similar to: help with parallel processing code

Displaying 20 results from an estimated 100 matches similar to: "help with parallel processing code"

2011 Oct 27
2
help with paste
Hello gurus, I have some variables, and i am creating combinations for analysis in the end i need these variables to be displayed like "LEPTIN+SAA+PTH". currently i am using loop to perform this. I would appreciate any pointers to do it without the loop. > mols=c("LEPTIN","SAA","PTH","sEGFR") > samples=mols[1:3] > samples [1]
2012 Feb 27
1
row names of a heatmap on left hand side
Hello R Gurus, I have a matrix for which I am doing a heatmap using heatmap.2. I want to put the rownames on the lefthand side instead of the right side of the heatmap. how can i put the rownames on left hand side: I have already tried axis but could not make it work ccc<-structure(c(1, 0.283300333755851, 0.237863231117007, 0.0148696794159253, -0.0780756406815149, -0.106161465097659,
2012 Mar 20
0
labeling rows in heatmap.2
how can i put the row labels on the left hand of the heatmap in heatmap.2? abnr<-structure(c(1, 0.678622097406395, 0.670294749652918, -0.0016314464654279, -0.000519068106572792, 0.199581999119988, -0.0106623494189115, 0.0840111691399559, -0.0461494399639137, 0.249279171677728, NA, 1, 0.757114062773504, 0.0352642759270137, -0.0255518450373996, 0.0943268190664674, -0.0536269679247722,
2010 Jun 13
1
Boxplot intervals combining names
Hi R users, This seems like a simple problem but I have searched nabble for the answer and can't seem to find it. All I want to do is produce a boxplot where I have two boxes for one Individual but on the xaxis I only have one tick mark centred between the boxes so I can add the Individuals' name. I have 30 IDs and have shown the code I use below for a couple of IDs, I figure the data
2011 Apr 13
1
error for ttest
Hello all, I have arranged my data as per Dennis's suggestion in this post http://www.mail-archive.com/r-help at r-project.org/msg107156.html. the posted code works fine but when I try to apply it to my data, i get "> u2 <- ddply(xxm, .(plateid, cytokine), as.data.frame.function(f)) Error in t.test.formula(conc ~ Self_T1D, data = df, na.rm = T) : grouping factor must have
2004 Jun 25
3
String manipulation
Hi, let's see, if someone can help my with this one: I have the string as follows: > str<-("one","two","three") Now I want to concatenate the items to one string, seperateted by space or something else, >str >"one, two, three" If possible without a loop. My actual goal ist to create string like >str.names >"female = names1, male
2009 Oct 30
2
Names of list members in a plot using sapply
Hi R users: I got this code to generate a graphic for each member of a lists. list1<-list(A=data.frame(x=c(1,2),y=c(5,6)),B=data.frame(x=c(8,9),y=c(12,6))) names1<-names(list1) sapply(1:length(list1),function(i) with(list1[[i]],plot(x,y,type="l",main=paste("Graphic of",names1[i])))) Is there a more elegant solution for not to use two separate lists? I would like to
2012 Jun 03
2
merging single column from different dataframe
Hi all, probably really simple to solve, but having no background in programming I haven't been able to figure this out: I have two dataframes like df1 <- data.frame(names1=c('aa','ab', 'ac', 'ad'), var1=c(1,5,7,12)) df2 <- data.frame(names2=c('aa', 'ab', 'ac', 'ad', 'ae'), var2=c(3,6,9,12,15)) Now I want merge
2013 Sep 05
2
binary symmetric matrix combination
Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1
2005 Oct 13
0
Maps package, coloration
I am having some trouble getting the colors correct on county maps using the maps package. I have a data.frame that contains coloration data for every county -- it also contains a variable 'mapm' which fits the 'state,county' format used in the mapping package. I use this to define colors from a range of 1:100. The problem is that the colors seem good for some states/counties,
2009 Aug 28
1
extracting pvalues from ttest
Hello list, I have a similar issue as this post http://tolstoy.newcastle.edu.au/R/e6/help/09/04/11438.html#options2 and I used the suggestion provided by Jorge with modifications to my data do.call(c,lapply(your_list_with_the_t_tests,function(x) x$p.value)) but I am getting the following error after excuting the code B<-by(eo,eo$PlateID, function(.sub) t.test(mcp1~Self_T1D,data=.sub,
2020 Oct 18
1
Lookup sid with libsmbclient (invoked from c# on mono)
Hi, There is some additional work to do (close lsa handle), but the far the code below is working. Let me share it with you in case somebody will have the same need that me. br Vincent LE TOUX using Microsoft.Win32.SafeHandles; using System; using System.ComponentModel; using System.Diagnostics; using System.Net; using System.Runtime.InteropServices; using System.Security.Principal; namespace
2012 Sep 14
0
rcdk package, fingerprint without name?
Hi! I have a question concerning the rcdk package: I generated a test sdf-file with 3 molecules in it and tried to perform clustering by fingerprints and plot the results in a dendogramm. This is what I did: mols <- load.molecules ("molecules.sdf") fp.list <- lapply (mols, get.fingerprint, 'maccs') fp.dist <- fp.sim.matrix (fp.list, method='tanimoto')
2011 Nov 07
2
help with formula for clogit
I would like to know if clogit function can be used as below clogit(group~., data=dataframe) When I try to use in above format it takes a long time, I would appreciate some pointers to get multiple combinations tested. set.seed(100) d=data.frame(x=rnorm(20)+5, x1=rnorm(20)+5, x2=rnorm(20)+5, x3=rnorm(20)+5, x4=rnorm(20)+5, x5=rnorm(20)+5, x6=rnorm(20)+5, x7=rnorm(20)+5, x8=rnorm(20)+5,
2011 Nov 08
2
match first consecutive list of capitalized words in string
Dear R-Helpers, this is my first post ever to a mailing list, so please feel free to point out any missunderstandings on my side regarding the conventions of this mailing list. My problem: Assuming the following character vector is given: names <- c("filia Maria", "vidua Joh Dirck Kleve (oo 02.02.1732)", "Bernardus Engelb Franciscus Linde j.u.Doktor referendarius
2004 Oct 11
0
ALERT - GroupShield ticket number OA480_1097509999_EXCH-03_1 was generated
Action Taken: The attachment was quarantined from the message and replaced with a text file informing the recipient of the action taken. To: nitingk at noida.hcltech.com <nitingk at noida.hcltech.com> From: openssh-unix-dev at mindrot.org <openssh-unix-dev at mindrot.org> Sent: -2017484928,29667242 Subject: Message could not be delivered Attachment Details:- Attachment Name: opg
2014 Oct 19
2
Checking if a matrix exists/is defined
Hello Can someone help me with the following, specifically in judging whether a matrix exists. I have trouble with the first line below. In this case, matrix obj$hessian exists and is 74 x 74. I receive the error message: Warning message: In all(w$hessian) : coercing argument of type 'double' to logical Thank you all. --- if (!all(obj$hessian)|OPG){ vb<-obj$gradientObs;
2013 Feb 17
2
Loop
Hi all, I want to execute a loop of a program: for (u in Timeframemin:Timeframe){} Imagine that Timeframemin<-10 Timefram<-10000 Is it posible to execute the loop but only proving from 10 to 10000 but jumping 10 each time, for example, execute for 10,20,30.....to Timeframe. Other question is, when a program is "heavy" and has a lot of loops to execute (how can I know where
2011 Nov 30
1
How can I pick a matrix from a function? (Out Product of Gradient)
Hi all, I would like to use optim() to estimate the equation by the log-likelihood function and gradient function which I had written. I try to use OPG(Out Product of Gradient) to calculate the Hessian matrix since sometime Hessian matrix is difficult to calculate. Thus I want to pick the Gradient matrix from the gradient function. Moreover, could R show the process of calculation on gradient
2008 Feb 21
3
applying a function to data frame columns
useR's, I want to apply this function to the columns of a data frame: u[u >= range(v)[1] & u <= range(v)[2]] where u is the n column data frame under consideration and v is a data frame of values with the same number of columns as u. For example, v1 <- c(1,2,3) v2 <- c(3,4,5) v3 <- c(2,3,4) v <- as.data.frame(cbind(v1,v2,v3)) uk1 <- seq(min(v1) - .5, max(v1) + .5,