Hello,
wonder if you could provide input on the following: please see toy example
below, wanted to see if there is a way to have restrictions on how the points
are dragged on the plot. More specifically I would like the points draggable
horizontally ONLY and have their y axis value remain
Thanks Bert, here it is in plain.
Hello,
please see if you have a thought on how to achieve the following:
we have:
df<-data.frame(a=Sys.Date()+1:10,
b=Sys.Date()+c(NA,NA,NA,rep(3,4),NA,NA,3),
c=Sys.Date()+c(NA,NA,NA,rep(9,4),NA,NA,9))
the idea I have
Hello,
please see if you have a thought on how to achieve the following:
we have:
df<-data.frame(a=Sys.Date()+1:10,
b=Sys.Date()+c(NA,NA,NA,rep(3,4),NA,NA,3),
c=Sys.Date()+c(NA,NA,NA,rep(9,4),NA,NA,9))
the idea I have difficulty wrapping my head around is to do the
Hello,
could you please provide your thoughts on what I may be missing? gamlss models
are supposedly supported by MuMIn yet this one fails:
library(MuMIn)
#this lm runs
linearMod <- lm(Sepal.Length ~ ., data=iris)
options(na.action = "na.fail")
res <-dredge(linearMod,beta = T, evaluate = T)
Hello All,
wonder if you have thoughts on a clever solution for this code:
df <- data.frame(a = c(6,1), b = c(1000,1200), c =c(-1,3))
#the caveat here is that the number of rows for df can be anything from 1 row
to in the hundreds. I kept it to 2 to have minimal reproducible
Works well! Thanks!
Andras
On Thursday, February 21, 2019, 8:47:51 AM EST, Jeff Newmiller
wrote:
Use ?merge instead of intersect.
On February 21, 2019 5:22:46 AM PST, Andras Farkas via R-help
wrote:
>Hello All,
>
>wonder if you have a suggestion for the following:
&
Hello All,
wonder if you have a suggestion for the following:
we have
a<-data.frame(ID=c(1,2,3,4,5,6,7),date=as.POSIXct(seq(as.Date('2011-01-01'),as.Date('2011-01-07'),by
= 1),format='%m/%d/%Y %H:%M'),z=rnorm(7,1,1))
Thanks Rui and Ivan, works perfectly...
Andras
On Monday, February 4, 2019, 4:18:39 PM EST, Rui Barradas
wrote:
Hello,
Like this?
Map('[', listA, lapply(listB, '*', -1))
Hope this helps,
Rui Barradas
Às 21:01 de 04/02/2019, Andras Farkas via R-help escreveu:
> Hello every
Hello everyone,
wonder if you would have a thought on a function for the following:
we have
a<-sample(seq(as.Date('1999/01/01'), as.Date('2000/01/01'), by="day"),5)
b<-sample(seq(as.Date('1999/01/01'), as.Date('2000/01/01'), by="day"), 4)
c<-sample(seq(as.Date('1999/01/01'),
keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Jan 6, 2019 at 5:16 AM Andras Farkas via R-help
wrote:
Hello Everyone,
would you be able to assist with some expertise on how to get the following
do
Hello Everyone,
would you be able to assist with some expertise on how to get the following
done in a way that can be applied to a data set with different dimensions and
without all the line items here?
we have:
id<-c(1,1,1,2,2,2,2,3,3,4,4,4,4,5,5,5,5)#length of unique IDs may differ of
cking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
> On Thu, Sep 13, 2018 at 12:11 PM Andras Farkas via R-help
> wrote:
> >
> > Dear All,
> >
> > I have data frame:
> > set.seed(123.456
Dear All,
I have data frame:
set.seed(123.456)
df <-data.frame(ID=c(1,1,2,2,2,3,3,3,3,4,4,5,5),
read=c(1,1,0,1,1,1,0,0,0,1,0,0,0),
int=c(1,1,0,0,0,1,1,0,0,1,1,1,1),
z=rnorm(13,1,5),
y=rnorm(13,1,5))
what I would like to achieve (as
Dear All,
once we run the following code, the results of the test will give us the
expected obvious, samples are from the common distribution...
library(kSamples)
u1 <- sample(rnorm(500,10,1),20,replace = TRUE)
u2 <- sample(rnorm(500,10,1),20,replace = TRUE)
u3 <-
Dear All,
wonder if you have some thoughts on running the with() function (and perhaps
including the pool() function to get the results?) in glmulti? In other words,
how to run glmulti with a data set that is produced by mice()?
publicly available code:
data <- airquality
data[4:10,3] <-
Dear All,
using the example from the help of summary.rms
library(rms)
n <- 1000# define sample size
set.seed(17) # so can reproduce the results
age<- rnorm(n, 50, 10)
blood.pressure <- rnorm(n, 120, 15)
cholesterol<- rnorm(n, 200, 25)
sex<-
Dear All,
probably a simple enough solution but don;t seem to be able to get my head
around it...example based on a publicly available data set:
mydata <- read.csv("https://stats.idre.ucla.edu/stat/data/binary.csv;)
mylogit <- glm(admit ~ gre + gpa + rank, data = mydata, family = "binomial")
Dear All,
could you please provide input on the following:
plot(1:10,main=paste("\n ","\nABCD","\n","\n","\n"),cex.main=1.3)
a<-500
b<-12
mtext(bquote(bold(.(formatC(1.2*a,decimal.mark=",",digits=2,format="f")))~ "
words "~bold(.(b))~" words"~"\n"~"\n"))
as you
t; b <- c(0:4,0:7)
>> f(a,b)
> [1] 1 1 1 1 1 2 2 2 2 2 2 2 2
>
>
>
>Bert Gunter
>
>"The trouble with having an open mind is that people keep coming along
>and sticking things into it."
>-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
Dear All,
wonder if you have thoughts on the following:
let us say we have:
df<-data.frame(a=c(1,2,3,4,5,1,2,3,4,5,6,7,8),b=c(0,1,2,3,4,0,1,2,3,4,5,6,7))
I would like to rewrite values in column name "a" based on values in column
name "b", where based on a certain value of column "b" the
> [1] 3 3
>
>
>
> Cheers,
>
> Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
&
Dear All,
wonder if you could please assist with the following
df<-data.frame(ID=c(1,1,1,2,2,3,3,4,4,5,5),samples=c("A","B","C","A","C","A","D","C","B","A","C"))
from this data frame the goal is to extract the value of 3 from the ID column
based on the logic that the ID=3 in the data frame
;- quantile(x, .1234))
pfun(q)
There are variations, e.g. the one-liner
approx(sort(x), seq(0,1,,length(x)), q)$y
-pd
> On 16 Jun 2017, at 01:56 , Andras Farkas via R-help <r-help@r-project.org>
> wrote:
>
> David,
>
> thanks for the response. In your respo
Never mind, I think i figured:
z<-df
apply(df,1,function(x) approx(sort(x[1:4]), seq(0,1,,length(x[1:4])), x[5])$y)
thanks again for the help
Andras Farkas,
On Friday, June 16, 2017 5:34 AM, Andras Farkas via R-help
<r-help@r-project.org> wrote:
Peter,
thanks,
Peter,
thanks, very nice, this will work for me... could you also help with setting up
the code to run the on liner "approx(sort(x), seq(0,1,,length(x)), q)$y" on the
rows of a data frame using my example above? So if I cbind z and res,
df<-cbind(z,res)
the "x" in your one liner would be
uch appreciate the help...
Andras Farkas,
On Thursday, June 15, 2017 6:46 PM, David Winsemius <dwinsem...@comcast.net>
wrote:
> On Jun 15, 2017, at 12:37 PM, Andras Farkas via R-help <r-help@r-project.org>
> wrote:
>
> Dear All,
>
> we have:
>
> t<-seq(0,24,1)
Dear All,
we have:
t<-seq(0,24,1)
a<-10*exp(-0.05*t)
b<-10*exp(-0.07*t)
c<-10*exp(-0.1*t)
d<-10*exp(-0.03*t)
z<-data.frame(a,b,c,d)
res<-t(apply(z, 1, quantile, probs=c(0.3)))
my goal is to do a 'reverse" of the function here that produces "res" on a data
frame, ie: to get the answer
Dear All
wonder if you could assist with the following
we have:
library(prophet)
library(dplyr)
abc<-c(0.3684693,0.4938679, 0.4429201,0.452598,0.4301452,0.4315169,
0.447026,0.496179,0.4045693,0.398533,
Dear All,
would you have some thoughts on how to extend the prediction interval lines to
beyond the "range of data"?
example:
y <-c(0.4316092,0.4156757,0.3517915,0.3669508,0.3899471,0.3964143,
0.4001074,0.3851003,0.4222451,0.375324,0.3652045,0.3376978,0.383012,
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