similar to: how to type long string

Hi, I'd like to pass a string to a function as the value of the argument. Here is my code. However, it doesn't work. Please help me with this. Thanks, Yu-Ling Wu ----------------------------------------------------- library(RODBC) odbcConnect("console") -> myConnect fun1 <- function(dd1) { h3 <- sqlQuery(myConnect, "select * from console where byday =

Thanks! This is helpful. Three more questions. 1. How to convert a string into a numeric value. For example, convert '20010122' to 20010122. 2. How to convert a string into a date value. For example, convert 20010122 to 2001 Jan 22, and get its day of the week, i.e. Monday? 3. How to handle date variables in a loop? How to make the following codes work? And how users should pass the

Hello all, I am trying to use xyplot to give a six panel plot and to highlight only points (in any panel) that meet a certain criterion. With the plot command I would do something like: plot.default(filein$Site,filein$circ.conc) points(filein$Site,filein$circ.conc,type="p", pch=ifelse(filein$p.value<5e-02,19,21)) I had thought I could just stick in the pch line from above into

Times from seq.POSIXt come out wrong in AEST timezone around Feb 29 every leap year before 1970 (on Windows XP). According to help(DateTimeClasses), this is handled by "our own C code". > x <- as.POSIXct("1968-02-27") # tz="AEST" > x.gmt <- as.POSIXct("1968-02-27", tz="GMT") > data.frame( GMT=seq(x.gmt, by="day",

In barplot for degree distribution x-axis is not seen. See the example below > g = barabasi.game(500, 0.4) > dd1 = degree.distribution(g) > plot(dd1, xlab="degree", ylab = "frequency") whereas barplot doesnot have any x-axis > barplot(dd1, xlab = "degree", ylab = "frequency") Please see the figures attached.

Given a vector;> ab = seq(0.5,1, by=0.1)> ab[1] 0.5 0.6 0.7 0.8 0.9 1.0 The euclidean distance between the vector elements is given by the lower triangular matrix > dd1 = dist(ab,"euclidean")> dd1    1   2   3   4   52 0.1                3 0.2 0.1            4 0.3 0.2 0.1        5 0.4 0.3 0.2 0.1    6 0.5 0.4 0.3 0.2 0.1 Convert the lower triangular matrix to a full

This is a follow-up to the thread ending with: http://rubyforge.org/pipermail/vpim-talk/2008/000120.html I too am in search of some ruby parser for icalendar which properly handles timezones on the datetimes in the icalendar RFC. As I understand it there are actually three types of times. 1) UTC times with a string form of yyyymmddThhmmssZ note the trailing Z indicates zulu time aka utc. 2)

Dear R-helpers, I have run the code below which I expected to make an object called dd1, but that object does not exist. So, in summary, my problem is that my function is meant to make an object (dd1), and it does indeed make that object (I know that the last line of the function prints it out) but then, after the function has run, the object has disappeared. It's late on a Friday so I may

Buenas tardes, He construido la función “myfun” al objeto de considerar aquellas persones que a partir de una determinada fecha de Apertura tienen como mínimo 65 años. Se tiene su fecha de nacimiento, su fecha de inicio en la institución y su fecha de salida de la misma. Doy vueltas al script y no acabo se saber cómo poder aplicar de un modo eficiente las instrucciones “if” ó bien “ifelse”, y me

Hi, I am plotting degree distribution of a graph using the function, library(igraph) dd1 = degree.distribution(G) plot(dd1, xlab = "degree", ylab="frequency") I would like to plot the mean of the distribution as a vertical line in the attached plot. Please let me know how to do this. Thanks, Kumar http://r.789695.n4.nabble.com/file/n3334375/cdata3_dd.png cdata3_dd.png --

Dear All, Suppose I have a program as below: Outside is a loop for simulation (with random generated data), inside there are several sapply()'s (10~100) over the data and something else, but these sapply's have to be sequential. And each sapply do not involve very intensive calculation (a few seconds only). So the outside loop takes minutes to finish one iteration. I guess the better way

I think that there is a bug in the as.POSIXct function on Windows. Here is what I get on Win2000, Pentium III machine in R 1.8.1. > dd1 <- ISOdatetime(2003, 10, 26, 0, 59, 59) > dd2 <- ISOdatetime(2003, 10, 26, 1, 0, 0) > dd2 - dd1 Time difference of 1.000278 hours Now, the 26th of October was the day that change to the standard time occurred, so I suspect that this has

Dear all, I have an xts object , t.xts with 4 columns: "v1" "DD1" "v2" "DD2" and created a data frame : t <- as.data.frame(t.xts) I would like to extract data and create a new data frame for when the values in column DD1 falls between 0 and 30 and extract the corresponding v1 value. How can I do this? Thanks. -- View this message in context:

Hi there, Problem :: When one tries to change one or some of the columns of a data.frame, R makes a copy of the whole data.frame using the '*tmp*' mechanism (this does not happen for components of a list, tracemem( ) on R-2.6.2 says so). Suggested solution :: Store the columns of the data.frame as a list inside of an environment slot of an S4 class, and define the '[',

Given a vector;> ab = seq(0.5,1, by=0.1)> ab[1] 0.5 0.6 0.7 0.8 0.9 1.0 The euclidean distance between the vector elements is given by the lower triangular matrix > dd1 = dist(ab,"euclidean")> dd1    1   2   3   4   52 0.1                3 0.2 0.1            4 0.3 0.2 0.1        5 0.4 0.3 0.2 0.1    6 0.5 0.4 0.3 0.2 0.1 Convert the lower triangular matrix to a full

I am trying to call a funtion within another function and I clearly am misunderstanding what I should do. Below is a simple example. I know lstfun works on its own but I cannot seem to figure out how to get it to work within ukn. Basically I need to create the variable "nts". I have probably missed something simple in the Intro or FAQ. Any help would be much appreciated. EXAMPLE

I am lousy at simple regex and I have not found a solution to a simple problem. I have a vector with some character values that I want to split. Sample data dd1 <- c( "XXY (mat harry)","XXY (jim bob)", "CAMP (joe blow)", "ALP (max jack)") Desired result dd2 <- data.frame( xx = c("XXY", "XXY", "CAMP",

No, I'm afraid I'm wrong. Something went wrong with my R session and gave me incorrect answers. After restarting, I continued to get the same error as you did with my supposed "fix." So just ignore what I said and sorry for the noise. -- Bert On Sat, Jul 8, 2023 at 8:28?AM Bert Gunter <bgunter.4567 at gmail.com> wrote: > Try this for your function: > >

Dear all, I just found this issue: dd1 = c(rep(NaN,82), rep(-1, 144), rep(1, 74)) xx = runmed(dd1, 21) -> R crashes reproducibly in R 3.4.3, R3.4.4 (Ubuntu 14.04/Ubuntu 16.04) With GDB: Program received signal SIGSEGV, Segmentation fault. swap (l=53, r=86, window=window at entry=0xc59308, outlist=outlist at entry=0x12ea2e8, nrlist=nrlist at entry=0x114fdd8, print_level=print_level at

I need to apply a yearly inflation factor to some wages and supply some simple sums by work category. I have gone at it with a brute force "for" loop approach which seems okay as it is a small dataset. It looks a bit inelegant and given all the warnings in the Intro to R, etc, about using loops I wondered if anyone could suggest something a bit simpler or more efficent? Example: