Displaying 20 results from an estimated 3000 matches similar to: "Efficient way to parse string and construct data.frame"
2007 Mar 01
3
Simplest question ever...
Let's say i have
a = c(1, 4, 5)
b = c(2, 6, 7)
and i have matrix m, what's an efficient way of access
m[1, 2], m[4, 6], m[5, 7]
like of course m[a, b] = is not going to do, but what's an expression that
will allow me to have that list?
Thanks!
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2007 Jun 28
2
restructuring matrix
Hi all,
let's say I have matrix
People Desc Value
Mary Height 50
Mary Weight 100
Fanny Height 60
Fanny Height 200
Is there a quick way to form the following matrix?
People Height Weight
Mary 50 100
Fanny 60 200
(Assuming I don't know the length of people/desc and let's say these are
characters matrix.. I tried
2003 May 31
5
parse on left hand side of R assignment
I keep finding myself in a situation where I want to calculate a
variable name and then use it on the left hand side of an assignment.
For example
iteration <- 1
varName <- paste("run",iteration,sep="")
myList$parse(text=varName) <- aColumn
I want to take some existing variable "aColumn" and use the name
"varName" name for it and put it into a
2001 Oct 18
2
Parsing for list components
How do I parse an identifier of a list component, e.g.
mylist$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.
2007 Aug 31
2
plotting
Hi, let's say I have data
x = c(1, 2, 10, 12)
y = c(100, -20, 50, 25)
if I go plot(x, y), then the default x-axis range goes from 1 to 12. Is
there a way to change it so that the axis looks like:
----|-----|-----|-----|----
1 2 10 12
This doesn't seem reasonable but let's say I want to plot intraday graph
with axis.POSIXct, my data is only from 8:30 to 4 every
2012 Jul 02
2
Constructing a list using a function...
Hi All
I have a dataframe:
myframe<-data.frame(ID=c("first","second"),x=c(1,2),y=c(3,4))
And I have a function myfun:
myfun<-function(x,y) x+y
I would like to write a function myfun2 that takes myframe and myfun
as parameters and returns a list as below:
mylist
$first
[1] 4
$second
[2] 6
Could you please help me with this? Doesn't seem like the
2008 May 29
2
creating library
Hi,
I'm able to create a library with R CMD INSTALL cmd, etc... I'm just
wondering.. is it possible that when the user says library(boo), it runs
some initialization code?
I have a dumb R file that is:
print(2)
boo <- function(x){}
when I R CMD INSTALL the library, I'm able to see 2 in my unix console..
but when I do
library(boo)
in R afterwards.. I don't see
2015 May 04
2
Define replacement functions
Hello
I tried to define replacement functions for the class "mylist". When I test them in an active R session, they work -- however, when I put them into a package, they don't. Why and how to fix?
make_my_list <- function( x, y ) {
return(structure(list(x, y, class="mylist")))
}
mylist <- make_my_list(1:4, letters[3:7])
mylist
mylist[['x']] <- 4:6
2017 Jun 15
4
is.null(mylist[1]) and is.null(mylist$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]
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 <-
2010 May 17
3
applying quantile to a list using values of another object as probs
Hi r-users,
I have a matrix B and a list of 3x3 matrices (mylist). I want to
calculate the quantiles in the list using each of the value of B as
probabilities.
The codes I wrote are:
B <- matrix (runif(12, 0, 1), 3, 4)
mylist <- lapply(mylist, function(x) {matrix (rnorm(9), 3, 3)})
for (i in 1:length(B))
{
quant <- lapply (mylist, quantile, probs=B[i])
}
But quant
2005 Mar 16
8
Summing up matrices in a list
Dear all,
I think that my question is very simple but I failed to solve it.
I have a list which elements are matrices like this:
>mylist
[[1]]
[,1] [,2] [,3]
[1,] 1 3 5
[2,] 2 4 6
[[2]]
[,1] [,2] [,3]
[1,] 7 9 11
[2,] 8 10 12
I'd like to create a matrix M<-mylist[[1]]+mylist[[2]]
[,1] [,2] [,3]
[1,] 8 12 16
[2,] 10 14 18
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
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
2012 Aug 28
3
Get variable data Reading from the list
Here i have a variable
MyVar <- data.frame(read.csv("D:\\Doc.csv"))
And now i am storing this variable name into a list.
MyList <- list()
MyList [length(MyList )+1]<- "MyVar"
Now what is the requirement is,
i need to call the variable name "MyVar" from the list "MyList " and get
the data.
2007 Jun 29
2
regexpr
Hi,
I 'd like to match each member of a list to a target string, e.g.
------------------------------
mylist=c("MN","NY","FL")
g=regexpr(mylist[1], "Those from MN:")
if (g>0)
{
"On list"
}
------------------------------
My question is:
How to add an end-of-string symbol '$' to the to-match string? so that 'M'
won't
2011 May 25
3
Accessing elements of a list
I have a list that is made of lists of varying length. I wish to create a
new vector that contains the last element of each list. So far I have used
sapply to determine the length of each list, but I'm stymied at the part
where I index the list to make a new vector containing only the last item
of each list
mylist =
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
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),
2004 Nov 01
5
make apply() return a list
Hi,
I have a dataframe (say myData) and want to get a list (say myList) that
contains a matrix for each row of the dataframe myData. These matrices are
calculated based on the corresponding row of myData. Using a for()-loop to do
this is very slow. Thus, I tried to use apply(). However, afaik apply() does
only return a list if the matrices have different dimensions, while my
matrices have