similar to: looking for a cleaner way to do something

Displaying 20 results from an estimated 900 matches similar to: "looking for a cleaner way to do something"

2009 Feb 18
2
understanding how R determines numbers and characters when creating a data frame
Hello R Users and Developers, I have a basic question about how R works. Over the past few years I have struggled when I try to generate a new data frame that I believe should contain numeric data in some columns and character data in others only to find everything converted to character data. Is there a general method to create data frames that contain the data in the desired format: numbers
2006 Aug 19
2
A matrix problem
Hi, I have a matrix with two columns. The first column means "indexes", the second one contents of those indexes. If I have a MATRIX like this, > MATRIX [,1] [,2] [1,] 1 3 [2,] 5 1 [3,] 2 1 [4,] 1 5 I'd like to get as a result vector the sums of these indexes, something like this: > c(8,1,0,0,1) How to do this? I did solved it this way, but is
2005 Oct 10
4
sip register incoming call contexts?
Sorry this is a bit of a newbie question, I've been at this for a few months and still have not quite figured this one out. I've been able to setup one itsp (incoming calls) (sip account) with a register line like this: register => nnnnnnn:ppppp@sip.provider.net -or- register => nnnnnnn:ppppp@sip.provider.net/nnn to come directly into an extension in the dialplan It seems that
2004 Sep 09
2
Skipping panels in Lattice
Dear all, I wish to generate a lattice boxplot which skips an empty cell in a design. I have trawled r-help, scruitinized xyplot(lattice) help page, and merrily reproduced examples of using skip from a couple of previous r-help queries and the example given in Pinheiro & Bates. But I must be missing something... Here's an example (running R 1.9.1 on Win2k): # generate some data df1
2007 Mar 03
1
function doesnt return/create object
hello, i have written a function to extract certain lines from a matrix. the result is a matrix with 6 cols, named dynamically according to the functions arguments. the problem is now, that i'm not able to return the resultmatrix for further use. the object is not being created. example from my console: getans(27,27) [...] [189,] 3969 161 27 1 0 1 [190,] 2142 87 27
2009 Feb 11
2
sorting a matrix by the column
this is a bad question but I can't figure it out and i've tried. if i sort the 2 column matrix , temp1, by the first column, then things work as expected. But, if I sort the 1 column matrix, temp2, then it gets turned coerced to a vector. I realize that I need to use drop=FALSE but i've put it in a few different places with no success. Thanks. temp1 <-
2009 Feb 11
2
error in my previous message
i'm sorry. i had an error in my previous code because i left out a letter in the rownames. while fixing that, i also found a solution. so i'm sorry for the confusion. below is my fix. temp2 <- matrix(rnorm(10),nc=1,nrow=10) rownames(temp2) <-
2006 Jan 04
3
matrix math
I am using R 2.1.1 in an windows XP environment. I have 2 dataframes, temp1 and temp2. Each dataframe has 20 variables (“cocolumns") and 525 observations (“rows”). All variables are numeric. I want to create a new dataframe that also has 20 columns and 525 rows. The values in this dataframe should be the sum of the 2 other dataframe. (i.e. temp1$column
2012 Oct 31
2
Aggregate Table Data into Cell Frequencies
R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the "temp" (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency ("Freq" | counts) of each time each "delayValue" occurs
2010 Nov 19
3
Sweave Dynamic Graph Question
i have a time Series of IBM closing px from 1/1/2000 to today I want to graph the time serie by dividing the graph by year and month all the monthly graphs with the same year will go to one page. so from 1/1/2000 to 11/19/2010. i will have 11 pages, and each page will have 12 graphs (jan to dec) except for 2010. I am able to do it in R, but when i use sweave, I can only print the last page.
2003 May 31
1
Minor problem with unwritable directories
Hi, I'm not a member of the list, but I think I've run across a (minor) bug. The rsync web page directed me to this list for bug reports. Anyway, I've described it below (including a simple test case with exact commands to reproduce and the output I get from those commands). I hope this helps. Let me know if any more information is needed. Elijah P.S. If it isn't a bug,
2006 Mar 03
1
NA in eigen()
Hi, I am using eigen to get an eigen decomposition of a square, symmetric matrix. For some reason, I am getting a column in my eigen vectors (the 52nd column out of 601) that is a column of all NAs. I am using the option, symmetric=T for eigen. I just discovered that I do not get this behavior when I use the option EISPACK=T. With EISPACK=T, the 52nd eigenvector is (up to rounding error) a
2011 Mar 09
2
Anomaly with unique and match
I stumbled onto this working on an update to coxph. The last 6 lines below are the question, the rest create a test data set. tmt585% R R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) # Lines of code from survival/tests/singtest.R > library(survival) Loading required package: splines
2008 Aug 26
1
parse and eval character vector
Dear R-help, I have a character vector, some elements will be numeric, some not, and some even empty. E.g.: temp1 <- c("abcd"," 2 ","") I'm only interested in the numeric elements, the rest I can just throw away. It is easy enough to loop through the vector: temp <- try(eval(parse(text=temp1[1])), silent=TRUE); class(temp) # try-error temp <-
2009 May 28
2
Replace is leaking?
Okay, someone explain this behaviour to me: Browse[1]> replace(rep(0, 4000), temp1[12] , temp2[12])[3925] [1] 0.4462404 Browse[1]> temp1[12] [1] 3926 Browse[1]> temp2[12] [1] 0.4462404 Browse[1]> replace(rep(0, 4000), 3926 , temp2[12])[3925] [1] 0 For some reason, R seems to shift indices along when doing this replacement. Has anyone encountered this bug before? It seems to crop up
2009 Jan 10
0
RMySQL CREATE TABLE error
Hi all- I am stumped. The code in A. returns errors at lines 14 and 15 and fails to load series1 and series2. However, in B., if temp1 and temp2 are called again (which returns a "Table exists" error; see lines 14-17 in B.) series1 and series2 load correctly. Any ideas? Also-I am open to any suggestions to improve the code as I am a horrific programmer. Thanks A. 1 >
2005 Nov 23
2
vector of permutated products
Given an x-vector with, say, 3 elements, I would like to compute the following vector of permutated products (1-x1)*(1-x2)*(1-x3) (1-x1)*(1-x2)*x3 (1-x1)*x2*(1-x3) x1*(1-x2)*(1-x3) (1-x1)*x2*x3 x1*(1-x2)*x3 x1*x2*(1-x3) x1*x2*x3 Now, I already have the correctly sorted matrix of permutations! So, the input looks something like: #input x<-c(0.3,0.1,0.2) Nx<-length(x) Ncomb<-2^Nx
2009 May 15
4
replace "%" with "\%"
Dear all, I'm trying to gsub() "%" with "\%" with no obvious success. > temp1 <- c("mean", "sd", "0%", "25%", "50%", "75%", "100%") > temp1 [1] "mean" "sd" "0%" "25%" "50%" "75%" "100%" > gsub("%",
2005 Dec 29
1
Problems with calloc function.
Hi all, I have a C code in Linux, it has 7 pointers and compile e run OK, but when I run in R happens problems with calloc function, it returns NULL. ############################################### > int *temp1,*temp2,*temp3,*temp4; temp1 = (int *)calloc(col,sizeof(int)); if(temp1 == NULL){ printf("\n\n No Memory1!"); exit(1); } temp2 = (int *)calloc(col,sizeof(int));
2011 Dec 12
1
Lagged values problem requiring short solution time
Hello, I am hoping someone can help tackle the problem below, for which I require a fast solution. It feels like there should be an elegant approach, but I am drawing blanks. Take a vector 'x' with random values > 0: x = runif(10,1,5) Assume some reasonably small positive value 'delta': delta = 0.75 The task is to find a solution vector 'y' of same length as