Displaying 20 results from an estimated 20000 matches similar to: "Assignment converts variable to factor"
2011 Jan 23
3
FUNC_ODBC and ARRAY
Gentlemen,
I have googled, searched the mailing list archives, and even spoke on
the IRC channel, but have not found an answer to the following
problem. I am attempting to retrieve multiple columns in an ODBC query
using ARRAY per the solutions offered by many individuals. My dialplan
code is as follows:
exten => _.,n,Set(ARRAY(var1,var2,var3)=${ODBC_LOOKUP(${KEYVAL})})
exten =>
2010 Jan 25
5
Data transformation
Dear all,
I have a dataset that looks like this:
x <- read.table(textConnection("col1 col2
3 1
2 2
4 7
8 6
5 10"), header=TRUE)
I want to rewrite it as below:
var1 var2 var3 var4 var5 var6 var7 var8 var9 var10
1 0 1 0 0 0 0 0 0 0
0 2 0 0 0 0 0 0 0 0
0 0 0 1 0 0
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
2011 Jun 17
4
profile plot in R
Hi friends,
I have a matrix with following format.
group var1 var2 .......varN
c1 group1 1.2399 1.4990....-1.4829
c2 group4 0.8989 0.7849.....1.8933
...
...
c100 group10 .....
I want to draw a profile plot
of each condition c1 to c100, which rows in above matrix and each line
representing a row should be uniquely colored according to the group(1
to 10).
I think this is simple task
2009 Jan 13
3
Comparing elements for equality
Suppose I have a dataframe as follows:
dat <- data.frame(id = c(1,1,2,2,2), var1 = c(10,10,20,20,25), var2 =
c('foo', 'foo', 'foo', 'foobar', 'foo'))
Now, if I were to subset by id, such as:
> subset(dat, id==1)
id var1 var2
1 1 10 foo
2 1 10 foo
I can see that the elements in var1 are exactly the same and the
elements in var2 are exactly
2012 Oct 17
2
loop of quartile groups
Greetings R users,
My goal is to generate quartile groups of each variable in my data set. I
would like each experiment to have its designated group added as a
subsequent column. I can accomplish this individually with the following
code:
brks <- with(data_variables,
cut2(var2, g=4))
#I don't want the actual numbers, I need a numbered group
data$test1=factor(brks,
2010 Nov 07
2
is this matrix symmetric
Hi,
I have this symmetric matrix, at least I think so.
col1 col2 col3
[1,] 0.20 0.05 0.06
[2,] 0.05 0.10 0.03
[3,] 0.06 0.03 0.08
or
structure(c(0.2, 0.05, 0.06, 0.05, 0.1, 0.03, 0.06, 0.03, 0.08
), .Dim = c(3L, 3L), .Dimnames = list(NULL, c("var1", "var2",
"var3")))
But isSymmetric() doesn't agree. Any comment? I am on R 2.10.1 Thanks.
Jun
2007 Apr 02
3
Create a new var reflecting the order of subjects in existing var
Dear R helpers
I have a data set sth like this:
set.seed(123);dat <- data.frame(ID= c(rep(1,2),rep(2,3), rep(3,3), rep(4,4),
rep(5,5)),
var1 =rnorm(17, 35,2),
var2=runif(17,0,1))
dat
ID var1 var2
1 1 33.87905 0.02461368
2 1 34.53965 0.47779597
3 2 38.11742 0.75845954
4 2 35.14102 0.21640794
5 2 35.25858 0.31818101
6 3 38.43013
2017 Sep 25
0
Sample of a subsample
For personal aesthetic reasons, I changed the name "data" to "dat".
Your code, with a slight modification:
set.seed (1357) ## for reproducibility
dat <- data.frame(var1=seq(1:40), var2=seq(40,1))
dat$sampleNo <- 0
idx <- sample(seq(1,nrow(dat)), size=10, replace=F)
dat[idx,"sampleNo"] <-1
## yielding
> dat
var1 var2 sampleNo
1 1 40
2017 Sep 25
1
Sample of a subsample
Hi David,
I was about to post a reply when Bert responded. His answer is good
and his comment to use the name 'dat' rather than 'data' is instructive.
I am providing my suggestion as well because I think it may address
what was causing you some confusion (mainly to use "which", but also
the missing !)
idx2 <- sample( which( (!data$var1%%2) & data$sampleNo==0 ),
2003 Aug 26
2
Simple simulation in R
Hello all
I have a feeling this is very simple......but I am not sure how to do
it
My boss has two variables, one is an average of 4 numbers, the other is
an average of 3 of those numbers i.e
var1 = (X1 + X2 + X3 + X4)/4
var2 = (X1 + X2 + X3)/3
all of the X variables are supposed to be measuring similar constructs
not surprisingly, these are highly correlated (r = .98), the question
is how
2012 Jan 10
4
Sum of a couple of variables of which a few have NA values
Dear everyone,
I have looked all over the internet but I cannot find a way to solve my problem.
In my data I want to sum a couple of variables. Some of these
variables have NA values, and when I add them together, the result is
NA
dat <- data.frame(
id = gl(5,1),
var1 = rnorm(5, 10),
var2 = rnorm(5, 7),
var3 = rnorm(5, 6),
var4 = rnorm(5, 3),
var5 = rnorm(5, 8)
)
dat[3,3] <- NA
dat[4,5]
2012 Jan 16
3
Select rows based on multiple comparisons
Dear all,
I have a data set in which the same unit appears 2 or 3 or 4 times. I need
to aggregate this data to maintain only one unit by row. But I need to do
that based on a comparison between the values of such units. I can't find a
function to help me on that. I appreciate any help. Below I provide an
example of what I want:
This is my data:
Units Var1 Var2 Var3
1 B 2
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),
2013 Dec 06
2
Using assign with mapply
I have a data frame whose first colum contains the names of the variables
and whose second colum contains the values to assign to them:
: kkk <- data.frame(vars=c("var1", "var2", "var3"),
vals=c(10, 20, 30), stringsAsFactors=F)
If I do
: assign(kkk$vars[1], kkk$vals[1])
it works
: var1
[1] 10
However, if I try with mapply
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))) }
)
)
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
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
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 <-
2010 Dec 20
2
Turning a Variable into String
I would like to know how to turn a variable into a string. I have tried
as.symbol and as.name but it doesnt work for what I'd like to do
Essentially, I'd like to feed the function below with two variables. This
works fine in the bit working out number of elements in each variable.
In the print(sprintf("OK with %s and %s\n", var1, var2)) line I would like
var1 and var2 to be