Às 17:43 de 25/03/2025, varin sacha via R-help escreveu:
Dear R-experts,
Here below a toy example not working. After some researches on the Net, still
don't get it !
Many thanks for your precious help.
#
library(knitr)
library(ggplot2)
library(pollster)
library(dp
Dear R-experts,
Here below a toy example not working. After some researches on the Net, still
don't get it !
Many thanks for your precious help.
#
library(knitr)
library(ggplot2)
library(pollster)
library(dplyr)
statut=c("married","not married","not
married","mar
a) Do be sure to read the Posting Guide... for example changes to contributed
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recommended forum for that contributed package. Also, you will avoid
mis-communication more reliably if you follow the guide recommendatio
@ Rui
that is the idea, but how do I apply this to a matrix with 200 columns?
I cannot write out the expression.
The colnames seem very messy, but they would be messy even under my
scheme with as little as 100 vars.
On Tue, 2025-03-25 at 06:15 +, Rui Barradas wrote:
> Às 15:28 de 24/03/2025,
Hi Everyone,
I trust this email finds you well, I am currently running an ODE model
using the plot.desolve method from the deSolve package. For context my
model is a box model with 75 boxes, therefore for each variable I will get
75 outputs. I am currently comparing the results of my model
to obs
Hello,
Sorry, much simpler:
poly(as.matrix(X), degree = 2L)
Hope this helps,
Rui Barradas
Às 13:58 de 25/03/2025, Rui Barradas escreveu:
Hello,
This seems to work and is independent of the number of columns. 'p' is
the output in my previous post.
f <- function(x, data = X) with(data,
Hello,
This seems to work and is independent of the number of columns. 'p' is
the output in my previous post.
f <- function(x, data = X) with(data, eval(parse(text = x)))
p2 <- poly(sapply(names(X), f), degree = 2L)
identical(p, p2)
# [1] TRUE
Hope this helps,
Rui Barradas
Às 13:42 de 2
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