similar to: Comparison of numeric and character vectors

Some help with how to re-label the vertical axis in a histogram would be appreciated. qplot(off.sc,weight=rel.freq,binwidth=.29,main="test Figure"+ylab("New from inside"))+ylab("New from outside")+ xlab("off.sc\nAggregated frequency plots for 17 equal intervals.") The code

Dear all, I have 100 files which are used as input.and I have to input the name of my files again and again.the name of the files are 1.out, 2.out......100.out. I want to know if there is anything like perl so that i can use something like this- for($f = 1; $f <= 100; $f++) { $file = $f.".out"; I have tried this thing in R but it does not work.Can somebody please help me.

Dear John Fox, I am using Snowleopard with a Intel Core 2 Duo processor. I have installed the R console and followed your steps at: http://socserv.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html but every time I try "library(Rcmdr)" it loads Tcl/tk infinitely. I have tried loading "library(tcltk)" by itself and it does the same thing. I have tried with X11 running to

ummm, Ista, it's 2^n. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Wed, Aug 23, 2017 at 8:52 AM, Ista Zahn <istazahn at gmail.com> wrote: > On Wed, Aug 23, 2017 at 11:33 AM, Christofer Bogaso >

Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between

On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > ummm, Ista, it's 2^n. ummm yes ughhhh. My point is, if the number of groups is large, check it before hand. If you can check it without embarrassing yourself in public like I did that's even better. Best, Ista > > Cheers, > Bert > > > Bert Gunter > > "The trouble

Hi guys, Are there any feasible methods in searching & finding non-ASCII characters in R? For example, from the following object, x <- mia. SzaÌmitaÌó The desired output is, x.out <- mia. SzaImitaIA Your help in resolving this would be greatly appreciated. [[alternative HTML version deleted]]

> On 23 Aug 2017, at 20:51 , Ista Zahn <istazahn at gmail.com> wrote: > > On Wed, Aug 23, 2017 at 12:35 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: >> ummm, Ista, it's 2^n. > > ummm yes ughhhh. > You didn't really say otherwise: sum(choose(n,0:n)) == 2^n by the binomial expansion of (1+1)^n (but you knew that) This points to a different

Hi! We have run a linear regression model with 3 explanatory variables and get the output below. Does anyone know what type of test the anova model below does and why we get so different result in terms of significant variables by the two tables? Thanks! /Sara > summary(model) Call: lm(formula = log(HOBU) ~ Vole1 + Volelag + Year) Residuals: Min 1Q Median 3Q Max

I know I cannot be the only one who is not able to install the XML package from CRAN (zip or tar file) Many packages depend on this XML package. Can someone help me either access the source for a good binary? I have received no assistance from the author/developer of the package. Steven Steven D. Randolph Associate Consultant Enterprise Capabilities - Business Intelligence & Analytics Eli

Untested, but I expect that setting the levels to be the same across the two factors levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6] and proceeding as before should be fine. Best, Ista On Jul 6, 2017 6:54 PM, "Gang Chen" <gangchen6 at gmail.com> wrote: Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp

On Tue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > Ista, et. al: efficiency? > (Note: I needed to correct my previous post: do.call() is required for > pmax() over the data frame) > > > x <- data.frame(matrix(runif(12e6), ncol=12)) > > > system.time(r1 <- do.call(pmax,x)) > user system elapsed > 0.049 0.000

On Thu, Dec 14, 2017 at 1:58 PM, Berend Hasselman <bhh at xs4all.nl> wrote: > >> On 14 Dec 2017, at 19:36, lily li <chocold12 at gmail.com> wrote: >> >> Hi R users, >> >> I have a question about reading from text files. The file has the structure >> below: >> >> Time Column1 Column2 >>

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

> On Dec 6, 2017, at 4:27 PM, Ashta <sewashm at gmail.com> wrote: > > Thank you Ista! Worked fine. Here's another (possibly more direct in its logic?): DM[ !ave(DM$x, DM$GR, FUN= function(x) {!length(unique(x))==1}), ] GR x y 5 B 25 321 6 B 25 512 7 B 25 123 8 B 25 451 -- David > On Wed, Dec 6, 2017 at 5:59 PM, Ista Zahn <istazahn at gmail.com> wrote:

Hi, I am using colsplit (package = reshape) to split all strings in a column according to the same patterns. Here an example: library(reshape2) df1 <- data.frame(x=c("str1_name2", "str3_name5")) df2 <- data.frame(df1, colsplit(df1$x, pattern = "_", names=c("str","name"))) This is nearly what I want but I want to remove the words

Hi Fellow R Users, I recently upgraded to Mac OS X 10.5 (Snow Leopard) and had some issues downloading and running R 2.10.1. I fixed the tcl/tk problem I was originally having but it was replaced with another. I run R out of the shell (terminal) and when I ask it to download.packages() it gives me the following message: *Loading Tcl/Tk interface ... done* *Error in file.info(x) : argument

Hi, I created a simple data frame with one factor and one numerical variable. The factor was actually a vector of names of techniques to trimm reaction time data. I want to create a plot that shows the value of F test for every trimming method. So the data frame has its trim factor (who has those labels

Dear professor Fox and R helpers, I have a quick question about the Anova function in the car package. When using the default "type II" SS I get results that I don't understand (see below). library(car) Data <- data.frame(y=rnorm(10), x1=factor(c(rep("a",4), rep("b",6))), x2 = factor(c(rep("j", 2), rep("k", 3), rep("j", 2),

This is what I was looking for. Thank you everyone! Sincerely, Milu <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> Mail priva di virus. www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail>