search for: all2

...I'm still trying to figure out the syntax of R to plot some distributions of some frequencies. I've managed to plot histograms from the data, but I would like to clean it up using xyplot from library(lattice). Unfortunately I cannot find a solution to my problem. Given a dataframe "all2", with numerical data in cols "year" and "standard" ("standard" being standardized data) >all2 Year standard 1 2001 0.034246575 2 2001 0.000000000 ... 141 2008 0.012820513 142 2008 0.230769231 I have plotted separate histograms for each year usin...

...ot;,"MHsub","SFD")) c<-list(c("MHpark","SFD","SFD")) d<-list(c("MultiFam","MultiFam","MultiFam","Duplex")) all<-list(a,b,c,d) names(all)<-c("5","6","7","8") all2<-list() for(i in 1:length(all)){ all2[[i]]<-table(unlist(all[i])) } names(all2)<-names(all) What i want to do is to put each of the elements of the list into a data frame where the element represents a row and each vector is accounted for, for all types within the entire list. So where t...

Dear all, i need some help with plotting. the specific problem is the following: #FYI a=100 b=95 d=94.5 e=70 all=c(a,b) all2=c(d,e) plot(all,type="b",col="blue",xlim=c(1,4),ylim=c(20,150)) lines(all2,type="o",col="yellow") this does work so far, but ... i´d like to have 4 intersects, just named by characters.. no scale please. the second problem is, i am trying to start the sec...

...r R-help subscribers, I have a quite stupid question about using lapply. I have the following function: create.gradient <- function(i){ colorgrad01<-color.scale(seq(0,1,by=0.01), extremes=c("red","blue")) tree1$edge[i,1] -> x tree1$edge[i,2] -> y print(x) print(y) all2[x] -> z all2[y] -> z2 round(z, digits = 2) -> z round(z2, digits = 2) -> z2 z*100 -> z z2*100 -> z2 print(z) print(z2) colorgrad<-colorgrad01[z:z2] colorgrad } Basically, I want to pick a partial gradient out of a bigger gradient (colorgrad01) for values that are on row i, fr...

..., 1, "sd") > csv <- read.csv(file="D:/Anne-Marie/Doctorat/puces ADN macrophages/puces rat/Annie Dube/Analyse/RaGene-1_0-st-v1.na27.rn4.transcript.header.DOS.csv",head=TRUE, sep=",") > all <- merge(csv, Pvalue, by.x="sample", by.y=0, all=T) > all2 <- merge(all, Mean1, by.x="sample", by.y=0, all=T) > all3 <- merge(all2, Mean2, by.x="sample", by.y=0, all=T) > all4 <- merge(all3, sd1, by.x="sample", by.y=0, all=T) > all5 <- merge(all4, sd2, by.x="sample", by.y=0, all=T) > all6...

Hi, I need to find the z-score of the data present in a speardsheet. The values needs to be calculated for each gene across the samples (refer the example). And, it should be a simple thing, but I am unable to do it right now ! The example re the structure of the spreadsheet is - # Example: MyFile <- read.csv( text= "Names,'Sample_1','Sample_2','Sample_3'

...function to maximize #NOTE: by default the solver minimizes the objective. therefore the original minus sign is not used obj=function(x){ all1=sum(mat%*%x) Hp=0 for (j in 1:BR){ p_b=mat[j,]%*%x/all1 Hp=Hp+p_b*log(p_b)*br[j] } return(Hp) } # the equality constraint function: equal1=function(x){ all2=sum(mat%*%x) sum_pterm=sum(mat[1:N,]%*%x/all2) sum_x=sum(x) return(c(sum_pterm,sum_x)) } # the right hand side for the constraint eqB=c(1,budget) # the inequality function inequal1=function(x){ all3=sum(mat%*%x) p_b=mat[1:BR,]%*%x/all3 return(as.vector(p_b)) } #the lower limit for inequalitie...

hello, I am using R1.2.2 on win98 with 64 meg of RAM I try to read a big csv file and get the error : a <- read.csv("c:/all2.csv",header=T) Error: vector memory exhausted (limit reached?) In addition: Warning message: Reached total allocation of 16Mb: see help(memory.size) Lost warning messages from there on, nothing works File-Exit, ^F4, q() , top right cross, the same message keep on appearing and the only sol...

...est STE AREA -0.229 V1 V1 -1.129 V1 V2 -2.084 V1 V4 -2.918 V1 V3 -0.548 V2 V1 -0.217 V2 V2 -1.937 V2 V3 -1.205 V2 V4 -4.126 V4 V1 -1.481 V4 V2 -2.070 V4 V3 -0.042 V4 V4 xyplot(est ~ ISITE, data = all2, groups = AREA, type = "b", auto.key = list(space = "right")) Ken Knoblauch -- Ken Knoblauch Inserm U371 Cerveau et Vision Dept. of Cognitive Neuroscience 18 avenue du Doyen Lépine 69500 Bron France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: +33 (0)6 84...

...V2 > -2.084 V1 V4 > -2.918 V1 V3 > -0.548 V2 V1 > -0.217 V2 V2 > -1.937 V2 V3 > -1.205 V2 V4 > -4.126 V4 V1 > -1.481 V4 V2 > -2.070 V4 V3 > -0.042 V4 V4 > > xyplot(est ~ ISITE, data = all2, groups = AREA, type = "b", > auto.key = list(space = "right")) > Ken Knoblauch -- Ken Knoblauch Inserm U371 Cerveau et Vision Dept. of Cognitive Neuroscience 18 avenue du Doyen L?pine 69500 Bron France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: +...

..."AsIs" > > oct <- gasoline$octane > > mars <- earth( oct ~ nir, nk=300, nfold=10, degree=3, trace=0 ) > > plotmo( mars ) Error: bad index (missing column in x?) > plotmo( mars, all1=T ) Error: bad index (missing column in x?) > plotmo( mars, all1=T, all2=T ) Error: bad index (missing column in x?) > > sessionInfo() R version 2.13.1 (2011-07-08) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Portuguese_Brazil.1252 LC_CTYPE=Portuguese_Brazil.1252 LC_MONETARY=Portuguese_Brazil.1252 [4] LC_NUMERIC=C L...

Hello, I was wandering if it is possible to perform on a dataframe called 'all' a shapiro wilk normality test for COUNTS by variable Group ACTIVITY? Could it be done using plyer? I saw an eg that applies to an array but not to a dataframe: lapply(split(dataset1$Height,dataset1$Group),shapiro.test) Any thoughts would be much appreciated. My dataframe is in shape: dat ACTIVIT