similar to: Extract/format/show for S4 objects

Is there any way to get the current memory used by R without running gc()? I'd like to include the memory usage in logging output, but calling gc() to get that information takes long enough to be noticeable (~ 6 s with ~ 20 GB of ram in use), which goes against the point of logging. Thanks, Johann

What is the current best package for manipulating HDF5 data files? I tried "hdf5" a long time ago, but I ran into memory problems. "h5r" is on CRAN now, and "rhdf5" is part of bioconductor. Ideally, I'd like to read simple vectors or tables, either the entire thing or a subset of rows. I don't need much writing support, but it would be nice. Compression is a

Using Achim's d this also works to generate z where FUN is a function used to transform the index column and format is also passed to FUN. z <- read.zoo(d, index = "time", FUN = as.yearqtr, format = "Q%q %Y") On Sun, Jan 28, 2018 at 4:53 PM, Achim Zeileis <Achim.Zeileis at uibk.ac.at> wrote: > On Sun, 28 Jan 2018, phil at philipsmith.ca wrote: > >> I

On Sun, 28 Jan 2018, phil at philipsmith.ca wrote: > I have a data set with quarterly time series for several variables. The > time index is recorded in column 1 of the dataframe as a character > vector "Q1 1961", "Q2 1961","Q3 1961", "Q4 1961", "Q1 1962", etc. I want > to produce line plots with ggplot2, but it seems I need to

Suppose I have a mixed-effects model where yij is the jth sample for the ith subject: yij= beta0 + beta1(age) + beta2(age^2) + beta3(age^3) + beta4(IQ) + beta5(IQ^2) + beta6(age*IQ) + beta7(age^2*IQ) + beta8(age^3 *IQ) +random intercepti + eij In R how can I get an F test against the null hypothesis of beta6=beta7=beta8=0? In SAS I can run something like contrast age*IQ 1,

I configured my * with the instructions found here http://www.voip-info.org/wiki/view/Asterisk+Google+Talk to work with gtalk. The Phone rings and connects - but no audio! I am using a self-compiled asterisk 1.4.2 There is a lot of output on the CLI but I can't make sense of it. Perhaps somebody can help? Michael Output from the CLI: JABBER: gtalk_account OUTGOING: <iq

I have financial data on a a set of firms, with a quarterly period (fundamental data). The data spans 10 years, and four quarters per year. The present file (.csv) reads the Date columns as "200706" for the second quarter of 2007; "199809" for the third quarter of 1997. Is there a way I can convert it to something like "2007 Q2", "1998 Q3"? I am aware of

Hello everyone, Thanks to all who offered suggestions after my initial query to the list. I've been busy trying to absorb the docs as fast as possible, but this little thing has me stumped for the moment. I'm creating a random sample of 100 item and saving it to a vector named 'iq'. I'd like to create a second vector (iq_prime, say) by adding 20 to each element of the

Hello everyone, Thanks to all who offered suggestions after my initial query to the list. I've been busy trying to absorb the docs as fast as possible, but this little thing has me stumped for the moment. I'm creating a random sample of 100 item and saving it to a vector named 'iq'. I'd like to create a second vector (iq_prime, say) by adding 20 to each element of the

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hello, I am not able to query Sybase IQ database from R using RODBC package. Check the below code R> sybaseiq.query <- function (sql, ..., as.is = FALSE) { connstr <- "Driver={Sybase IQ};UID=ajadhav2;PWD=*****;ServerName=PSGSESHR01A_IQ;CommLinks=SharedMemory,TCPIP{host=psgseshr01_iq.sg.csfb.com;port=3051}" chan <- odbcDriverConnect(connstr)

Hi, I'd like to make a time series at an annual frequency. > a<-xts(x=c(2,4,5), order.by=c("1991","1992","1993")) Error in xts(x = c(2, 4, 5), order.by = c("1991", "1992", "1993")) : order.by requires an appropriate time-based object > a<-xts(x=c(2,4,5), order.by=1991:1993) Error in xts(x = c(2, 4, 5), order.by =

Dear list, i am a student of psychology and have to do a multilevelanalysis on some data. About that i have one general and one specific question. This is what i have copied from the help-file on lme: data(bdf) fm <- lme(langPOST ~ IQ.ver.cen + avg.IQ.ver.cen, data = bdf, random = ~ IQ.ver.cen | schoolNR) summary(fm) after summary(fm) i get the following error:

Another known issue, fixed with https://github.com/dovecot/core/commit/ca4c2579f0456072bdb505932a9cf7602e42afd2.patch Aki > On 24 January 2019 at 20:54 Maciej Milaszewski IQ PL <maciej.milaszewski at iq.pl> wrote: > > > Hi > Like bumerang. > > 2.2.36 works fine on debian10 but connect by telnet like: > > Connected to 46.xxx.xxx.xxx. > Escape character is

Hi Thenx. Dump core: Core was generated by `dovecot/auth'. Program terminated with signal SIGSEGV, Segmentation fault. #0? 0x00007fa910394c7a in ?? () from /lib/x86_64-linux-gnu/libc.so.6 (gdb) bt full #0? 0x00007fa910394c7a in ?? () from /lib/x86_64-linux-gnu/libc.so.6 No symbol table info available. #1? 0x00005604e6f98bc8 in password_scheme_register_crypt () at password-scheme-crypt.c:144

Hi all, I am trying to figure out the formula used by R's Spearman rho (using cor(method="spearman")) because I can't seem to get the same value as by calculating "by hand". Perhaps I'm using "cor" wrong, but I don't know where. Basically, I am running these commands: > y=read.table(file="tmp",header=TRUE,sep="\t") >

Dear R users, I'm estimating a two-level regresion model but having difficuly in finding a good R syntax example. Let me use an example to explain what I'm doing. The dependent variable is math score (mathscore). Predictors are at two levels. At the student level, they are household income (hinc) and IQ (IQ). At the school level, there is a dummy variable indicating if the school is a

Hi all - (this is posted to r-help and R-SIG-MAC) OSX 10.3.7, R 2.0.1, lme4/Matrix/latticeExtra latest, fresh install of R. MASS loaded (or not). I am getting an error message for the print() and summary() commands with all lme models I try and run in lme4 (GLMM's work fine). Using the example from the lme help, summary and print produce the following errors, despite the model being

Dear Madam or Sir I am writing you hoping, that you can help me with a problem concerning the output of regressions done with the function lme in R. I would need the standard deviations for intercepts and predictors, but in the output I can only find those for the intercepts. Could it be, that this is my fault? (I am just a beginner with R and multilevel modeling). I am sorry to annoy you with

Hi all Assume I have a data set xx; Group: 1=group1 ?, 2=group2 IQ: ?1= High, 0 =low fit <- glm(IQ ~group, data = xx, family = binomial()) summary(fit) Results ?????? ????????????Estimate Std. Error z value Pr(>|z|) (Intercept) -2.55456??? 0.210 -12.273? < 5e-16 *** group????????? 0.36180 ?????0.076?? 3.952 ????5.24e-05 *** the odd ratio = exp(0.36180 )= 1.435912 My question