similar to: cbind and automatic type conversion

When I add the --with-pam option in my configure command I get the following error: checking for pam_start in -lpam... no configure: error: Can't build with PAM support: libpam not found PAM is most certainly installed on the machine: ii libpam-modules 0.79-5 Pluggable Authentication Modules for PAM ii libpam-runtime 0.79-5 Runtime support

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

On Sat, 2004-09-11 at 12:49, Jeff Cohen wrote: > For the heck of it I tried upgrading to gcc 3.4.2 (from 3.3.3). It > didn't make a difference. So here are the failures for llvm-test. All > diffs are against the "native" output. > > ===================== MultiSource/Applications/sgefa > > cbe failed differently from jit/llc. First cbe: > > 84c84

On Sat, 11 Sep 2004 13:53:11 -0700 Reid Spencer <reid at x10sys.com> wrote: > On Sat, 2004-09-11 at 12:49, Jeff Cohen wrote: > > > > ===================== MultiSource/Applications/sgefa > > > sgefa is a known XFAIL. See the nightly test results over the last > several months. Actually, you should compare your test results with the > 1.3 release test results

I have 2 zoo objects - 1) Interest rate spread between 10-YR-US-Treasury and 2-YR-US-Treasury (object name = sprd) 2) S&P 500 index (object name = spy) > str(spy) ?zoo? series from 1976-06-01 to 2011-03-31 Data: num [1:8791] 99.8 100.2 100.1 99.2 98.6 ... Index: Class 'Date' num [1:8791] 2343 2344 2345 2346 2349 ... > str(sprd) ?zoo? series from 1976-06-01 to 2011-03-31

For the heck of it I tried upgrading to gcc 3.4.2 (from 3.3.3). It didn't make a difference. So here are the failures for llvm-test. All diffs are against the "native" output. ===================== MultiSource/Applications/sgefa cbe failed differently from jit/llc. First cbe: 84c84 < One-Norm(A) ---------- 8.879153e+02. --- > One-Norm(A) ---------- 8.879156e+02.

I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:1000000,1] y = table2[1:1000000,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- -------------

I've spent some time looking at klibc today, and find myself in a state of some puzzlement. The 0.79 snapshot does not build against 2.5.68; it fails almost immediately, apparently due to problems with include files. I'm hesitant to try patching anything to get it building, because I can find no sign of a CVS or BK repo that I can check against, and I don't want to replicate

Dear all Is there an easy way to display only one half (top-right or bottom-left) of a correlation matrix? > require(Hmisc) > rcorr(as.matrix(mtcars[ , 1:4])) mpg cyl disp hp mpg 1.00 -0.85 -0.85 -0.78 cyl -0.85 1.00 0.90 0.83 disp -0.85 0.90 1.00 0.79 hp -0.78 0.83 0.79 1.00 n= 32 P mpg cyl disp hp mpg 0 0 0 cyl 0 0 0 disp 0 0

Dear list, First my information: platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 3.1 year 2006 month 06 day 01 svn rev 38247 language R version.string Version 2.3.1 (2006-06-01) Now my question: How is it possible that a command in an R script is not

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

Hi, the command line with 'text' should be: text(-8,-8, expression(R^2 * " = 0.62, r = 0.79, N = 161"), cex = 2 ) Best, Kimmo su, 2024-02-04 kello 17:16 +0100, Jibrin Alhassan kirjoitti: > Here is the script I used to plot the graph indicating the text I > wanted to > insert. The line in the script that I have issues with is: text(-8,- > 8, > "R^2=?

Here is the script I used to plot the graph indicating the text I wanted to insert. The line in the script that I have issues with is: text(-8,-8, "R^2= 0.62", r = 0.79, N = 161", cex = 2 R^2= 0.62 is not producing R squared = 0.62. Thanks. Sys.setenv( TZ="GMT" ) dt <- read.table("CLMXAPTY_sim", col.names = c("FDcli", "FDapt"))

Dear R Users, Could you please help me on the following issue? I have a real large yearly data set. For each year I have 365 flow values. Some of the flow values are not known and that’s why you will see NA written in those slots. I wanted to know, is there a way that I can estimate those values? I tried approx command but it seems least helpful for the kind of issue I am up against.

Hi guys, one of the questions where you need a real human instead of a search engine, so it would be great if you could help. I have a matrix of z-scores which I would like to filter, sometimes columnwise, sometimes rowwise. Data looks like this: Allstar hsa.let.7a hsa.let.7a.1 hsa.let.7a.2 2 0.87 0.79 -0.57 1.07 3 0.67 -1.14 -0.78 -0.95 4

Dear r-helpers, why do I get an output in the first iteration of the for-loop which contains the string values of the input vector, and how can I avoid that? Here's the output (only line 1 is wrong) latentVariable Indiv Group 1 rPlanning rIterat rTDD 2 rPlanning 0.79 0.84 3 rIterat 0.79 0.83 4 rTDD 0.9 0.96 5 rStandup 0.83 0.82 6

Hi, how to write out data frame in format? for examples, I wrote out a data using 'write.table(siredata, file='siredata.txt', row.names=F, quote=F)'. it produced data alike this: 2882 1 0 0 0 0 L1600481991910012 L1600011988880196 2883 0.79 0.21 0 0 21 L1622881993930001 L1600481991910012 2884 0.84 0.16 0 0 23 L1622881993930005 L1600481991910012 2885 0.9 0.1 0 0 23

Hi list, I'd like to make a factor with seven 1s and three 2s using the factor() function. That is, 1 1 1 1 1 1 1 2 2 2 I will then bind this factor to the matrix below using cbind.data.frame(). 0.56 0.48 0.22 0.59 0.32 0.64 0.26 0.60 0.25 0.38 0.24 0.45 0.56 0.67 0.78 0.97 0.87 0.79 0.82 0.85 I am new to R and have been using various manuals and have made many attempts without

If we are on the subject of byte compilation, let me bring a couple of examples which have been puzzling me for some time. I'd like to know a) if the compilation will likely to improve the performance for this type of computations, and b) at least roughly understand the reasons for the observed numbers, specifically why x[i]<- assignment is so much slower than x[i] extraction. The loops

Dear all ?Hmisc::rcorr states that it takes as main argument "a numeric matrix". But is it normal that it fails in such an ugly way on a data frame? (See below.) If the function didn't attempt any conversion to a matrix, I would have expected it to state that in the error message that it didn't accept 'data.frame' objects in its input. Also, I vaguely remember having used