Displaying 20 results from an estimated 10000 matches similar to: "wildcard operator"
2006 Feb 17
2
creating 3-way tables for mantelhaen.test
Hi R users
I have serveral binary variables (e.g., X1, X2, X3, X4, X5, X,6, and X7) and
one continuous variable (e.g., Y1).
I combined these variables using data.frame()
mydata <- data.frame(X1,X2,X3,X4,X5,X6,X7,Y1)
after that, I sorted this data.frame
rank.by.Y1<-order(mydata[,8])
sorted.mydata<-mydata[rank.by.Y1,]
after that, I replaced Y1's values with values ranging from 1
2007 Aug 26
3
subset using noncontiguous variables by name (not index)
Hi All,
I'm using the subset function to select a list of variables, some of
which are contiguous in the data frame, and others of which are not. It
works fine when I use the form:
subset(mydata,select=c(x1,x3:x5,x7) )
In reality, my list is far more complex. So I would like to store it in
a variable to substitute in for c(x1,x3:x5,x7) but cannot get it to
work. That use of the c function
2011 Feb 25
1
help please ..simple question regarding output the p-value inside a function and lm
Dear R community members and R experts
I am stuck at a point and I tried with my colleagues and did not get it out.
Sorry, I need your help.
Here my data (just created to show the example):
# generating a dataset just to show how my dataset look like, here I have x
variables
# x1 .........to X1000 plus ind and y
ind <- c(1:100)
y <- rnorm(100, 10,2)
set.seed(201)
P <-
2012 Oct 20
2
Help with programming a tricky algorithm
Hi All,
I'm a little stumped by the following problem. I've got a dataset with
the following structure:
idxy ix iy country (other variables)
1 1 1 c1 x1
2 1 2 c1 x2
3 1 3 c1 x3
. . . . .
3739 55 67 c7 x3739
3740 55 68 c7 x3740
where ix and
2012 Jul 23
3
How to do the same thing for all levels of a column?
Dear all,
I am a R beginner, and I am looking for a way to do the same thing for all
levels of a column in a table.
Basically, I have a bunch of protein sequences composed of different amino
acid residues, and each residue is represented by an uppercase letter. I
want to calculate the ratio of different amino acid residues at each
position of the proteins. Here is an example table:
Proteins
2020 Sep 10
5
aplicar codigo
Hola:
Como dice Carlos, algo así, por ejemplo:
transforma <- function(df) sapply(df, function(x)
ifelse(x%in%c("x1","x2","x3"),
"prueba1",ifelse(x%in%c("x4","x5","x6"),"prueba2",x)))
> transforma(df1)
col1
[1,] "prueba1"
[2,] "prueba1"
[3,] "x11"
[4,]
2013 May 29
3
bootstrap
Hi,
You might need to check library(boot).? I have never used that before.? So, I can't comment much.? It is better to post on R-help list.? I had seen your postings on Nabble in the past.? Unfortunately those postings were not accepted in R-help.? You have to directly post at ? r-help at r-project.org after registering at:
https://stat.ethz.ch/mailman/listinfo/r-help
?
2005 Oct 05
8
R crashes for large formulas in lm() (PR#8180)
Full_Name: Hallgeir Grinde
Version: 2.1.1
OS: Windows XP
Submission from: (NULL) (144.127.1.1)
While using lm(y~(x*z*c*...*v)^2) R crashes/closes if the numbers of variables
are at least 8.
2020 Sep 10
3
aplicar codigo
Hola,
me gustar?a hacer algo como en el siguiente ejemplo
A un df a?adirle una columna que es la transformaci?n de otra,
en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1
todo lo que sea x4,x5,x6 lo llamo prueba 2
el resto de x las dejo como est?n.
Ser?a algo as?
col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5',
2007 Jan 30
2
R and S-Plus got the different results of principal component analysis from SAS, why?
Dear Rusers,
I have met a difficult problem on explaining the differences of principal
component analysis(PCA) between R,S-PLUS and SAS/STATA/SPSS, which wasn't
met before.
Althought they have got the same eigenvalues, their coeffiecients were
different.
First, I list my results from R,S-PLUS and SAS/STATA/SPSS, and then show
the original dataset, hoping sb. to try and explain it.
2020 Sep 10
2
aplicar codigo
Yo copio y pego este código y me sale correctamente. Se me ocurre que
pueda deberse a la versión de R ¿cuál usas?
El 10/09/2020 a las 17:51, Samura . escribió:
> Gracias por las respuestas.
>
> Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal.
> Ahora con el código de Marcelino tampoco me sale.
>
> col1 <- c('x1', 'x2', 'x11',
2013 Apr 13
1
how to add a row vector in a dataframe
Hi,
Using S=1000
and
simdata <- replicate(S, generate(3000))
#If you want both "m1" and "m0" #here the missing values are 0
res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1})
res1[,997:1000]
#????? [,1]???????? [,2]???????? [,3]???????? [,4]???????
#x1??? Numeric,3000 Numeric,3000
2005 Jun 29
2
quick way to construct formula
Dear R users,
I have a data with 1000 variables named "x1", "x2", ..., "x1000", and
I want to construct a formula like this format:
~x1+x2+...+x1000+x1:x2+x1:x3+x999:x1000+log(x1)+...+log(x1000)
That is: the base variables followed by all interaction terms and all
base feature log-transformations. I know I can use several paste
functions to construct it. But is
2011 Oct 31
3
How to get Quartiles when data contains both numeric variables and factors
When data contains both factor and numeric variables, how to get quartiles
for all numeric variables?
n <- 100
x1 <- runif(n)
x2 <- runif(n)
x3 <- x1 + x2 + runif(n)/10
x4 <- x1 + x2 + x3 + runif(n)/10
x5 <- factor(sample(c('a','b','c'),n,replace=TRUE))
x6 <- factor(1*(x5=='a' | x5=='c'))
data1 <- cbind(x1,x2,x3,x4,x5,x6)
data
2008 Dec 22
1
sem package fails when no of factors increase from 3 to 4
#### I checked through every 3 factor * 3 loading case.
#### While, 4 factor * 3 loading failed.
#### the data is 6 factor * 3 loading
require(sem);
cor18<-read.moments();
1
.68 1
.60 .58 1
.01 .10 .07 1
.12 .04 .06 .29 1
.06 .06 .01 .35 .24 1
.09 .13 .10 .05 .03 .07 1
.04 .08 .16 .10 .12 .06 .25 1
.06 .09 .02 .02 .09 .16 .29 .36 1
.23 .26 .19 .05 .04 .04 .08 .09 .09 1
.11 .13 .12 .03 .05 .03
2010 Apr 19
2
How to pass a list of parameters into a function
Does anyone know how to pass a list of parameters into a function?
for example:
somefun=function(x1,x2,x3,x4,x5,x6,x7,x8,x9){
ans=x1+x2+x3+x4+x5+x6+x7+x8+x9
return(ans)
}
somefun(1,2,3,4,5,6,7,8,9)
# I would like this to work:
temp=c(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9)
somefun(x1=1,x2=2,temp)
# OR I would like this to work:
temp=list(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9)
2006 Aug 31
3
what's wrong with my simulation programs on logistic regression
Dear friends,
I'm doing a simulation on logistic regression model, but the programs can't
work well,please help me to correct it and give some suggestions.
My programs:
data<-matrix(rnorm(400),ncol=8) #sample size is 50
data<-data.frame(data)
names(data)<-c(paste("x",1:8,sep="")) #8 independent variables,x1-x8;
#logistic regression model is
2010 Apr 03
1
"moving product", by rows, of a dataframe?
What is a "good" way to calculate the "moving product", for each row of a dataframe, where I wish to be able to specify the length of the moving product? Depending on my needs, I'd like to be able to specify the "length" over which to calculate the moving average (in this example, length=3).
For example, if I have a dataframe with 20 rows and 6 columns, and I
2005 Oct 05
1
Ad: Re: R crashes for large formulas in lm() (PR#8180)
Dette er en melding med flere deler i MIME-format.
--=_alternative 004613C000257091_=
Content-Type: text/plain; charset="US-ASCII"
And some more informastion I forgot.
R does not crash if I write out the formula:
set.seed(123)
x1 <- runif(1000)
x2 <- runif(1000)
x3 <- runif(1000)
x4 <- runif(1000)
x5 <- runif(1000)
x6 <- runif(1000)
x7 <- runif(1000)
x8 <-
2012 Oct 12
7
ifelse reformulation
Hi, i'm trying to simplify some R code but i got stucked in this:
test<-data.frame(cbind(id,x1,x2,x3,x4,x5,x6,x7))
test
> test
id x1 x2 x3 x4 x5 x6 x7
1 1 36 26 21 32 31 27 31
2 2 45 21 46 50 22 36 29
3 3 49 47 35 44 33 31 46
4 4 42 32 38 28 39 45 32
5 5 29 42 39 48 25 35 34
6 6 39 31 30 37 46 43 44
7 7 41 40 25 23 42 40 24
8 8 27 29 47 34 26 38 28
9 9 25 35 29 36