = 0, to = 10)
I know I can change variable mu to x, but is there a way to tell R that mu
is the va
m of the screen
>
> On Mon, Jan 12, 2015 at 4:31 PM, Duncan Murdoch
> wrote:
> > On 12/01/2015 12:00 PM, C W wrote:
> >> I use R on Mac, and I use RStudio on Windows. That's my opinion.
> >>
> >> I have one problem.
> >>
> >> When
I use R on Mac, and I use RStudio on Windows. That's my opinion.
I have one problem.
When I use R on Mac. The function plot() gives a graph that's cut off.
For example, try
plot(rnorm(100)
I believe there should be space below "index" on x-axis.
Why is that?
Thanks,
Mike
On Mon, Jan 12, 2
. Playing
> Research Engineer (Solar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
Hi list,
I have trying to calculate the covariance/correlation of three elements. I
have vector say,
v <- c(700, 800, 1000)
I want to have a 3 by 3 correlation matrix, meaning cor(v1, c2), cor(v1,
c3), cor(v2, v3), etc...
So far I get,
> cor(v)
Error in cor(v) : supply both 'x' and 'y' or a mat
Dear list,
I am curious about how R computes the autocorrelation? What formula does
it use? Theoretical or sample ACF?
I have tried to type acf in the R console, but I don't understand what it
is actually doing besides the basic if...else... statement. Could someone
help me read this output?
>
That's what I am looking for. Thanks for being so helpful, I appreciate it
very much.
On Sun, Feb 23, 2014 at 10:16 PM, Gabor Grothendieck <
ggrothendi...@gmail.com> wrote:
> On Sun, Feb 23, 2014 at 9:06 PM, C W wrote:
> > Is there a way to rbind this? Do I have to us
Is there a way to rbind this? Do I have to use use a package like 'zoo'
and merge()?
p <- predict(model, data.frame(t = 1, q = factor(1, 1:4))
rbind(tsdat, p)
On Sun, Feb 23, 2014 at 8:43 PM, C W wrote:
> I guess this is for anyone in the future.
>
> > predict(m
I guess this is for anyone in the future.
> predict(model, data.frame(t = 1, q = factor(1, 1:4))
This would be the answer. Thanks again, Gabor!
Mike
On Sun, Feb 23, 2014 at 8:23 PM, Gabor Grothendieck wrote:
> On Sun, Feb 23, 2014 at 7:40 PM, C W wrote:
> > Gabor,
> > Yo
, Feb 23, 2014 at 7:25 PM, C W wrote:
> > Gabor,
> > Let me change newdata since it's confusing.
> >
> > Suppose I want to predict, year 1990, and quarter 2.
> >> newdata <- data.frame(c(1990, 1, 0, 0)
> >
> > Since Q1 is a baseline, we will onl
ieck wrote:
> On Sun, Feb 23, 2014 at 6:56 PM, C W wrote:
> > Gabor,
> > I want the new data to be this,
> > newdata <- data.frame(c(1, 0, 0, 0))
> >
>
> Its not clear what this means. There are two input variables so we
> must specify two inputs.
> For e
394 0.20732358 -0.19547465 -1.05329759 -0.32826061 0.16938691
19 20
-0.23341132 -1.09123426
Warning message:
'newdata' had 4 rows but variables found have 20 rows
On Sun, Feb 23, 2014 at 6:49 PM, Gabor Grothendieck wrote:
> On Sun, Feb 23, 2014 at 6:34 PM, C W wrot
Hello list,
Does predict.lm() take ts object. I am aware that it requires data to come
in data.frame
Here's the time series regression.
y = t + Q1 + Q2 + Q3 + Q4
Here's my R code,
> dat <- rnorm(20)
> tsdat <- ts(dat, start=c(1900, 1), freq=4)
> q <- as.factor(rep(1:4, 5))
> t <- 1:20
> lm(ts
Hello,
I don't know how to use predict.lm() for ts object.
Here's the time series regression.
y = t + Q1 + Q2 + Q3 + Q4
Here's my R code,
> dat <- rnorm(20)
> tsdat <- ts(dat, start=c(1900, 1), freq=4)
> q <- as.factor(rep(1:4, 5))
> t <- 1:20
> lm(tsdat~t+q)
Call:
lm(formula = tsdat ~ t + q)
uot;)
>
> In your example, what you compute is not exactly a moving average, but in
> can be computed with something like the following.
>
> s <- (seq_along(dat) - 1) %/% 3
> sapply(split(dat, s), mean)
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> Em 17-0
Hi list,
How do I calculate a moving average without using filter(). filter() does
not seem to give weighted averages.
I am looking into apply(), tapply,... But nothing "moves".
For example,
dat<-c(1:20)
mean(dat[1:3])
mean(dat[4:6])
mean(dat[7:9])
mean(dat[10:12])
etc...
I understand the poi
Thanks Duncan. I have no knowledge of SAS, though many in industry use it.
I hope R would expand its usage to more at the industry level.
Mike
On Fri, Sep 20, 2013 at 12:17 PM, Duncan Murdoch
wrote:
> On 20/09/2013 11:52 AM, C W wrote:
>
>> Dear R community,
>> I am havin
Dear R community,
I am having trouble changing the tick marks on y-axis to every 5 units? I
have the following:
x <- c(12, 16, 6, 23, 27, 8, 5, 19, 23, 13, 16, 8)
y <- c(29, 29, 23, 34, 38, 24, 22, 34, 36, 27, 33, 27)
plot(x, y, pch=19)
Should I change ylim=c(0,40), and then use axis()?
I kep
On Sat, Jul 20, 2013 at 1:50 PM, C W wrote:
> Thanks, you guys are correct, I had different data.
> But why I get length 5 and 6, should only be 1 to 3.
>
> Full R code :
>
> vec <- c(1, 2, 3, 4, 5, 6, 7, 8, 13, 1, 2, 3, 5, 7, 8, 10, 12, 13, 14,
> 15, 1, 2, 3, 5, 6, 10, 12,
t;>>> Sent: Sat, 20 Jul 2013 12:36:55 -0400
>>>> To: b...@xs4all.nl
>>>> Subject: Re: [R] How to search for a sequence(and its combination)
>>>> inside
>>>> a vector?
>>>>
>>>> Hi Berend
>>>> I am looking for a
table,
>> # of times one element (out of 1, 2, 3) showed up, two elements, and all
>> three.
>>
>> I am trying, don't know if this works:
>>
>>> aa <- rle(a)
>>> b <- aa$lengths[aa$values]
>>> table(b)
>> b
>> 1 3
>> 3 12
end Hasselman wrote:
>
> On 20-07-2013, at 18:05, C W wrote:
>
>> Hi R list,
>>
>> I have a sequence repeating 1:15 . Some numbers are deleted. I want
>> to find how many times 1, 2, 3 appeared.
>> Basically, I want to "grab" the beginning of the sequen
Hi R list,
I have a sequence repeating 1:15 . Some numbers are deleted. I want
to find how many times 1, 2, 3 appeared.
Basically, I want to "grab" the beginning of the sequence and tally it up.
R code:
> vec <- c(1, 2, 3, 4, 5, 6, 7, 8, 13, 1, 2, 3, 5, 7, 8, 10, 12, 13, 14,
15, 1, 2, 3, 5, 6,
e the \begin{document} \end{document} tags.
>
>
>
> Best
>
> Simon
>
> On Jul 18, 2013, at 10:22 PM, C W wrote:
>
>> Thanks, Simon. I would never figured it out!
>>
>> I apologize if I sound frustrated, because I am.
>>
>> @package aut
at 4:19 PM, Berend Hasselman wrote:
>
> On 18-07-2013, at 22:09, C W wrote:
>
>> http://tex.stackexchange.com/questions/85154/knitr-with-texworks/85165#85165
>>
>> In step 3: "add the executable file (step 3)".
>>
>> What is the executabl
http://tex.stackexchange.com/questions/85154/knitr-with-texworks/85165#85165
In step 3: "add the executable file (step 3)".
What is the executable file? Locate package knitr directory path in R?
Mike
On Thu, Jul 18, 2013 at 3:56 PM, C W wrote:
> Actually, I see it at the b
Actually, I see it at the bottom. Sorry!
On Thu, Jul 18, 2013 at 3:44 PM, C W wrote:
> Hi Simon,
> I am on OS X Lion, I have TeXworks, I don't have knitr as an option.
>
> How do I install that into TeXworks? Seems like I have to something
> in terminal?
>
> Mike
>
xamples/knitr-minimal.Rnw
>
>
> Hope this helps
>
>
> Best
>
> Simon
>
>
>
>
>
> On Jul 18, 2013, at 8:52 PM, C W wrote:
>
>> How do you create a .Rnw file, in R or LaTex? I don't think any
>> tutorial mentions it.
>>
>>
com/product/isbn/9781482203530
>
> Regards,
> Yihui
> --
> Yihui Xie
> Phone: 206-667-4385 Web: http://yihui.name
> Fred Hutchinson Cancer Research Center, Seattle
>
>
> On Thu, Jul 18, 2013 at 11:13 AM, C W wrote:
>> Hi everyone,
>>
>> I am using pack
Hi everyone,
I am using package knitr, FIRST TIME. I don't have access to RStudio.
Read through Yihui's page, didn't find it helpful. Stuck on terms
Rnw, GFM (GitHub Flavored Markdown). Never used Sweave, so the
reference is not helping.
Is there a simple step-by-step example WITHOUT RStudio?
; scale <- apply(x, 2L, f)
> x <- sweep(x, 2L, scale, "/", check.margin = FALSE)
> }
> }
> else if (is.numeric(scale) && length(scale) == nc)
> x <- sweep(x, 2L, scale, "/", check.margin = FALSE)
>
t;scaled:center")<-NULL
> attr(x1,"scaled:scale")<-NULL
> str(x1)
> #num [1:15, 1:10] -0.2371 -0.5606 -0.8242 1.5985 -0.0164 ...
> A.K.
>
>
>
>
> - Original Message -
> From: C W
> To: r-help
> Cc:
> Sent: Tuesday, July 16, 2013 3
Hi list,
I am using scale() to standardize a distribution? But why does it
give me attributes attached to the data? I just want a standardized
matrix, that is all.
library(mvtnorm)
> x <- rmvnorm(15, mean=rep(50, 10))
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[,
Hi R community,
What package would you recommend to download pictures and descriptions
of the pictures?
I have looked at package XML and RCurl so far. Is this what everyone uses?
Thanks,
Mike
__
R-help@r-project.org mailing list
https://stat.ethz.ch/m
I asked a similar question earlier in the year,
http://r.789695.n4.nabble.com/How-to-stop-set-seed-besides-exiting-out-of-R-td4661717.html
I liked this solution from William,
> rm(list=".Random.seed", envir=globalenv())
Mike
On Thu, Jun 13, 2013 at 5:27 PM, Michael Weylandt
wrote:
>
>
> On Jun
How about
?intersect
> a<-1:5;b<-1:9
> a
[1] 1 2 3 4 5
> b
[1] 1 2 3 4 5 6 7 8 9
> intersect(a, b)
[1] 1 2 3 4 5
I haven't used this in simulation, so I don't know how fast it is.
-Mike
On Tue, May 28, 2013 at 2:05 PM, arun wrote:
> Hi,
> You could use:
> which(a%in%b)
> #[1] 1 2 3 4 5
>
> a1<-
gt; So, thee error is in the predict.multnet function. If you peak at that
> function, you see where the function falls apart. It seems that the
> function wants a0 to be a matrix but in this example it is a vector. I'm
> not familiar enough with the package to offer advice on how
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