Displaying 20 results from an estimated 40000 matches similar to: "apply ( , , table)"
2004 Feb 06
3
column names in matrix vs. data frame in R 1.8
Is the difference in behavior below, introduced in 1.8, inconsistent or,
at least, undesirable? I couldn't find this in the NEWS.
On the one hand,
> a <- matrix (1:4, nrow=2)
> a <- data.frame (a)
> names (a) <- c("break","next")
> names (a)
[1] "break" "next"
On the other,
> a <- matrix (1:4, nrow=2)
> dimnames(a)
2010 Aug 03
2
How to name matrices from a list with lapply ?
Dear list,
I have a list of matrices :
i1 <- matrix(1:10, nrow = 2, ncol = 5)
i2 <- matrix(11:20, nrow = 2, ncol = 5)
j <- list(i1 = i1, i2 = i2)
I would like to attribute names to each dimension, for each matrix,
as follows :
$i1
B1 B2 B3 B4 B5
A1 1 3 5 7 9
A2 2 4 6 8 10
$i2
B1 B2 B3 B4 B5
A1 11 13 15 17 19
A2 12 14 16 18 20
However, I have to use lapply function and attribute
2003 Mar 05
2
reserved words documentation
I wanted a data frame component to be named "next", for example:
> m <- data.frame (matrix (0, nrow=2, ncol=2))
> names (m) <- c("prev", "next")
> m
prev next
1 0 0
2 0 0
But "next" being reserved prevents $ indexing without quotes:
> m$next
Error: syntax error
> m$"next"
[1] 0 0
Although they are mostly
2010 Aug 23
3
extracting p-values from Anova objects (from the car library)
Dear all,
is there anyone who can help me extracting p-values from an Anova object
from the car library? I can't seem to locate the p-values using
str(result) or str(summary(result)) in the example below
> A <- factor( rep(1:2,each=3) )
> B <- factor( rep(1:3,times=2) )
> idata <- data.frame(A,B)
> fit <- lm( cbind(a1_b1,a1_b2,a1_b3,a2_b1,a2_b2,a2_b3) ? sex,
2005 Jan 13
4
zero index and lazy evaluation in ifelse()
I don't understand this behavior:
> a <- c(0, 1, 2, 3)
> b <- c(1, 2, 3, 4)
> ifelse (a == 0, 0, b[a])
[1] 0 2 3 1
rather than the desired 0 1 2 3. Thanks for any explanation.
2013 Sep 04
2
Random products of rows in a matrix
Hello everybody,
Without any loop and any package,
I would like to return N products of M rows in a matrix A :
Today, I managed to do it with a loop :
B <- matrix(NA, ncol = ncol(A), nrow = 0)
for (i in 1 : N) B <- rbind(B, apply(A[sample(1 : nrow(A), M, replace = T),
], 2, prod))
Do you have a solution ?
Thank you in advance !
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2013 Feb 23
1
how to calculate left kronecker product?
For an application, I have formulas defined in terms of a left Kronecker
product of matrices,
A,B, meaning
A \otimes_L B = {A * B[i,j]} -- matrix on the left multiplies each
element on the right.
The standard kronecker() function is the right Kronecker product,
A \otimes_R B = {A[i,j] * B} -- matrix on the right multiplies each
element on the left.
The example below shows the result of
2012 Apr 19
3
How to "flatten" a multidimensional array into a dataframe?
Hi,
I have a three dimensional array, e.g.,
my.array = array(0, dim=c(2,3,4), dimnames=list( d1=c("A1","A2"),
d2=c("B1","B2","B3"), d3=c("C1","C2","C3","C4")) )
what I would like to get is then a dataframe:
d1 d2 d3 value
A1 B1 C1 0
A2 B1 C1 0
.
.
.
A2 B3 C4 0
I'm sure there is one function to do
2004 Mar 30
4
rank() vs SAS proc rank
SAS proc rank has ties options of high and low that would allow
producing ranks of the type found in the sports pages, e.g.,
rank (c(1,1,2,2,2,2,3)) == 1 1 3 3 3 3 7
Could R support these ties.methods?
2012 Jun 02
2
Add a dim to an array
Dear list,
I'm trying to add a new dim to a multidimensional array. My array looks
like this
a1 <- array(1:8, c(2, 2, 2))
dimnames(a1) <- list(A = c("A1", "A2"),
B = c("B1", "B2"),
D = c("D1", "D2"))
I would like to add a new dim 'group' with the value "low".
2013 Nov 21
1
how can I import a number of datsets in a folder in my working directory to a list in R
Hi,
Suppose, if I create 15 files in my working directory.
set.seed(48)
lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)})
?D <-dir()
D1 <- D[order(as.numeric(gsub("\\D+","",D)))]
D1
?res <- t(sapply(D1,function(x) {x1<-
2017 Jun 06
2
surprisingly, S4 classes with a "dim" or "dimnames" slot are final (in the Java sense)
Hi,
It's nice to be able to define S4 classes with slots that correspond
to standard attributes:
setClass("A1", slots=c(names="character"))
setClass("A2", slots=c(dim="integer"))
setClass("A3", slots=c(dimnames="list"))
By doing this, one gets a few methods for free:
a1 <- new("A1", names=letters[1:3])
2009 Feb 06
1
Linear model: contrasts
Hey,
I am modelling a linear regression Y=X*B+E. To compute the effect of ?group? the B-values of the regressors/columns that code the interaction effects (col. 5-8 and col. 11-14, see below) have to be weighted with non-zero elements within the contrast "Group 1" minus "Group 2" (see below). My first understanding was that the interaction effects add up to zero in each group.
2005 Jun 03
2
rearrange data
Dear all:
I have this:
A1 B1 C1 D1 E1
A2 B2 C2 D2 E2
A3 B3 C3 D3 E3
And I want this
A1 E1
B1 E1
C1 E1
D1 E1
A2 E2
B2 E2
C2 E2
D2 E2
A3 E3
B3 E3
C3 E3
D3 E3
Example:
m<- matrix(1:15,nrow=3,byrow=T)
m
v<- unlist(list(t(m[,1:4])))
u<- rep(c(5,10,15),c(4,4,4))
data.frame(v,u)
This is the result I want but I would like to learn a simpler way to do it.
Any clue?
2013 Mar 19
1
How can I eliminate a loop over a data.table?
I've two data.tables as shown below:
***
N = 10
A.DT <- data.table(a1 = c(rnorm(N,0,1)), a2 = NA))
B.DT <- data.table(b1 = c(rnorm(N,0,1)), b2 = 1:N)
setkey(A.DT,a1)
setkey(B.DT,b1)
***
I tried to change my previous data.frame implementation to a
data.table implementation by changing the for-loop as shown below:
***
for (i in 1:nrow(B.DT)) {
for (j in nrow(A.DT):1) {
if
2004 Jul 08
2
Getting elements of a matrix by a vector of column indice s
See if the following helps:
> m <- outer(letters[1:5], 1:4, paste, sep="")
> m
[,1] [,2] [,3] [,4]
[1,] "a1" "a2" "a3" "a4"
[2,] "b1" "b2" "b3" "b4"
[3,] "c1" "c2" "c3" "c4"
[4,] "d1" "d2" "d3" "d4"
[5,]
2006 Sep 11
5
Successive Graphs
Hello! I have written an R script on a Windows platform where I
calculate eight result matrices I plot using matplot. I would like to
display the resulting plots successively, rather than simultaneously,
and I was wondering if anyone could point me in the right direction as
to how to do this. The graphs pop up in this manner by default when I
run my script in S-PLUS, with tabs separating them so I
2003 Apr 04
3
creating function bodies using body()
I'm having trouble figuring out how to create a function using "body<-"
(). The help file for body() says that the argument should be a list of
R expressions. However if I try that I get an error:
> tmpfun <- function(a, b=2){}
> body(tmpfun) <- list(expression(z <- a + b),expression(z^2))
Error in as.function.default(c(formals(f), value), envir) :
2007 Aug 07
1
Naming Lists
Hi
Im pretty new to R and I have run in to a problem. How do I name all the
levels in this list.
Lev1 <- c("A1","A2")
Lev2 <- c("B1","B2")
Lev3 <- c("C1","C2")
MyList <- lapply(Lev1,function(x){
lapply(Lev2,function(y){
lapply(Lev3,function(z){
paste(unlist(x),unlist(y),unlist(z))
})})})
I would like to name the different
2012 May 04
7
Breaking up a Row in R (transpose)
I have the following:
Time A1 A1 B1 B1 C1 C2
x y x y x y
0 5 6 6 7 7 9
1 3 4 4 3 9 9
2 5 2 6 4 7 4
I want to change it to the following:
0 1 2
x y x y x y
A1 5 6 3 4 5 2
B1