similar to: need help

here's my question: suppose I have a matrix: mt<-matrix(1:12,ncol=6) now I have a vector vt<-c(1,2,2,2,1,2) which means I want to get: the 1st row for column1; the 2nd row for column2; the 2nd row for column3; the 2nd row for column4; ... that what I want is this vector: 1,4,6,8,9,12 Does anyone know how to do this fast? I know I can use for-loop to travel all columns,but

I'm trying to analyze the following data set (sample): "ID" "adj.P.Val" "logFC" "Gene.symbol" "1419156_at" "5.32e-12" "2.6462565" "Sox4" "1433575_at" "5.32e-12" "3.9417089" "Sox4" "1428942_at" "2.64e-11"

My box: Slackware-current, Xfree 4.4. ERROR as follows: gcc -I. -I../../../src/include -I../../../src/include -I/usr/X11R6/include -I/usr/local/include -DHAVE_CONFIG_H -D__NO_MATH_INLINES -mieee-fp -fPIC -g -O2 -c dataentry.c -o dataentry.lo In file included from dataentry.c:31: /usr/X11R6/include/X11/Xlib.h:1390: error: parse error before "_Xconst"

--T4sUOijqQbZv57TR Content-Type: text/plain; charset=us-ascii Content-Disposition: inline Dear R, I would like to report what I think is a bug in R. I am running R within emacs on a Digital AlphaStation. See the version information at the end of my R session for details. I also attach a copy of the file that is read in the `read.table' command. Here's my R session, with a few

Hi all, I am hoping someone can help me out with this: If I have dataframe of years and ages and the first column and first row are filled with leading values: Df<- age1 age2 age3 Yr1 1 0.4 0.16 Yr2 1.5 0 0 Yr3 0.9 0 0 Yr4 1 0 0 Yr5 1.2 0 0 Yr6 1.4 0 0 Yr7 0.8 0 0 Yr8 0.6 0 0 Yr9 1.1 0 0 Now the rest of the cells need to be filled according to the previous year and age

Hi, I am using levelplot, and would like remove from each panel (condition) its unused x levels. e.g. Remove from panel vs=1 the cyl level=8. data(mtcars) levelplot(mpg~factor(cyl)*factor(gear)|factor(vs)) Thanks for your help, Ronny -- View this message in context: http://r.789695.n4.nabble.com/Lattice-levelplot-remove-unused-levels-per-panel-tp4656087.html Sent from the R help mailing

The mt address is about to be used more, make sure it's set appropriately. Reported-by: Emil Velikov <emil.l.velikov at gmail.com> Tested-by: Emil Velikov <emil.l.velikov at gmail.com> Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> Cc: "10.2 10.3" <mesa-stable at lists.freedesktop.org> --- src/gallium/drivers/nouveau/nv50/nv84_video.c | 2 ++ 1 file

Good morning RGuru's I have a data frame of 575 columns.? I want to extract only those columns that are numeric(double) or integer to do some machine learning with.? I have searched the web for a couple of days (off and on) and have not found anything that shows how to do this.?? Lots of ways to extract rows, but not columns.? I have attempted to use "(x == y)" indices extraction

Hi, I tried this: # extract date from the time stamp dt1 <- cbind(as.Date(dt$EndDate, format="%m/%d/%Y"), dt$EnergykWh) head(dt1) colnames(dt1) <- c("date", "EnergykWh") and my dt1 becomes these, the dates are replace by numbers. dt1 <- cbind(as.Date(dt$EndDate, format="%m/%d/%Y"), dt$EnergykWh) dput(head(dt1)) colnames(dt1) <-

i have converted my data into date format like below: > day=as.Date(originaldate,"%m/%d/%Y") > day[1:5] [1] "2008-04-12" "2011-07-02" "2011-09-02" "2008-04-12" "2008-04-12" I wish to select only those observations from 2007 to 2009, how can I select from this list? [[alternative HTML version deleted]]

You rang sir? library(tidyverse) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat1 <- data.frame(xx , yr1, yr2, y3) dat1 %>% select(!starts_with("yr")) or for something a bit more exotic as I have been trying to learn a bit about the "data.table package library(data.table) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat2 <- data.table(xx , yr1, yr2, yr3) dat2[, !names(dat2)

Can lme or lmer fit a plain regular fixed effects anova? Ie a model without a random effect, or have there be at least one random effect in order for these functions to work? Trying to run such, (1) without specifying a random effect produces an error, (2) specifying that there is no random effect does not produce the same output as an anova run in lm(); (2b) specifying that there is no

> dt1[ vapply(dt1, FUN=is.numeric, FUN.VALUE=NA) ] a c 1 1 1.1 2 2 1.0 ... 10 10 0.2 Bill Dunlap TIBCO Software wdunlap tibco.com On Fri, Apr 29, 2016 at 9:19 AM, Carl Sutton via R-help < r-help at r-project.org> wrote: > Good morning RGuru's > I have a data frame of 575 columns. I want to extract only those columns > that are numeric(double) or integer to do

Dear all, I have this set of data. I would like to sum the EnergykWh according date sequences. > head(dt1,20) StationName date time EnergykWh 1 PALO ALTO CA / CAMBRIDGE #1 1/14/2016 12:09 4.680496 2 PALO ALTO CA / CAMBRIDGE #1 1/14/2016 19:50 6.272414 3 PALO ALTO CA / CAMBRIDGE #1 1/14/2016 20:22 1.032782 4 PALO ALTO CA / CAMBRIDGE #1 1/15/2016 8:25 11.004884 5

Is this what you are after? library(tidyverse) library(lubridate) input <- structure(list(StationName = c("PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE #1", "PALO ALTO CA / CAMBRIDGE

There seems to be a bug in seq.Date such that it will not allow the user to pass in length.out =3D NULL, despite the fact that this is the = default argument. For example: > dt1 <- as.Date("2004-12-31") > dt2 <- as.Date("2005-12-31") > seq.Date(dt1, dt2, length.out =3D NULL, by =3D "month") Error in seq.Date(dt1, dt2, length.out =3D NULL, by =3D

I'm trying to simulate trend data over a five year period. I want different trend profiles...the simplest being a linear trend. I've been using the following code: patBdta1 <- NULL for(i in 1:100) patBdta1 <- rbind(patBdta1,c(yr1= mean(rbinom(50,1,.50)), yr2 =mean(rbinom(50,1,.51)), yr3 =mean(rbinom(50,1,.52)),

Dear Friends, I have been trying to save multiple 3x3 (mfrow=c(3,3) graphics inside a loop using tiff figure format (not using PDF or savePlot functions) with no success. Could you please help? Here is a simplified example code: dat=data.frame (ID=rep(1:10,each=10),IDV=rep(seq(1:10),times=10)) dat$DV <- with(dat, 50+15*IDV) dat=dat[order(dat$ID,dat$IDV),] for(i in 1:10){ dt1 =

Hello I have a string which contains microseconds, can anyone help on constructing this in to a time object, with the microseconds, that I can take to a ZOO file? Thanks Sean > UK[1,3] [1] "17:09:53.824" > UK[1,1] [1] "2007-12-11 00:00:00" > mydates <- paste( substr(UK[,1], 1, 10), UK[,3]) > mydates[1] [1] "2007-12-11 17:09:53.824" >

R-Devel, I store and retrieve a large amount of financial data (millions of rows) in a PostgreSQL database keyed by date (and represented in R by class Date). Unfortunately, I frequently find that a great deal of processing time is spent converting dates from character representations to Date class representations in R, presumably because strptime is not fast for large vectors (>10,000