Hi,
While learning how to implement XGBoost in R I came across below case and want
to know how to go about it.
Outcome variable: continous
independent features: mix of categorical and continuous
nrow(train_set): 8523
Since, XGBoost natively supports only numeric features, I applied one hot
enc
I'm trying to develop a markov chain transition matrix to simulate an
infectious disease model. I've got a much larger matrix that I'm working
with but here's the code for a toy version of the model:
library("markovchain")
byRow <- TRUE
#Parameters
pop <- 1000
b1 <- 0.095
b2 <- 0.048
b3 <
Thanks Jeff, I've already added color and lwd to make it visible. I know it
is not the optimal thing to do, but I was required.
Cheers.
On Tue, Feb 9, 2016 at 3:11 AM, Jeff Newmiller
wrote:
> R by default puts the axes at the edge of the plot, not at x=0 and y=0,
> for the reason that doing oth
Hi, Jeff et al.:
On 2/8/2016 9:52 PM, Jeff Newmiller wrote:
> plotxy(y1~x1, XY, xlim=c(0, max(XY$x1)))
Yes, Thanks.
Is there a way to do this from within "plotxy", so I can call
"plotxy" as I call "plot"?
Thanks,
Spencer
> --
> Sent from my phone. Please excu
plotxy(y1~x1, XY, xlim=c(0, max(XY$x1)))
--
Sent from my phone. Please excuse my brevity.
On February 8, 2016 7:17:57 PM PST, Spencer Graves
wrote:
>I'm getting an interesting error:
>
>
> > plotxy <- function(x, ...){
>+ plot(x, ...)
>+ }
> > XY <- data.frame(x1=1:3, y1=4:6)
> > plotxy(y1~x1
I'm getting an interesting error:
> plotxy <- function(x, ...){
+ plot(x, ...)
+ }
> XY <- data.frame(x1=1:3, y1=4:6)
> plotxy(y1~x1, XY, xlim=c(0, max(x1)))
Show Traceback
Rerun with Debug
Error in eval(expr, envir, enclos) : object 'x1' not found
The following work:
plotxy(y1~x
Hi Jean,
Many laptops perform complicated workarounds to accommodate the virtual
keypads on the keyboard. Sometimes a sticky shift key (e.g. NumLock, Caps
Lock) will mess up keyboard entry if it is on. Just a wild guess.
Jim
On Mon, Feb 8, 2016 at 7:16 PM, MAURICE Jean - externe <
jean-externe.m
Hi Rosa,
This looks like a mix of a R problem and a statistical problem. For the
statistical part I think you are going to need to find a statistician/adviser
to sit down with and workout what you want to measure and whether it is
possible.
For the R part it might be more useful to ask specif
Dear All,
I have a dataframe of 1000 rows and 4 columns. Each row represents a pair
of vectors (in my case a pair of genes) while the columns represent the
following
estimate.A: spearman correlation coefficient of gene[i] and gene[j]
expression across 120 samples from cancer type A
prob.A: Probab
> On Feb 8, 2016, at 12:45 PM, Göran Broström wrote:
>
> Thanks Marc, but see below!
>
> On 2016-02-08 19:26, Marc Schwartz wrote:
>>
>>> On Feb 8, 2016, at 11:26 AM, Göran Broström wrote:
>>>
>>> I have a data frame with dates as integers:
>>>
summary(persons[, c("foddat", "doddat")])
Dear all,
I’m conducting a met analysis and I usually use Revman, bur as I’m trying to
use R more and more, I would like to conduct the met analysis here, in R
(R-studio).
One off my problems, I think, is that:
1st. it’s the first time :)
2. I only have data for 1 arm as you can see on the data
Hi,
I've been taught that if I want to nest random factor A into B in an lme
model, the syntax is as follows: lme(x~y+B,random=~1|B/A).
In the case of my data, matters seem to be complicated by the fact that B
is a categorical variable with only 2 levels. When I run the lme with the
above syntax,
Thanks Marc, but see below!
On 2016-02-08 19:26, Marc Schwartz wrote:
On Feb 8, 2016, at 11:26 AM, Göran Broström wrote:
I have a data frame with dates as integers:
summary(persons[, c("foddat", "doddat")])
foddat doddat
Min. :1679 Min. :1800
1st Qu.:1876090
> On Feb 8, 2016, at 11:26 AM, Göran Broström wrote:
>
> I have a data frame with dates as integers:
>
> > summary(persons[, c("foddat", "doddat")])
> foddat doddat
> Min. :1679 Min. :1800
> 1st Qu.:18760904 1st Qu.:18810924
> Median :19030426 Median :19091227
I have a data frame with dates as integers:
> summary(persons[, c("foddat", "doddat")])
foddat doddat
Min. :1679 Min. :1800
1st Qu.:18760904 1st Qu.:18810924
Median :19030426 Median :19091227
Mean :18946659 Mean :19027233
3rd Qu.:19220911 3rd Qu.:19
Hi,
Although you did not provide any reproducible example, it seems you
store the same type of values in your data.frames. If this is true, it
is much more efficient to store your data in an array:
mylist <- list(a = data.frame(week1 = rnorm(24), week2 = rnorm(24)),
b = data.fr
Or
do.call( rbind, list.of.df )
from base R (without some of the robust behaviors that ldply implements).
--
Sent from my phone. Please excuse my brevity.
On February 8, 2016 7:33:54 AM PST, Ulrik Stervbo
wrote:
>Hi Wolfgang,
>
>I'm not sure exactly what you want, but the ldply in the package
R by default puts the axes at the edge of the plot, not at x=0 and y=0, for the
reason that doing otherwise makes the plot harder to read. To see this,
consider:
plot( c( -4, 0, 4 ), c( 0, 1, 1 ), type="s", xlab="x", ylab="y", axes=FALSE,
xlim=c( -5, 5 ), ylim=c( -2, 2 ), lwd=2 )
axis( side=1,
Hi Wolfgang,
I'm not sure exactly what you want, but the ldply in the package plyr can
help you make a data.frame from a list of data.frames:
library(plyr)
dfa <- data.frame(cola = LETTERS[1:5], colb = c(1:5))
dfb <- data.frame(cola = LETTERS[1:5], colb = c(1:5))
df.lst <- list(dfa.name = dfa,
On 2016-02-08 14:11, 于慧琳 wrote:
Hi guy,I have some problems when plotting survival curves for the
survfit object.My survfit model is as
follows:fit6<-survfit(Surv(Survival_time_year_,event)~Race_1_Desc,data=data6)Race_1_Desc
is a indicator variable which has 0 and 1 values.However, in my
datase
Here is another approach:
> interval <- c("1902-1931", "1912-1930", "1902-1951")
> sapply(strsplit(interval, "-"), function(x) mean(as.numeric(x)))
[1] 1916.5 1921.0 1926.5
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77
HI,
I am new to R and English is not my natural language.
I work on 4 laptops that are supposed to be the same. Win7-64. R version 3.1.0.
I can use readline to enter a value from keyboard on three of them but not on
the fourth : R doesn't stop on the readline instruction and acts as if the
resp
Hello,
I have a list of 7 data frames, each data frame having 24 rows (hour of
the day) and 5 columns (weeks) with a total of 5 x 24 values
I would like to combine all 7 columns of week 1 (and 2 ...) in a
separate data frame for hourly calculations, e.g.
> apply(new.data.frame,1,mean)
In some wa
Hi guy,I have some problems when plotting survival curves for the survfit
object.My survfit model is as
follows:fit6<-survfit(Surv(Survival_time_year_,event)~Race_1_Desc,data=data6)Race_1_Desc
is a indicator variable which has 0 and 1 values.However, in my dataset, all
of the Race_1_Desc are of
Hi
you have got one answer, here is another
interval <- c("1902-1931", "1912-1930", "1902-1951")
(as.numeric(substr(interval, 1,4))+as.numeric(substr(interval,
nchar(interval)-3, nchar(interval/2
[1] 1916.5 1921.0 1926.5
It works only if positions of years in intervals is first and last 4
On 08/02/2016 5:09 AM, catalin roibu wrote:
Dear R users!
I have a data frame with first column is an interval (1902-1931) and I want
to find the middle of the interval? Is there a possibility to do that in R?
Thank you very much!
Best regards!
I don't know of a single function that does th
Dear R users!
I have a data frame with first column is an interval (1902-1931) and I want
to find the middle of the interval? Is there a possibility to do that in R?
Thank you very much!
Best regards!
--
-
-
Catalin-Constantin ROIBU
Lecturer PhD, Forestry engineer
Forestry Faculty of Suceav
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