similar to: Efficient way to parse string and construct data.frame

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! -- View this message in context: http://www.nabble.com/Simplest-question-ever...-tf3329894.html#a9258932 Sent from the R help mailing list
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