search for: steps

...was allready implemented (I couldn't find any on R topic), and if there is models and manners to make it more efficiently, add errors check, options, etc. At Berkeley's http://mathsite.math.berkeley.edu/smp/smp.html MathSite there is a very straighfoward example of the algorithm and its steps. My implementation follow the principle: 1. All men (or women) seeks for their best partner. 2. If there is no tie in a column (or row), stop. 3. If there is a tie, removes the worst-partners-tied and seek again the second-best (till n-best) alternative. The function is working right up to 6x6...

This patch introduces an optimized assembler version of md5_process(), the inner loop of MD5 checksumming. It affects the performance of all MD5 operations in rsync - including block matching and whole-file checksums. Performance gain is 5-10% depending on the specific CPU. Originally created by Marc Bevand and placed in the public domain, later integrated into OpenSSL. This is the original

...o, has_many, has_and_belong_..., etc... However I''m stillnot sure if what I''m trying to do will work *well* with ActiveRecord. So enough talk, here''s what i''m trying to do. +-------+ +-------------+ +-----+ |clients|----------|clients_steps|-------------|steps| +-------+ +-------------+ +-----+ |id | |client_id | |step_id | |completed_at | +-------------+ |...

We have a networked HP OfficeJet All-In-One. I have the scanner working with the CentOS 5.3, except xsane is ignoring SANE_DEFAULT_DEVICE. scanimage does however take this environment variable just fine: server1.wendellfreelibrary.org% scanimage -L No scanners were identified. If you were expecting something different, check that the scanner is plugged in, turned on and detected by the

Hey gang, I find myself struggling with multiple step definitions in cucumber all the time. I regularly test the contents of my flash[:notice] in steps, since I think that''s part of the behaviour of the app. So, while implementing a new feature, I ran into the following error: /opt/local/lib/ruby/gems/1.8/gems/cucumber-0.1.15/bin/../lib/cucumber/ step_mother.rb:81:in `regexp_args_proc'': Multiple step definitions match &quot...

Hello all, I'm stuck with a strange issue with writing jpegs of plots to a folder in a loop. This works: for (step in 1:length(steps)) { jpeg(filename=paste("frame_",sprintf("%05d",step),".jpg",sep="")) plot(steps[[step]]) dev.off() } But if I use qplot to generate the plot (which is my aim): for (step in 1:length(steps)) { jpeg(filename=paste("frame_",sprintf("%05d...

...at ~ rs13479085 + rs13480057 + rs13482096 + rs8254221 Df Deviance AIC <none> 233.82 841.18 rs13480057:rs13475933 6 215.95 851.12 (***) Another problem is that the final model seems to be the first model that satisfies (bAIC >= AIC + 1e-07) if steps haven't used up, rather than the one before that. Please see below. > formula(step2.bic.out) pheno.dat ~ rs13479085 + rs13480057 + rs13482096 + rs8254221 + rs13482096:rs13483699 > step2.bic.out$anova Step Df Deviance Resid. Df Resid. Dev AIC 1 NA NA 298 233.8...

Dear R users, I recently stumbled upon problems of reproducibility while running GAM analyses in different R and gam package versions. In the example below, a small dataset is created in which the y and x1 variables are 100% correlated. The intents of this example were primarily for regression testing and, secondarily, to evaluate how the gam algorithm behaves under extreme/limit conditions. I

Hi all, Could you please help me? I am having the following weird problem when debugging R programs using "browser()": In my function, I've inserted a "browser()" in front of Step 1. My function has 3 steps and at the end of each step, it will print out the message "Step i is done"... However, after I hit <ENTER> when the program stopped before Step 1 and entered into the debugging mode, it not only executed the next line(i.e. the Step 1), but also all the (many) remaining lines in th...

Full_Name: Jerome Asselin Version: 1.3.1 OS: MacOS 9.2 Submission from: (NULL) (142.103.173.46) The step() function attempts to calculate the deviance of fitted models even if does not really need it. As a consequence, the step() function gives an error when it is used with coxph(). (There is currently no method to calculate the deviance of coxph() fits.) The code below gives an example of how

I'm trying to get the following section of code to work, I think the problem is being caused by the assignment of data to the lm function not evaluating to "train" in the parent environment but I can't seem to figure out how to do this. fitmodel <- function(trial,data) { wrap.lm <- function(formula,data,...) { cat("in wrap lm",NROW(data),"\n");

...I have been trying to get the sub function to work but can't for the life of me figure it out. I tried looking at ?sub but I think there is something that I am just not understanding. Example failed attempts > sub("\\","/","G:\Compensation Audits_Reports\Step Audit\Steps off Step.accdb") Error: '\C' is an unrecognized escape in character string starting "G:\C" > sub("\\","/","G:\Compensation Audits_Reports\Step Audit\Steps off Step.accdb",fixed=TRUE) Error: '\C' is an unrecognized escape in character...

Hi, How can I have a common set of steps that all my stories share? i.e. My stories often start out looking like this: Given a user Joe Given a user Jordan then: Given("a user $username") do |username| @users ||= {} @user_sessions ||= {} @users[username] = create_user(:username => username) @user_sessions[username...

Hi R users I am trying to run a non linear parameter optimization using the function optim() and I have problems regarding memory allocation. My data are in a dataframe with 9 columns. There are 656100 rows. >head(org_results) comb.id p H1 H2 Range Rep no.steps dist aver.hab.amount 1 1 0.1 0 0 1 100 0 0.2528321 0.1393901 2 1 0.1 0 0 1 100 0 0.4605934 0.1011841 3 1 0.1 0 0 1 100 4 3.4273670 0.105...

...at ~ rs13479085 + rs13480057 + rs13482096 + rs8254221 Df Deviance AIC <none> 233.82 841.18 rs13480057:rs13475933 6 215.95 851.12 (***) Another problem is that the final model seems to be the first model that satisfies (bAIC >= AIC + 1e-07) if steps haven't used up, rather than the one before that. Please see below. > formula(step2.bic.out) pheno.dat ~ rs13479085 + rs13480057 + rs13482096 + rs8254221 + rs13482096:rs13483699 > step2.bic.out$anova Step Df Deviance Resid. Df Resid. Dev AIC 1 NA NA 298 233.8...

Dear All, I need CentOS 4.4 and RHEL4 Update 3 Step-by-Step Text Mode Installation Screen Shots. Please send me the following Installation Screen Shots Details or link Regards -S.Balaji

Hi, I'm working with step.gam in gam package. I'm interested both in spline and lowess functions and when I define all the models that I'm interested in I get something like that: > gam.object.ALC<-gam(X143S~ALC,data=dane,family=binomial) >

...5.8840 1011.7679 1011.7679 Step 2 1 -486.0691 974.1382 977.0581 Step 5 2 -484.8520 973.7040 979.5440 Step 6 3 -483.4018 972.8036 981.5636 Step 7 4 -483.3801 974.7602 986.4401 Step 8 5 -483.2287 976.4573 991.0572 Step 9 6 -483.1112 978.2224 995.7423 first of all, why the number of steps between the above 2 outputs are different? I confirmed with coxph() that the numbers (log.p.lik, AIC, BIC) on the 1st row of summary(fit.a) are from a NULL Cox model, i.e. a model with only an intercept. Then how Step 1 in the output of summary(fit.a) is corresponding to "Step 1" in the o...

Hi, I'm playing around with stepwise regression, using the step function, and believe I have found a bug (or at least, a strong case for improvement): > ex <- data.frame(y=rnorm(100),x=rnorm(100)) > l <- lm(y ~ x, data=ex) > step(l) [output is correct] > rm(ex) > step(l) Start: AIC=11.79 y ~ x Df Sum of Sq RSS AIC - x 1 0.120 108.221 9.900

Hi there, I'd like to create a heatmap from my matrix with a) a defined color range (lets say from yellow to red) b) using striking colors above and below a certain threshold (above = "green", below = "blue") Example matrix (there should be a few outliers generated...) + simple levelplot without outliers marked: library(lattice) my.mat <- matrix(rnorm(800), nrow =