Now we're getting somewhere. I suspect that each Excel sheet looks
something like this:
YearTongaSamoaFiji
200821.2 32.0 18.7
...
201723.7 31.9 19.3
# in the above there are three columns (countries) and ten rows
# import this sheet as "MPG3"
nrows<-nrow(M
Hi
When I do multiple plots of similar data I usually put the plots into the
multipage pdf file
pdf("somename.pdf")
for (i in columns) {
p<-ggplot(something)
print(p+geom_point(size=4)+stat_smooth(se=F, span=0.5, size=1.3)+
facet_grid(al2o3~teplota, labeller="label_both"))
or
plot(something)
Yes, it works for me.
Eik Vettorazzi , 21 Kas 2018 Çar, 00:19 tarihinde şunu
yazdı:
> How about this:
>
> library(rootSolve)
> f1<-function(x)5/((1+x)^1) + 5/((1+x)^2) + 5/((1+x)^3) + 105/((1+x)^4) -105
> uniroot.all( f1,c(-1e6,1e6))
>
> [1] -1.9881665 0.0363435
>
> Cheers
>
>
> Am 20.11.2018 um
For your first question, yes, you will need to adjust the number of "x"
values to match the number of "y" values. You can use the "nrow" function
to get that number. I don't really know what the abscissa scale is on your
plots, I just made up the data I used.
If you are comparing countries, you ma
Hello Sir,
Thank you very much. I will try it out and will let you the result.
The no. of rows varies per sheet by a different number of observations. Due
to different no. of rows or observations, I separated the columns in
different sheets.
*Will a different number of rows create a problem for
I assume that you are importing the Excel sheets separately. When you
import a sheet, you can get the number of columns with this:
ncol()
Using the data frame "mpg" that I created:
ncolumns<-ncol(mpg)
ncolumns
[1] 38
You can then substitute "ncolumns" each time you import another sheet. How
you
You are right. Specifically, I need to predict the mean and median time to
failure from a coxph model and several parametric models using new data.
Thanks.
El lun., 5 nov. 2018 a las 7:11, Therneau, Terry M., Ph.D. (<
thern...@mayo.edu>) escribió:
> First, type='expected' gives the expected cumu
Hello Sir,
Thanks, now I understood and will check them out.
One more thing I want to ask that I have 1 excel file with multiple (i.e.
12 sheets). Each sheet contains different number of columns, for instance,
1st sheet contains 38 columns, 2nd sheet contains 10 columns, Third 2
columns, 4th 1 co
Hi Subhamitra,
1. Here I manufacture some data so that the example is "reproducible", that
is anyone can run the code and get the same output that I do. Yes,
veh1...veh38 are the names of the variables.
2. Here I join the 38 variables I created into a data frame, which I think
is the input for yo
Hello Sir,
Thanks, I'll check them out.
But, I am not understanding 2 points of your suggestion.
1. In the line,* "*for(i in 1:38) assign(paste0("veh",i),rep(sam
ple(10:35,1),10)+runif(10,-4,*4))", *what veh, rep(sample(10:35,1),10)
+runif(10,-4,4)) indicate? Here veh indicates columns right?
*2
Hi Subhamitra,
As Bert noted, you are mixing base and grid graphics. Here is a simple
way to get a plot like what you described. It will probably take more
work to find what you actually do want and discover how to get it.
for(i in 1:38) assign(paste0("veh",i),rep(sample(10:35,1),10)+runif(10,-4,4
How about this:
library(rootSolve)
f1<-function(x)5/((1+x)^1) + 5/((1+x)^2) + 5/((1+x)^3) + 105/((1+x)^4) -105
uniroot.all( f1,c(-1e6,1e6))
[1] -1.9881665 0.0363435
Cheers
Am 20.11.2018 um 13:09 schrieb Engin Yılmaz:
Dea(R)
I try to solve one equation but this program did not give me real r
You need to do some studying! ggplot is built on the grid graphics system,
which is separate from the base graphics system. The par() function is part
of the *base* graphics system and so ignored by ggplot.
Others may offer you solutions using the "faceting" functionality of
ggplot. But you really
Dear R users,
I have one excel file with 5 sheets. The no. of columns vary for each
sheet. The 1st sheet consists of 38 columns. So, I want to plot 38 separate
line charts and arrange them in par(mfrow = c(4, 10)) order. Please suggest
me how to do this. I have tried with the following code by run
A bit pedestrian, but you might try
pf <- function(x){5/((1+x)^1) + 5/((1+x)^2) + 5/((1+x)^3) + 105/((1+x)^4) -105}
uniroot(pf,c(-10,10))
curve(pf, c(-10,10))
require(pracma)
tryn <- newton(pf, 0)
tryn
pf(0)
pf(0.03634399)
yc <- c(-105, 5,5,5,105)
rooty <- polyroot(yc)
rooty
rootx <- 1/rooty - 1
r
Dea(R)
I try to solve one equation but this program did not give me real roots
for example
yacas("Solve( 5/((1+x)^1) + 5/((1+x)^2) + 5/((1+x)^3) + 105/((1+x)^4) -105
==0, x)")
gave me following results
How can I find real roots?
expression(list(x == complex_cartesian((1/42 - ((1/63 -
((root(733945
Well, this is not an elegant (or robust) solution, but this would work
for the example you give, at least:
starttime <- as.POSIXct("2018-11-20 23:01:18") # Just pick a random date
format(starttime + c(0:4), format = "%T")
There are probably better ways. :)
--
Regards,
Bjørn-Helge Mevik
signat
On Tue, 20 Nov 2018, Knut Krueger writes:
> I have an dataframe from with a given time format:
>
> "23:01:19"
>
> to change some given data:
>
> x=data.frame
> ("Y"=c(1:5),"TIME"=c("23:01:18","23:01:18","23:01:18","23:01:18","23:01:18"))
>
> I need to change the time increasing in seconds
>
> x=d
Thanks a lot!
Berend Hasselman , 20 Kas 2018 Sal, 12:02 tarihinde şunu
yazdı:
>
>
> R package Ryacas may be what you want.
>
> Berend
>
>
> > On 20 Nov 2018, at 09:42, Engin Yılmaz wrote:
> >
> > Dea(R)
> >
> > Do you know any system solver in R ?
> >
> > For example, in matlab, is very easy
> >
Hi everyone,
Collateral is now available on CRAN. If you've used purrr's side effect
capturing functions before, you'll love collateral: it provides you with
map variants that automatically wrap your mapped function in safely() or
quietly() and provide nicely printed output, allowing you to quickl
I have an dataframe from with a given time format:
"23:01:19"
to change some given data:
x=data.frame
("Y"=c(1:5),"TIME"=c("23:01:18","23:01:18","23:01:18","23:01:18","23:01:18"))
I need to change the time increasing in seconds
x=data.frame
("Y"=c(1:5),"TIME"=c("23:01:18","23:01:19","23
R package Ryacas may be what you want.
Berend
> On 20 Nov 2018, at 09:42, Engin Yılmaz wrote:
>
> Dea(R)
>
> Do you know any system solver in R ?
>
> For example, in matlab, is very easy
>
> syms a b c x eqn = a*x^2 + b*x + c == 0; sol = solve(eqn)
>
> How can I find this type code in
Dea(R)
Do you know any system solver in R ?
For example, in matlab, is very easy
syms a b c x eqn = a*x^2 + b*x + c == 0; sol = solve(eqn)
How can I find this type code in R (or directly solver)?
*Since(R)ely*
Engin YILMAZ
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