similar to: Missing users (who otherwise work fine)

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, >>> junk1 <- junk[, rev(seq_len(10), ] so that junk[1,1,1 ] = junk1[1,10,1] junk[1,2,1] = junk1[1,9,1] etc. The genesis of this is the program is downloading data from a variety of sources on (time, altitude, lat, lon) coordinates, but all

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

My error. Clearly I did not do enough testing. z <- array(1:24,dim=2:4) > all.equal(f(z,1),f2(z,1)) [1] TRUE > all.equal(f(z,2),f2(z,2)) [1] TRUE > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" # Your earlier example > z <- array(1:120, dim=2:5) >

>>>>> David L Carlson <dcarlson at tamu.edu> >>>>> on Wed, 2 May 2018 21:43:52 +0000 writes: > Typo: dat[[z]] should be x[[z]]: > > x2 <- do.call(rbind, lapply(names(x), function(z) > data.frame(type=z, x[[z]]))) > x2 > type x y > 1 A 1 3 > 2 A 2 4 > 3 B 5 7 > 4 B 6 8 > >

?? > z <- array(1:24,dim=2:4) > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" In fact, > dim(f(z,3)) [1] 2 3 4 > dim(f2(z,3)) [1] 3 4 2 Have I made some sort of stupid error here? Or have I misunderstood what was wanted? Cheers, Bert Bert Gunter

Hi,all: I want to perform multiple correspondance analysis via MCA{FactoMineR}. The data is in the attachment. My code: dat<-read.delim("e:\\mydata.txt",header=T) MCA(dat,quanti.sup=7,quali.sup=1:6) Error in `[.data.frame`(tab, , i) : undefined columns selected My question: Why does the error happen? Many thanks. Best. -------------- next part -------------- An embedded and

> On 24 Oct 2017, at 22:45 , David L Carlson <dcarlson at tamu.edu> wrote: > > You left out all the most important bits of information. What is yo? Are you trying to assign a data frame to a single column in another data frame? Printing head(samples) tells us nothing about what data types you have, especially if the things that look like text are really factors that were created

Thank you so much David, here is correction: d1=suppressWarnings(read.csv("/Users/anamaria/Downloads/B1.csv", stringsAsFactors=FALSE, header=TRUE)) d1$X <- NULL d2=as.matrix(sapply(d1, as.numeric)) pdf("~/graph.pdf") b<-barplot(d2, legend= c("SYCL", "CUDA"), beside= TRUE,las=2,cex.axis=0.7,cex.names=0.7,ylim=c(0,80), col=c("#9e9ac8",

No html!, Copy the list using Reply-All. The data frame group_PrivateLabel does not contain variables called Product_Name or Region. David C From: nguy2952 University of Minnesota <nguy2952 at umn.edu> Sent: Friday, June 1, 2018 2:13 PM To: David L Carlson <dcarlson at tamu.edu> Subject: Re: [R] Regroup and create new dataframe Hi David, your example is perfect! I am still

I think you mean between column 1 and 2 of A? Why is 36003918 not included? It is clearly between 35838396 and 36151202 in the first row of A. My earlier solution should work fine. Just create a new matrix AX that has the columns switched so that the start is always column 1 and use that to identify the ones you want to select. That way you are not modifying B. This will be faster than checking

Hi David,Thank you for the example.When I try to use the cat function, I get an error cat(attr<-getAttractors(net, type="asynchronous"))Error in cat(attr <- getAttractors(net, type = "asynchronous")) : argument 1 (type 'pairlist') cannot be handled by 'cat' Please let me know, if I have used the function in right way?. Thank you Priya ?

Hi Peter, Thanks for contributing such a great answer. Can you please provide a pointer to the documentation where it explains why dd$B <- s and dd["B"] <- s have such different behavior? (I am perfectly happy if you write the explanation but if it saves you time to point to some reference that works fine for me.) Regards, Eric On Wed, Oct 25, 2017 at 2:27 PM, Peter Dalgaard

We often refer requesters to the Posting Guide and chide them for not reading it. Recently I had occasion to re-read the Posting Guide which is for all R lists not just R-help. The word "reproducible" does not appear anywhere in the guide. The closest it comes is the following suggestion: "Sometimes it helps to provide a small example that someone can actually run."

Typo: dat[[z]] should be x[[z]]: x2 <- do.call(rbind, lapply(names(x), function(z) data.frame(type=z, x[[z]]))) x2 type x y 1 A 1 3 2 A 2 4 3 B 5 7 4 B 6 8 David C -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of David L Carlson Sent: Wednesday, May 2, 2018 3:51 PM To: William Dunlap <wdunlap at tibco.com>; Kevin E.

Here is an alternative approach using apply(). Note that with apply() you are reversing rows or columns not indices of rows or columns so apply(junk, 2, rev) reverses the values in each column not the column indices. We actually need to use rev() on everything but the index we are interested in reversing: f2 <- function(a, wh) { dims <- seq_len(length(dim(a))) dims <-

I'm not sure how you are incorporating time period into your data structure. Typically we are looking at plots or assemblages as the rows and taxa as the columns. Time period adds a third dimension that could be added as blocks of rows. For example, depending on the resolution of your data, one approach would be to have up to 8 rows for each locality: Loc1.1000, Loc1.2000, . . . Loc1.8000. If

You left out all the most important bits of information. What is yo? Are you trying to assign a data frame to a single column in another data frame? Printing head(samples) tells us nothing about what data types you have, especially if the things that look like text are really factors that were created when you used one of the read.*() functions. Use str(samples) to see what you are dealing with.

Responses should be copied to r-help using ReplyAll. You are still sending html formatted emails. If you are using Microsoft Outlook, click the Format Text tab and select ?Aa Plain Text?. No one has asked you to reveal the data set, only to create one with a similar structure. Is the data I sent reasonably close? What should it look like after it is transformed? David C From: nguy2952