Displaying 20 results from an estimated 2000 matches similar to: "smart way to turn a vector into a matrix"
2010 Mar 11
2
Can't convert list to matrix properly
Hi guys, here is a list of names that I have:
MyList:
> myList<-list("A", "B","C","D")
> myList
[[1]]
[1] "A"
[[2]]
[1] "B"
[[3]]
[1] "C"
[[4]]
[1] "D"
I want to turn this list into a matrix of 1 row and 4 columns with those
four components (A, B, C, D) so here is what I do:
myDataFrame <-
2006 Jul 27
2
Vector extracted from a matrix. How can I specify dimensions in as.matrix?
Transpose vector extracted from a matrix
Hello,
I am doing a recursive analysis that uses every line (vector) of a matrix in
a loop. In the model, I need to transpose those vectors that are extracted
from a matrix.
Using simple vectors (no matrix involved) the transpose function works fine:
simplevector <-matrix(1:3,3,1)
tsimplevector <-t(simplevector) #transposed
dim(simplevector)
2010 Apr 05
1
Deleting many list elements in one time
Hi guys, here is a simple thing I want to do but it doesn't work:
I have a vector of the element indexes that I want to delete called index
so when I write myList[[index]] <- NULL to delete these elements here is
what I get:
Error in myList[[index]] <- NULL :
more elements supplied than there are to replace
Isn't it possible to delete multiple elements?
-----
Anna Lippel
--
2010 Apr 19
2
Using split and then unsplit
Hello everyone,
I use the split function splitting with the f function on a 3 columns and
more than 100 000 rows data frame. Once it's split I have a list of data
frames still with 3 columns and n rows. I manipulate those list elements and
get a list of data frames still with 3 columns but less rows. So when I
unsplit it, I get an error as I use the same factor function I used to split
( f in
2010 Mar 03
5
filtering signals per day
Hello R lovers,
I have a vector of dates and signals. I want to filter the signals per day
in a way that only the first signal of the day remains like this:
Dates Signals Filtered Signal
2006-11-02 0 0
2006-11-02 1 1
2006-11-02 0 0
2006-11-02 1 0
2006-11-02 1
2010 Jan 29
7
Simple question on replace a matrix row
Hello, I have a matrix mat1 of dim [1,8] and mat2 of dim[30,8], I want to
replace the first row of mat2 with mat1, this is what I do:
mat2[1,]<-mat1 but it transforms mat2 in a list I don't understand, I want
it to stay a matrix...
-----
Anna Lippel
--
View this message in context: http://n4.nabble.com/Simple-question-on-replace-a-matrix-row-tp1427857p1427857.html
Sent from the R help
2010 Mar 26
3
Using lapply with two lists
Hello guys,
I have a list L1 of matrix. I have another list L2 with the same number of
elements representing the row of the L matrix that I want to delete
(L1[[i]][-L2[[i]],]) but I can't do this with lapply as it iterates through
L1 (first argument) and not L2. Any idea?
-----
Anna Lippel
--
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2010 Jan 29
3
Applying a function on each columns of a matrix
Hello everyone, I have the following matrix
[,1] [,2] [,3] [,4]
[1,] 0.002809706 0.0063856960 0.0063856960 0.011749681
[2,] 0.004893124 0.0023118418 -0.0005122951 -0.014646465
[3,] 0.003547897 0.0063355297 0.0030410542 0.011403953
[4,] 0.004838299 -0.0040383645 -0.0090406831 -0.011027569
[5,] 0.035648755 0.0334815590 0.0380977404
2010 Mar 25
4
Convert number to Date
Hello, I have a date value in excel: 1/4/2010 which in number format gives me
40182. When I read this with read.xls from R I get same 40182 so what I do
is that I use the as.Date() function but here is what the as.Date() function
returns me:
> as.Date(40182)
[1] "2080-01-06"
Why don't I get the same thing as in excel?
Thanks!
-----
Anna Lippel
--
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2009 Nov 16
5
Writing a data frame in an excel file
Hello, I am having trouble by using the write.table function to write a data
frame of 4 columns and 7530 rows. I don?t know if I should just use a
sep="\n" and change the .xls file into a .csv file. Thanks in advance
-----
Anna Lippel
new in R so be careful I should be asking a loooooooot of questions!:teeth:
--
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2015 May 04
2
Define replacement functions
Hello
I tried to define replacement functions for the class "mylist". When I test them in an active R session, they work -- however, when I put them into a package, they don't. Why and how to fix?
make_my_list <- function( x, y ) {
return(structure(list(x, y, class="mylist")))
}
mylist <- make_my_list(1:4, letters[3:7])
mylist
mylist[['x']] <- 4:6
2009 Oct 25
3
NULL elements in lists ... a nightmare
I can define a list containing NULL elements:
> myList <- list("aaa",NULL,TRUE)
> names(myList) <- c("first","second","third")
> myList
$first
[1] "aaa"
$second
NULL
$third
[1] TRUE
> length(myList)
[1] 3
However, if I assign NULL to any of the list element then such
element is deleted from the list:
> myList$second <-
2007 Oct 20
1
Getting at what a named object represents in a function...
Hi,
I'm pretty new to R.
I have an object (say a list) and I I have a function that I call on
various columns in that list (excuse terminology if it's wrong/ambiguous).
Imagine its like this (actual values are unimportant) and called mylist:
>mylist
A B
1 5
2 5
3 6 4 8
5 0
I have a function:
foo = function(param){
#modify list A or B values depending on
2004 May 10
2
Lists and outer() like functionality?
Hi,
I'm have a list of integer vectors and I want to perform an outer()
like operation on the list. As an example, take the following list:
mylist <- list(1:5,3:9,8:12)
A simple example of the kind of thing I want to do is to find the sum
of the shared numbers between each vector to give a result like:
result <- array(c(15,12,0,12,42,17,0,17,50), dim=c(3,3))
Two for() loops is the
2011 Apr 05
1
Help in splitting a list
Dear R users,
Let's say I have a list with components being 'm' matrices (as exemplified
in the "mylist" object below). Now, I'd like to subset this list based on an
index vector, which will partition each matrix 'm' in 2 sub-matrices. My
questions are:
1. Is there an elegant way to have the results shown in mylist2 for an
arbitrary number of matrices in mylist?
2017 Jun 15
4
is.null(mylist[1]) and is.null(mylist$a) returns different values
Hi
I have a list :
mylist <- list( a = NULL, b = 1, c = 2 )
> mylist[1]
$a
NULL
> is.null(mylist[1])
[1] FALSE
> is.null(mylist$a)
[1] TRUE
why? I need to use mylist[1]
2005 Jan 30
3
trellis graphics in loops
I have this awkward problem with trellis (lattice). I am trying to
generate some plots through loops but the .eps file is empty. When I
generate them in a list and print them outside the loop all is fine. this
is an example below:( nothing shows up in foo.eps, but all show up in
foo1.eps)
R vesion 2.0.1, lattice version 0.10-16, on a debian 2.6.8-1 kernel.
X <- data.frame(x=rnorm(10000),
2001 Oct 18
2
Parsing for list components
How do I parse an identifier of a list component, e.g.
mylist$mycomponent
or
mylist[[1]] ?
Parse does not do the job, e.g.
parse(text="mylist$mycomponent")
returns an expression with just one term, instead of "mylist", "$",
"mycomponent".
What I need is a way to extract the list name (e.g. "mylist"), given
an identifier of a component.
2010 Sep 04
4
Please explain "do.call" in this context, or critique to "stack this list faster"
I've been doing some consulting with students who seem to come to R
from SAS. They are usually pre-occupied with do loops and it is tough
to persuade them to trust R lists rather than keeping 100s of named
matrices floating around.
Often it happens that there is a list with lots of matrices or data
frames in it and we need to "stack those together". I thought it
would be a simple
2011 Apr 03
1
Help in splitting ists into sub-lists
Dear List,
Let's say I have a list whose components are 2 matrices (as exemplified in
the "mylist" object below). I'd like to create a list with components being
4 matrices based on an logical index vector. is there a way to simplify what
I'm doing to obtain the results in "mylist2"? I'd like something that would
work on an arbitrary number of elements in