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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.

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....

...x city wave 1 New York 1 1 Los Angeles 1 1 New York 2 1 Los Angeles 2 I've started writting a simple function to do this, but it seems like a common situation and I'm wondering if there are any packages or functions that might make this easier. Thanks! Ista -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org

i am trying to fit a linear model with both continuous covariates and factors. When fitted with the intercept term the first level of the factor is treated by R as intercept and the estimate of the effects of remaining levels(say i th level) are given as true estimate of i th level - estimate of 1st level.can any please help me? thanks in advance..... -- View this message in context:

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

Hi everyone, Is there a way to make R save the workspace output (just the results, not the objects themselves) as you go? I'm running analysis that takes a long time to run and I want to be able to interrupt it without losing all the output to date. Is there an alternative to putting "save.image()" commands after every couple lines of code? Best, Tatyana

I am using ggplot2 to make a boxplot that overlays a scatterplot: pp = qplot(time, error, data=times, size=I(1), geom="jitter", main=title, ylab="Error (min)", xlab="Time before ON (min)", alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) print(pp2 + geom_boxplot(alpha=.5,

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 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 with...

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...

> 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...

...on't feel that I understand the test. Questions: 1) What are "max", "quad" and "scalar"? (book/article references would be appreciated) 2) Can I use arbitrary test statistics with coin? For example, can I test the independence of the variances using coin? Thanks, Ista -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org

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 <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2...

...t;a" > Inf [1] FALSE This came as a surprise to me (although it is quite possibly a trivial issue), and I'd appreciate any information about why R considers character vectors to even be comparable to numeric vectors, and why it considers characters to have very large values. Thanks! Ista

Hello I simply want to plot two variables against one 'year' variable in qplot. Is any way of doing this without reshaping data in long format and using facet function afterwards? Thank you Denis

> 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: &g...

Hi David, Ista and all, I have one related question Within one group I want to keep records conditionally. example within group A I want keep rows that have " x" values ranged between 15 and 30. group B I want keep rows that have " x" values ranged between 40 and 50. group C I want k...

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 Lilly & Company 317-276-2593-office 317-506-0484-mobile [[alternative HTML version deleted]]

Hello, Here are three vectors to give context to my question below: *id <- c(1,1,1,1,1,2,2,2,3,3,3)) month <- c(1, 1, 2, 3, 6, 2, 3, 6, 1, 3, 5) value <- c(10, 12, 11, 14, 16, 12, 10, 8, 14, 11, 15)* and I want to plot "value" over "month" separately for each "id". Before I can do that, I need to section both month and value, based on ID. I create a

Dear R Users, I would like to know is there any package to create a plot like this? http://dl.dropbox.com/u/5409929/cs1160521f01.gif X axis is categorical. And the positions of the points are corresponding to the frequency. (similar to violinplot) Thank you. Regards, CH -- CH Chan