Displaying 20 results from an estimated 1000 matches similar to: "Treat Variable as String and a String as variables name"
2007 Jun 14
2
function with xyplot
Hi,
I'm a new user trying to switch from SAS, so sorry for the beginner's
question: Suppose I have a dataframe DF that contains variables X,Y,Z. I am
trying to write a function like this:
myplot <- function(varname){xyplot(varname ~ Y, group = Z, data = DF)}.
The problem is then how to enter X into my function. If I write myplot("X")
I get an error because the argument is a
2003 Oct 08
2
Generating automatic plots
Hello,
I have been trying to write a small program to generate automatic plots. What
I want is to draw boxplots for some variables in my data frame, contrasting
with a variable called 'missing' that has value 1 if some variable in a
concrete case has at least one missing value, in order to check if the cases
that don't enter in my analysis are biased.
The first attempt was to
2016 Dec 13
2
syntax difference clusterExport in parallel and snow
We got some errors and eventually figured out that
parallel::clusterExport second argument is "varlist" while in
snow::clusterExport it is "list".
The user had loaded parallel first, but did something else which
inadvertently loaded snow, then clusterExport failed because we had
"varlist" and not "list".
Are these different on purpose?
pj
--
Paul E.
2005 Feb 08
5
How to get variable names in a function?
Hello,
applying a function to a list of variables I face the following problem:
Let's say I want to compute tables for several variables. I could write a
command for every single table, like
bravo<-c(1,1,2,3,5,5,5,);charly<-c(7,7,4,4,2,1)
table(bravo); table(charly)
> table(bravo); table(charly)
bravo
1 2 3 5
2 1 1 3
charly
1 2 4 7
1 1 2 2
The results are two tables with the
2010 Feb 18
1
variable substitution
Hi
I would like to write a script that reads a list of variable names. These variable names are some of the column headers in a data.frame. Then I want do a for-loop to execute various operations on the specified variables in the data.frame, but can't figure out how to do the necessary variable substitution. In bash (or C) I would use "$var", but there seems to be no equivalent in
2007 Oct 22
3
Elasticity in Leslie Matrix
Dear R-users,
I would like to calculate elasticities and sensitivities of each parameters
involved in the following transition matrix:
A <- matrix(c(
sigma*s0*f1, sigma*s0*f2,
s, v
), nrow=2, byrow=TRUE,dimnames=list(stage,stage))
The command "eigen.analysis" avaliable in package "popbio" provides
sensibility matrix and elasticity matrix
2010 Feb 25
3
variable substitution in for loops
Friends
I can't quite find a direct answer to this question from the lists, so here goes:
I have several dataframes, 200+ columns 2000+ rows. I wish to script some operations to perform on some of the variables (columns) in the data frames not knowing what the column number is, hence have to refer by name. I have variable names in a text file "varlist". So, something like this:
2012 Nov 24
1
Bootstrap lmekin model
Hi,I use the 'lmekin' model of the 'kinship' package of R in order to estimate heritability. I want to estimate the confidence interval of the variance coefficient and so I should use a bootstrap simulation. The pedigree file has 1386 subjects so I create a kinship matrix [1386*1386].This is the code of R I use:
kfit2 <- lmekin(IT~1+AGE +(1|ID), dati1,
2009 Oct 11
1
How do you test if a number is in a list of numbers?
Hi,
I want to subset a data frame if one of the variables matches any in a list.
I could of course do something like this:
subset(dataset, var == 1 | var == 2 | var ==3)
but that's tedious.
I tried
varlist = c(1,2,3,4)
subset(dataset, any(var == varlist))
but it doesn't work because 'any' doesn't go row-by-row and hence always
returns TRUE.
Is there any simple way to do this?
2013 Feb 18
3
foreach loop, stata equivalent
Hi! I'm a recent convert from Stata, so forgive my ignorance.
In Stata, I can write foreach loops (example below)
foreach var of varlist p1-p14 {
foreach y of varlist p15-p269 {
reg `var' `y'
}
}
It's looping p1-p15, p1-p16...., p1-p269, p2-p15, p2-p16,... p2-p269,...
variable pairs.
How can I write something similar in R?
I 'tried' understanding the
2005 Oct 03
2
grob questions
If I run the following example from:
http://www.stat.auckland.ac.nz/~paul/grid/doc/grobs.pdf
> grid.newpage()
> pushViewport(viewport(w = 0.5, h = 0.5))
> myplot <- gTree(name = "myplot", children = gList(rectGrob(name = "box",
+ gp = gpar(col = "grey")), xaxisGrob(name = "xaxis")))
> grid.draw(myplot)
>
2005 Mar 09
3
function in order to plot the same graph to postscript and pdf
Hi,
I've written a function in order to plot the same graph in a postcript and in
a pdf file. Unfortunately, the second graph is always empty, i.e.:
plot.both <- function{myplot, filename}{
pdf(file=paste(filename, ".pdf", sep=""))
myplot
dev.off()
postscript(file=paste(filename, ".eps", sep=""))
myplot
dev.off()
}
yields in a
2011 Apr 15
1
no solution yet, please help: extract p-value from mixed model in kinship package
I am making the question clear. Please help.
> Dear R experts
>
> I was using kinship package to fit mixed model with kinship matrix.
> The package looks like lme4, but I could find a way to extract p-value
> out of it. I need to extract is as I need to analyse large number of
> variables (> 10000).
>
> Please help me:
>
> require(kinship)
>
> #Generating
2009 Apr 04
2
help with formula and data= argument
Sorry for posting this twice, but I still have not solved this problem
and am hoping for some assistance.
I am attempting to write a function that is flexible enough to respond
to the user providing a formula (with a data= argument) or not (similar
to plot(x,y) versus plot(y~x,data=data)). I have found a method to work
with this in a simple case but am having trouble determining how to
2008 Aug 08
2
gridBase and new.page() / grid.newpage()
Hello all,
I'm trying to write a function using the gridBase package. I'd like
to push several base subplots to a larger plot constructed with grid.
However, I'm having trouble getting consistent results when running
the function when the plotting window (quartz) is closed, when it is
left open and the plot function is repeated to the same window, and
when the output is saved to a
2009 Sep 02
1
get function to return object "name"?
Dear list,
I've written a function that plots subjects. Something like:
myplot <- function(subject) { plot(subject) }
Subjects are vectors, e.g. ...
s1 <- c(200,200,190,180)
... and plotting them works fine, e.g. ...
myplot(s1)
Now I want to have "s1" etc appear in the plot title, but I don't know
how to refer to this generically (the object "name"? I tried
2007 Aug 04
3
using loops to create multiple images
I have a data.frame with ~100 columns and I need a barplot for each column
produced and saved in some directory.
I am not sure it is possible - so please help me.
this is my loop that does not work...
vars <- list (substitute (G01_01), substitute (G01_02), substitute (G01_03),
substitute (G01_04))
results <- data.frame ('Variable Name'=rep (NA, length (vars)),
2009 Sep 14
2
How to set default plotting colors by treatment?
Dear R-helpers,
I have a number of dataframes that looks something like this:
mydfr <- data.frame(treatment=c(rep("A",3),rep("B",3)), Xmeas=1:6,
Ymeas=c(2,4,3,3,5,6))
# except with many more variables, which I plot all the time colored
by treatment to quickly check things:
with(mydfr, plot(Xmeas, Ymeas, pch=19, col=c("blue","red")[treatment]))
# I
2010 Dec 17
2
Nested layout()
Hello,
Is it possible to call a graphing function that uses layout() multiple times and layout those outputs ? Here's a minimal example :
myplot <- function()
{
layout(matrix(1:2, nrow=1), widths = c(1, 1))
plot(1:10)
plot(10:1)
}
layout(matrix(1:2), heights = c(1, 2))
myplot()
myplot()
--------------------------------------
Dario Strbenac
Research Assistant
Cancer Epigenetics
Garvan
2008 Sep 25
2
ggplot, qplot in loop
Dear List,
yes, me again trying to work with qplot ;-)
I would like to make several single plots within a loop, like this
(simplified and so on...):
trials <- c("A","B","C")
mycolours <- ("wheat","darkolivegreen","lightgreen",