similar to: Fancy MySQL footwork

Displaying 20 results from an estimated 2000 matches similar to: "Fancy MySQL footwork"

2008 Jan 06
0
Some repo fancy footwork?
I am looking at K12LTSP-EL5 which is Centos 5 based. Supposedly their disc2 maps to Centos 1of6 and so forth. I have a repo of the Centos 5.1 isos: /centos/5/os/i386 The question is can I put the K12LTSP disc 1 in a directory: /centos/k12ltsp-el5/i386 then with some logical link magic make this directory appear to have all 7 CDs worth of content? Or put the k12 disc 1 somewhere else
2009 Jul 30
0
randomized block design analysis PROBLEM
Dear All user, Hello, I'm a student and I have some trouble with the experimental (columns-experiments) design of my project. I use a randomized block design with 4 treatments including a control. For each treatment, I use 3 replicates and 3 blocks. The treatments are: -T1 = COD (300 mg/Lit) COD=chemical oxygen demand -T2 = COD (200 mg/Lit) -T3 = COD (100 mg/Lit) -T4 = COD (0 mg/Lit) as
2012 Jul 06
2
[LLVMdev] Excessive register spilling in large automatically generated functions, such as is found in FFTW
Hi, I've noticed that LLVM tends to generate suboptimal code and spill an excessive amount of registers in large functions, such as in those that are automatically generated by FFTW. LLVM generates good code for a function that computes an 8-point complex FFT, but from 16-point upwards, icc or gcc generates much better code. Here is an example of a sequence of instructions from a 32-point
2006 Jan 21
0
Means from balanced incomplete block design
The code below is intended to analyse a textbook example of a balanced incomplete block design: # # Data taken from pp. 219-230 in # Cox, D.R. (1958) Planning of Experiments. John Wiley and Son, Inc. New York. 308 pp. # day <- factor(rep(1:10, each = 3)) T <-
2010 Nov 23
2
[LLVMdev] Unrolling an arithmetic expression inside a loop
Hello, I've been redirected from cfe-dev, as code optimizations in clang are done in llvm layer. I'm investigating how optimized code clang generates, and have come across such an example: I have two procedures: void exec0(const int *X, const int *Y, int *res, const int N) { int t1[N],t2[N],t3[N],t4[N],t5[N],t6[N]; for(int i = 0; i < N; i++) { t1[i] = X[i]+Y[i];
2017 Jun 04
0
New var
# read.table is NOT part of the data.table package #library(data.table) DFM <- read.table( text= 'obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004 ',header = TRUE, stringsAsFactors = FALSE) # cleaner way to compute D DFM$start <- as.Date( DFM$start, format="%m/%d/%Y" ) DFM$end
2010 Jun 02
1
compute the associate vector of distances between leaves in a binary non-rooted tree
Hello. I'd like to compute the associate vector of distances between leaves in a binary non-rooted tree. The definition of a distance between two leaves in a binary non-rooted tree is the number of edges in the path joining the two leaves. I've tried the ape package but I'm unable to find this vector. For example, using rtree(5,rooted=F) I've obtained the following tree: $edge
2002 Mar 13
0
rpart error with 0-frequency factor levels (with partial fix) (PR#1378)
(I'm sending to r-bugs because rpart is one of the recommended packages and is always installed. I'm also sending it directly to Dr. Ripley, as the maintainer.) rpart working as a classifier does not work (produces no splits) when the class indicator has no instances of one of the factor levels, as long as the factor level is not the final level. I have at least a partial fix, which I
2017 Jun 04
0
New var
Since the number of choices is small (6), how about this? Starting with Jeff's initial DFM: DFM <- structure(list(obs = 1:6, start = structure(c(16467, 14710, 13152, 13787, 15126, 12696), class = "Date"), end = structure(c(17167, 14975, 13636, 13879, 15340, 12753), class = "Date"), D = c(700, 265, 484, 92, 214, 57), bin = structure(c(6L, 3L, 5L, 1L, 3L, 1L), .Label
2017 Jun 03
0
New var
Ii is difficult to provide useful help, because you have failed to read and follow the posting guide. In particular: 1. Plain text, not HTML. 2. Use dput() or provide code to create your example. Text printouts such as that which you gave require some work to wrangle into into an example that we can test. Specifically: 3. Have you gone through any R tutorials?-- it sure doesn't look like
2010 Aug 14
1
Simple problem with lm/predict
Hi all, I've got an xts time series with monthly OHLC Dow Jones industrial index data from 1980 to present, the data is in stored in x. I've done an OLS fit on the data in 1982::1994 and stored it in extrapolate1 (x[,4] contains the closing value for the index). > t3 <- seq(1980,1994,length = length(x["1980::1994",4])) > t4<-t3^2 > extrapolate1 <-
1999 Nov 23
0
rbind problem
In the new version 0.90.0, rbind won't take a vector and a matrix from me, but works OK if I coerce the vector to a matrix. The following was run on a new session (i.e., no prior work), and has been duplicated on 2 machines: an SGI running Irix 6.5, and an Intel box running Red Hat Linux 6.0. Either case works fine in version 0.65.1. > t3 <- c(.5, .5) > t4 <-
2017 Jul 07
2
Error in v64i32 type in x86 backend
also i further run the following command; llc -debug filer-knl_o3.ll and its output is attached here. by looking at the output can we say that legalization runs fine and the error is due to instruction selection/ pattern matching which is not yet implemented? so do i need to worry and try to correct it at this stage or should i move forward to implement instruction selection/ pattern matching?
2017 Jun 03
4
New var
Hi all, I have a data set with time interval and depending on the interval I want to create 5 more variables . Sample data below obs, Start, End 1,2/1/2015, 1/1/2017 2,4/11/2010, 1/1/2011 3,1/4/2006, 5/3/2007 4,10/1/2007, 1/1/2008 5,6/1/2011, 1/1/2012 6,10/15/2004,12/1/2004 First, I want get interval between the start date and end dates (End-start). obs, Start , end, datediff
2015 Jan 16
1
S3 generic method dispatch on promises
Dear R friends I wanted a function to make a simple percent table that would be easy for students to use. The goal originally was to have a simple thing people would call like this pctable(rowvar, colvar, data) and the things "rowvar" and "colvar" might be names of variables in data. I wanted to avoid the usage of "with" (as we now see in the table help). Then
2024 Jan 30
1
linear programming in R | limits to what it can do, or my mistake?
Apart from the fact that the statement "such that t1+t2+t3+t4=2970 (as it must)" is not correct, the LP can be implemented as follows: library(lpSolve) LHS <- rbind( c(0,0,0,0, 1, 0, 0,0), c(1,0,0,0,-1, 1, 0,0), c(0,1,0,0, 0,-1, 1,0), c(0,0,1,0, 0, 0,-1,1), cbind(-diag(4),diag(4)), c(0,0,0,0,0,1,0,0), c(0,0,0,0,0,0,1,0), c(0,0,0,0,0,0,0,1) ) RHS <-
1999 Nov 23
2
rbind problem (PR#338)
In the new version 0.90.0, rbind won't take a vector and a matrix from me, but works OK if I coerce the vector to a matrix. The following was run on a new session (i.e., no prior work), and has been duplicated on 2 machines: an SGI running Irix 6.5, and an Intel box running Red Hat Linux 6.0. Either case works fine in version 0.65.1. > t3 <- c(.5, .5) > t4 <-
2017 Jun 03
2
New var
Thank you all for the useful suggestion. I did some of my homework. library(data.table) DFM <- read.table(header=TRUE, text='obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004',stringsAsFactors = FALSE) DFM DFM$D =as.numeric(difftime(as.Date(DFM$end,format="%m/%d/%Y"),
2017 Nov 05
2
What pattern string corresponds to CopyToReg?
Hmm, okay. Then what's the problem being reported here? I'm not sure what I'm supposed to do with "LLVM ERROR: Cannot select: t1: i16 = Constant<127>".BTW, the function is: ; ModuleID = 'return.c' source_filename = "return.c" target datalayout = "E-m:e-p:16:16:16-i1:16:16-i8:16:16-i16:16:16-i32:16:16-i64:16:16-S16-n16" target triple =
2012 Sep 18
1
chunk row to new table/file
I have big .csv file. I would like to filter that file into a new table. For example, I have .csv file as below: f1 f2 f3 f4 f5 f6 f7 f9 f10 f11 t1 1 0 1 0 1 0 0 0 0 1 t2 1 0 0 0 0 1 1 1 1 1 t3 0 0 0 0 0 0 0 0 0 0 t4 1 0 0 0 1 0 0 0 0 0 t5 0 0 0 0 0 0 0 0 0 0 t6 0 0 0 0 0 0