Displaying 20 results from an estimated 20000 matches similar to: "Turning a Variable into String"
2010 Dec 19
3
monthly median in a daily dataset
Hello,
I have a multi-year dataset (see below) with date, a data value and a flag
for the data value. I want to find the monthly median for each month in this
dataset and then plot it. If anyone has suggestions they would be greatly
apperciated. It should be noted that there are some dates with no values and
they should be removed.
Thanks
Emily
> print ( str(data$flow$daily) )
2006 Jan 13
2
find mean of a list of timeseries
Can someone please give me a clue how to 're'write this so I dont need
to use loops.
a<-ts(matrix(c(1,1,1,10,10,10,20,20,20),nrow=3),names=c('var1','var2','var3'))
b<-ts(matrix(c(2,2,2,11,11,11,21,21,21),nrow=3),names=c('var1','var2','var3'))
2012 Nov 15
1
Extracting list names within a looped function
Hello all,
I have the following problem:
1) A list was defined as 'a'
a <- list("var1"=c(100:1), "var2"=c(1:100), "var3"=rnorm(100))
2) a function 'foo' was defined that extracts the variable name assigned to x using the deparse(substitute()) functionality. This name will then be used within the function to generate specific output files, etc.
2009 Sep 22
3
converting a character vector to a function's input
Hi all, I have been trying to solve this problem and have had no luck so far.
I have numeric vectors VAR1, VAR2, and VAR3 which I am trying to cbind. I also have a character vector "VAR1,VAR2,VAR3". How do I manipulate this character vector such that I can input a transformed version of the character vector into cbind and have it recognize that I'm trying to refer to my numeric
2005 Apr 29
2
Automating plot labelling in custom function in lapply() ?
Dear List,
Consider the following example:
dat <- data.frame(var1 = rnorm(100), var2 = rnorm(100),
var3 = rnorm(100), var4 = rnorm(100))
oldpar <- par(mfrow = c(2,2), no.readonly = TRUE)
invisible(lapply(dat,
function(x) {
plot(density(x),
main = deparse(substitute(x))) }
)
)
2010 Feb 12
2
Access dataframe with variable name in function
Sorry guys, but I have another one:
I want to write a function that returns a certain column of a
dataframe. The function accepts two argument: the dataframe and the
name of the column, but the column is not given as a "string" but as
a variable name.
EXAMPLE
----------------------
> testdata
start stop censor groupvar var1 var2
1 0 1 0 1
2008 Jul 16
2
Group level frequencies
Dear List,
I have Multi-level Data
i= Indivitual Level
g= Group Level
var1= First Variable of interest
var2= Second Variable of interest
and I want to count the frequency of "var1" and "var2" on the group
level.
I found a way, but there must be a much simpler way.
data.ml <-
data.frame(i=c(1:8),g=as.factor(c(1,1,1,2,2,3,3,3)),var1=c(3,3,3,4,4,4,4
,4),
2011 Jan 02
1
Please, need help with a plot
Please, I wonder if someone knows how to add the
less than or equal to symbol in the plot generated by the code below:
var1<-c('age <= 3','age <= 7','age <= 10','age <= 11','age <= 20','age <=
25','age <= 30','age <= 45','age <= 50','age < 55','age >= 55')
2004 Aug 17
5
Bug in colnames of data.frames?
Hi,
I am using R 1.9.1 on on i686 PC with SuSE Linux 9.0.
I have a data.frame, e.g.:
> myData <- data.frame( var1 = c( 1:4 ), var2 = c (5:8 ) )
If I add a new column by
> myData$var3 <- myData[ , "var1" ] + myData[ , "var2" ]
everything is fine, but if I omit the commas:
> myData$var4 <- myData[ "var1" ] + myData[ "var2" ]
the name
2014 Aug 21
2
pregunta
Buenas noches Javier y José,
Estoy en contra de usar attach(), asi que propongo la siguiente alternativa
con with():
# paquete
require(epicalc)
# los argumentos en ... pasan de epicalc:::cc
# ver ?cc para mas informacion
foo <- function(var1, var2, var3, ...){
or1 <- cc(var1, var2, ...)
or2 <- cc(var1, var3, ...)
list(or1 = or1, or2 = or2)
}
# datos
x <-
2006 Mar 02
1
CCF and Lag questions
I am new to R and new to time series modeling.
I have a set of variables (var1, var2, var3, var4, var5) for which I have
historical yearly data.
I am trying to use this data to produce a prediction of var1, 3 years into
the future.
I have a few basic questions:
1) I am able to read in my data, and convert it to a time series format
using 'ts.'
data_ts <- ts(data, start = 1988, end =
2012 Jul 01
2
list to dataframe conversion-testing for identical
HI R help,
I was trying to get identical data frame from a list using two methods.
#Suppose my list is:
listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2))
#Creating dataframe using cbind
dat1<-data.frame(do.call("cbind",listdat1))
colnames(dat1)<-c("Var1","Var2","Var3")
#Second dataframe conversion
2008 May 02
2
Coercing by/tapply to data.frame for more than two indices?
Dear Colleagues,
Apologies for a long email to ask what I feel may be a very simple
question; I figure it's better to overspecify my situation.
I was asked a question, recently, by a colleague in my department
about pre-aggregating variables, i.e., computing the mean of defined subsets
of a data frame. Naturally, I thought of the 'by' and 'tapply' functions, as
2009 May 20
1
Comparing spatial distributions - permutation test implementation
Hello everyone,
I am looking at the joint spatial distribution of 2 kinds of organisms
(estimated on a grid of points) and want to test for significant
association or dissociation.
My first question is: do you know a nice technique to do that,
considering that I have a limited number of points (36) but that they
are repeated (4 times)? I did GLMs to test for correlations between
the
2010 Mar 18
1
Using a function to consolidate variables
Dear List,
I'm getting the error: object of type 'closure' is not subsettable
And am not sure how to get around the problem. I've included two
short code sets below. One that shows what I want to do and works,
but without using the function much, and another that tries to use the
function but causes the error.
# THIS WORKS AND SHOWS WHAT I'D LIKE TO DO
a <- c(1,2,3)
b
2002 Jul 09
3
Assignment converts variable to factor
Hello,
I would like to know if this behaviour in R is as expected.
I have a data frame 'dat' with column var1 being character (and not
factor). Then I create a new column 'var2' by:
> dat[,"var2"] <- dat$var1;
Column var2 is now a factor.
But if I do:
> dat$var2 <- dat$var1;
Then column var2 is character (and not factor).
I don't want to have
2008 Apr 25
1
Summarize data for MCA (FactoMineR)
Hi :-)
I'm new to R and I started using it for a project (I'm the CS guy in a group
of statisticians helping them find out how to solve issues as they come out).
This is my first post to the list and I am starting to learn R.
Well, they were used to doing MCA analysis in other programs where the data
seems to be preprocessed automatically before running MCA.
So, they need to process a
2014 Apr 11
2
crear variable en base a nombre de columnas que tienen un 1
Carlos, en principio si sería algo así, sólo que en vez de quedarme con
todas las columnas var1 a var4 tuviera sólo 3, ya que en mis datos no
hay ningún caso que tenga el valor 1 en más de 3 variables..
Había llegado a una solución (mucho menos elegante que usando reshape),
que implicaba un for sobre las filas.
Jorge, creo que tu solución me vale.
Muchas gracias a los dos..
Saludos
El
2002 Nov 29
2
Obtaining the variable names of a glm object
Is names(model1$coef) what you're looking for?
-----Original Message-----
From: Kenneth Cabrera [mailto:krcabrer at epm.net.co]
Sent: 29 November 2002 10:36
Cc: R-help at stat.math.ethz.ch
Subject: [R] Obtaining the variable names of a glm object
Hi, R users!
Suppose I make a model like this:
2014 Aug 21
2
pregunta
Estimados
Estoy entrenando hacer funciones que respondan a comandos,
en esta caso en la salida gráfica se observa que dice : Exposure=var3 y
outcome=var 1
quisiéramos que se reflejan los nombres de la base de datos : var1=estado,
var2=cake, var3=chocolate
Espero haberme explicado adecuadamente
Adjunto tabla con datos
####################################
#Comando que llama