Please, I'm trying to put the right plot higher and centered on the left
values but I don't achive.
I would appreciate so much your help
El 6 jun. 2017 22:37, "Pedro páramo" escribió:
> Hi all,
>
> I have this code, but the marginal distribution plot doesn´t appear
> aligned with the left plot.
Dear All,
When I tried to run the following code (taken from the *spdplyr* package
vignettes):
library(spdplyr)
library(maptools)
data(wrld_simpl)
worldcorner <- wrld_simpl %>%
mutate(lon = coordinates(wrld_simpl)[,1], lat =
coordinates(wrld_simpl)[,2]) %>%
filter(lat < -20, lon > 60) %>%
Hi,
I have a hierarchy of such classes. Subject has a list of measurements. Let
assume I have a list of such 'Subject' RC's.
Can I use dplyr to navigate from Subject to the list of measurement RC's and
filter and group data ? dplyr should
be able to call the methods on these RC's to operate on
Hi,
I am completely new to GARCH models and trying to fit a multivariate time
series model using DCC GARCH model and forecast it.
The data looks like this:
> head(datax)
x vibration_x Speed
1 2017-05-16 17:53:00 -0.132 421.4189
2 2017-05-16 17:54:00 -0.296 129
Hi Pedro,
As a one-off, you just shove the coordinates around a bit:
par(mar=c(11,0,6,6))
barplot(fhist$counts,axes=FALSE, space=0,horiz=TRUE,col="lightgray",
ylim=c(0,24))
However, I don't think that this plot illustrates quite what you think it does.
Jim
On Wed, Jun 7, 2017 at 4:01 PM, Pedr
On Wed, 7 Jun 2017 at 17:29 Shige Song wrote:
> Dear All,
>
> When I tried to run the following code (taken from the *spdplyr* package
> vignettes):
>
> library(spdplyr)
> library(maptools)
> data(wrld_simpl)
>
> worldcorner <- wrld_simpl %>%
> mutate(lon = coordinates(wrld_simpl)[,1], lat =
>
If you want the Julian date, you could use Bert's index on the original data
frame:
Daily[out$Q, ]
Date wyrQ
4 1911-04-04 1990 5.097032
6 1911-04-06 1991 6.569508
9 1911-04-09 1992 4.445745
15 1911-04-16 1993 3.001586
18 1911-04-28 1994 3.369705
Another way to get that index
Hi all,
In checking my R codes, I encountered the following problem. Is there a
way to fix this?
I tried to specify options(digits=). I did not fix the problem.
Thanks so much for your help!
Hanna
> cdf(pmass)[2,2]==pcum[2,2][1] FALSE> cdf(pmass)[2,2][1] 0.758>
> pcum[2,2][1] 0.75
Hello,
See FAQ 7.31
And try ?all.equal.
Hope this helps,
Rui Barradas
Em 07-06-2017 15:32, li li escreveu:
Hi all,
In checking my R codes, I encountered the following problem. Is there a
way to fix this?
I tried to specify options(digits=). I did not fix the problem.
Thanks so much for
Hi All,
I try to do a scatterplot for a bunch of variables. I plot a dependent
variable against a bunch of independent variables:
-- cut --
graphics::plot(
v01_r01 ~ v08_01_up11,
data = dataset,
xlab = "Dependent",
ylab = "Independent #1"
)
-- cut --
It is tedious to repeat the stateme
Hi,
Check the FAQ 7.31
https://cran.rstudio.com/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
And read the posting guide too...
https://www.r-project.org/posting-guide.html
HTH,
Ivan
--
Dr. Ivan Calandra
TraCEr, Laboratory for Traceology and Controlled Experiments
MON
Thank you!
2017-06-07 10:38 GMT-04:00 Ivan Calandra :
> Hi,
>
> Check the FAQ 7.31
> https://cran.rstudio.com/doc/FAQ/R-FAQ.html#Why-doesn_0027t-
> R-think-these-numbers-are-equal_003f
>
> And read the posting guide too...
> https://www.r-project.org/posting-guide.html
>
> HTH,
> Ivan
>
> --
> Dr
For a data frame, we can add an additional row or column easily. For
example, we can add an additional row of zero and an additional row of
column as follows.
Is there an easy and similar way to add zeros in each dimension? For
example, for
array(1:12, dim=c(2,2,3))?
Thanks for your help!!
Han
Aggregate can do both which.max and group length calculations, but the
result ends up as a matrix inside the data frame, which I find cumbersome
to work with.
Daily <- read.table( text =
" Date wyrQ
1911-04-01 1990 4.530695
1911-04-02 1990 4.700596
1911-04-03 1990 4.898814
1911-04-
> On Jun 7, 2017, at 8:33 AM, li li wrote:
>
> For a data frame, we can add an additional row or column easily. For
> example, we can add an additional row of zero and an additional row of
> column as follows.
>
> Is there an easy and similar way to add zeros in each dimension? For
> example, f
Please read
https://www.lifewire.com/how-to-send-a-message-in-plain-text-from-gmail-1171963
Re your question: easy is in the eye of the beholder.
a <- array( 1:12, dim = c( 2, 2, 3 ) )
b <- array( 0, dim = dim( a ) + 1 )
b[ 1:2, 1:2, 1:3 ] <- a
If you want to do a lot of this, it could be wrapp
Hi...
I have a dataframe with n columns and n rows. I need to find how many rows
contains zero raw read count across all column.
Thanks
--
*Yogesh Gupta*
*Postdoctoral Researcher*
*Department of Biological Science*
*Seoul National University*
*Seoul, South Korea*
[[alternative HTML ver
Hi ,
I am trying to open vignette DESeq2 but getting below error:
> vignette("DESeq2")
> START /usr/bin/evince "/usr/lib64/R/library/DESeq2/doc/DESeq2.pdf"
Cannot parse arguments: Cannot open display:
xdg-open: no method available for opening
'/usr/lib64/R/library/DESeq2/doc/DESeq2.pdf'
--
*Yog
Thanks! That is the method I am using now but I was checking whether there
is a simpler option. I will look into the abind package too.
Hanna
2017-06-07 12:25 GMT-04:00 Jeff Newmiller :
> Please read
> https://www.lifewire.com/how-to-send-a-message-in-plain-text
> -from-gmail-1171963
>
> Re yo
A) You are not making reproducible examples. Try out the package "reprex"
to help you recognize when you are forgetting details.
B) I suspect your problem is not understanding formulas. The first thing
that comes to my mind is using a version of the plot function that does
not use formulas for
Sounds like your operating system does not have a default PDF viewer, or
does not have X-windows installed.
On Wed, 7 Jun 2017, Yogesh Gupta wrote:
Hi ,
I am trying to open vignette DESeq2 but getting below error:
vignette("DESeq2")
START /usr/bin/evince "/usr/lib64/R/library/DESeq2/doc/DES
I don't understand "zero raw read count across all column".
Please read the Posting Guide (mentioned below) which points out that this
is a plain text mailing list (your email gets damaged to varying degrees
if you don't set your sending format to plain text), and you need to make
your example
To get the number of TRUE in logical vector 'x',
tabulate(x, 1)
can be used. Actually, it gives count of 1, but TRUE as integer is 1. Since R
3.4.0, it gives a correct answer, too, when the count is 2^31 or more.
__
R-help@r-project.org mailing list --
Bad idea!
In R3.3.3 it doesn't even work:
> y1 <- tabulate(x,1)
Error in tabulate(x, 1) : 'bin' must be numeric or a factor
## This does:
> y1 <- tabulate(as.integer(x),1)
But it's more inefficient than just using sum(), even discounting the
as.integer() conversion:
> y1 <- tabulate(as.intege
On Tue, 6 Jun 2017, Morway, Eric wrote:
Using the dataset below, I got close to what I'm after, but not quite all
the way there. Any suggestions appreciated:
Daily <- read.table(textConnection(" Date wyrQ
1911-04-01 1990 4.530695
1911-04-02 1990 4.700596
1911-04-03 1990 4.898814
1
I hope this is the appropriate list for this type of question
Consider the dataset below:I have a column DOC with values from 3 to 101and
those are the values that I want to show on my x axis, howeverI only get 3,
3.1, 3.2 and so on. I tried to change those values with xlim(3, 101) but I
getthe
You might try
matplot()
example:
x <- matrix(rnorm(30), ncol=3)
## plot a dependent variable (1:10) against a bunch of independent variables
(the three columns of x)
matplot(x , 1:10, type='b')
## or a bunch of dependent variables (the three columns of x) against an
independent variable (1:
On 07/06/17 21:08, Yogesh Gupta wrote:
Hi...
I have a dataframe with n columns and n rows. I need to find how many rows
contains zero raw read count across all column.
(1) You should probably have a *matrix*, not a data frame.
(2) Have you ever heard of the idea of providing a *reproducible
Suppose I have a file (named "tmp.rmd") containing:
---
title: Test
---
```{r example, echo=FALSE, results='asis'}
tmp <- data.frame(a=1:5, b=letters[1:5])
print( knitr::kable(tmp, row.names=FALSE))
```
And I render it with:
rmarkdown::render('tmp.rmd', output_format=c('html_document','pdf_d
Ah, yes, of course.
tabulate(x, 1)
doesn't work, too, in R 3.4.0. Sorry, I didn't actually try.
I thought of an alternative when TRUE count is 2^31 or more. sum(x) returns NA
with a warning. sum(as.numeric(x)) works, but requires a quite large memory.
I need to have all elements of a matrix multiplied by a weight before
being post-multiplied by itself, as shown in the forst block of codes
below. I can also multiply the matrix by the square root of the weight
and then take the outer product.
Actually, what I need is this. Denote each row of t
I hope this is the appropriate list for this type of question
Consider the dataset below:I have a column DOC with values from 3 to 101and
those are the values that I want to show on my x axis, howeverI only get 3,
3.1, 3.2 and so on. I tried to change those values with xlim(3, 101) but I
getthe
Is this a question? You seem to have three possible calculations, have already
implemented two of them (?) and it is unclear (to me) what you think the right
answer for any of them is supposed to be.
--
Sent from my phone. Please excuse my brevity.
On June 7, 2017 8:50:55 PM PDT, Steven Yen w
OK Thanks. Your response made me think. Here (the last line) is what I need:
set.seed(76543211)
w<-1:10; w
a<-matrix(rpois(20,2),nrow=10); a
t(w*a)%*%a
On 6/8/2017 12:09 PM, Jeff Newmiller wrote:
> Is this a question? You seem to have three possible calculations, have
> already implemented two o
Fine, except that you already seen to have a very compact solution if that
really is what you are looking for. What am I missing?
--
Sent from my phone. Please excuse my brevity.
On June 7, 2017 9:16:48 PM PDT, Steven Yen wrote:
>OK Thanks. Your response made me think. Here (the last line) is
I would like test AICc as a criteria for model selection for a glm using
stepAIC() from MASS package.
Based on various information available in WEB, stepAIC() use
extractAIC() to get the criteria used for model selection.
I have created a new extractAIC() function (and extractAIC.glm() and
e
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