similar to: r data structures

Hi All, I noticed the following: pip = array(1:6, dim = c(3,2)) dim(pip) [1] 3 2 pup = pip[1,] dim(pup) NULL I bet there is a *good* reason why one row of an array is *dimensionless*, but it's highly inconvenient for my purpose, i.e. to use apply() after an array goes through a number of logical steps and is redimensioned, sometimes to one single row. How do I keep dim(pup) to 1 2?

> 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) >

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 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

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 <-

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

hi, i've setup a LDAP server with account information, and compiled samba with ldap support. everything works great, except for the password changes i still have to run two seprate commands ( passwd, smbpasswd ) to change a users password. i've tried to put the pam_smbpasswd.so module into system-auth, but that does work? any pointers? thanks adriaan putter

?? > 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

On 04/28/2015 05:13 PM, Laine Stump wrote: > Bah. Your debug symbols aren't enough to give a full stack trace - what > I was really looking for was those things marked as ??. > > However, your debug output shows several things leading up tothe > qemuProcessStop that might have been the reason for it being called: > > 1) virStorageFileBackendFileReadHeader() complains

Full_Name: Haobo Ren Version: 2.1.1 OS: Windows 2000 Submission from: (NULL) (192.11.226.116) When running constrOptim, there is error message Error: subscript out of bounds

On 04/22/2015 08:19 AM, Laine Stump wrote: > If you're compiling yourself, then you should be all set to run under > gdb. libvirt-debuginfo is just a separate subpackage that contains all > the symbol and line number info from the build so that backtraces in > gdb make sense. Try attaching gdb to the libvirtd process and do > something like "thread apply all bt" - if

Dear subscribers, I have made a simulation using loops rather than apply, simply because the loop function seems more natural to me. However, the current simulation takes forever and I have decided - finally - to learn how to use apply, but - as many other people before me - I am having a hard time changing habits. My current problem is: My current code for the loop is: distances <-

Can R do a repeated measures MANOVA and tell what dimensionality the statistical variance occupies? I have been using MATLAB and SPSS to do my statistics. MATLAB can do ANOVAs and MANOVAs. When it performs a MANOVA, it returns a parameter d that estimates the dimensionality in which the means lie. It also returns a vector of p-values, where each p_n tests the null hypothesis that the mean

Hello, I am running a few simulations for clinical trial anlysis. I want some help regarding the following. We know trhat as the sample size increases, the variance should decrease, but I am getting some unexpected results. SO I ran a code (shown below) to check the validity of this. large<-array(1,c(1000,1000)) small<-array(1,c(100,1000)) for(i in

----- Forwarded Message ----- From: vramaiah at neo.tamu.edu To: "bernhard pfaff" <bernhard.pfaff at pfaffikus.de> Sent: Friday, November 11, 2011 9:03:11 AM GMT -06:00 US/Canada Central Subject: Use of R for VECM Hello Fellow R'ers I am a new user of R and I am applying it for solving Bi-Variate (Consumption and Output) VECM with Co-Integration (I(1)) with three lags on

Martin Devera wrote: > If you read the manual, the algorithm will not work correctly > with {,c}burst < MTU ... > devik > I just tried to change {,c}burst to 1600, or leaving them by default but no visible result. here is the latest tc -s -d class show dev eth0 class htb 1:101 parent 1:1 prio 0 rate 40Kbit ceil 40Kbit burst 1599b/8 mpu 0b cburst 1599b/8 mpu 0b quantum 512 level

Dear All, I am using the pmst function from the sn package (version 0.4-0). After inserting the example from the help page, I get non-trivial answers, so everything is fine. However, when I try to extend it to higher dimension: xi <- alpha <- x <- rep(0,27) Omega <- diag(0,27) p1 <- pmst(x, xi, Omega, alpha, df = 5) I get the following result: >p1 [1] 0 attr(,"error")

I installed package plotrix because reading its vignette it looks like it can help me solve a "legend" problem. The package instaleed correctly on my Mac OS/X 10.5.8 But I cannot reproduce the examples centered on function "lgendg". > library(plotrix) > plot(0.5,0.5,xlim=c(0,1),ylim=c(0,1),type="n", + main="Test of grouped legend function") >

Dear All, Probably a very basic question, but can't seem to work my way around it. I want to which row has the maximum value. But what if the row names do not correspond with the row numbers. In the example below, you'll see that the max of example is row 4, but the name of row 4 is "9". How do I get R to return "9" as value, instead of 4. example <-