search for: pop2

Hello all, I have two matrices, pop and pop2, each the same number of rows and columns that I want to compare for equality. I am concerned about efficiency in this operation. I've tried a few things without success so far. Doing something simple like: if (pop==pop2) { cat('equal') } else { cat('NOT equal') } results in...

Hi, I want to make a barplot with the following datasets: I have a file as following: name chr position A1 A2 pop1 pop1 pop2 pop2 I have calculated a measure using all the values in columns "pops", the values are saved in a vector. Now I want to make a barplot using the values in this vector as the y axis and the values in the column "position" in the x-axis. thank you for your help. [[alternativ...

http://dovecot.org/test/ - Several mbox fixes, upgrade recommended for test68 mbox users - Possibly fixes some IMAP hangs where Dovecot just stopped replying - Fixed delay-newmail workaround. It was badly broken before. And somewhat off topic advertisement: I got a bit distracted from Dovecot a week ago when a guy started mailing me about wanting to write an irssi2 client as a project to

...of view) thing that progeny of each plant (i.e. family) was exposed to both treatments. Taking the artificial example one could easily do it with car-package: library(car) dat <- data.frame(Family = 1:60, # Plant family name Pop = rep(c("Pop1","Pop2","Pop3"), each=20), # Population name Cond1 = rnorm(60, 15, 1), # obtained values at experimental conditions 1 Cond2 = rnorm(60, 20, 1)) # experimental conditions 2 # rearrange data data.wide <- data...

..., below is my code: population=read.csv("nz2.csv") population names(population) nz=readShapeSpatial((sprintf('nz-geodetic-marks',tmp_dir))) nz$land_distr mapnames = nz$land_distr mapnamesnew=as.character(mapnames) popnames =as.character(mapnamesnew) pop =population$People pop2=as.numeric(pop) popdich = ifelse(pop2 < 100000, "red", "blue") popnames.lower = tolower(popnames) cols=popdich[match(mapnames,popnames.lower)] plot(nz,fill=TRUE,col=cols,proj="GCS_NZGD_2000") thanks in advance for any assistance Kind regards Iverson -- V...

...9;ing finds me nothing. When I try to write pop(...) I get pop1 <- function(vector_arg) { + length(vector_arg) -> lv + vector_arg[1] -> ret + vector_arg <<- vector_arg[2:lv] + ret + } > > pop1(s) [1] 1 > print(s) [1] 1 2 3 4 5 i.e., no side effect on the argument pop2 <- function(vector_arg) { + length(vector_arg) -> lv + vector_arg[1] -> ret + assign("vector_arg", vector_arg[2:lv]) + return(ret) + } > > pop2(s) [1] 1 > print(s) [1] 1 2 3 4 5 ditto :-( What am I missing? * Is there already a stack API for R (which I would e...

...somewhat less than numerate. I need to label the y-axes of a multhist with the y-axis labeled not as counts but as percentage of a population. Plotting the standard histogram is in a way fine, all I need is to: -- have a left-handside y-axis labels for pop 1 and a right-handside y-axis labels for pop2 -- replace the counts in each axis with population percentages (easy to calculate, but how to stick them there?) Any suggestion would be gratefully received. F -- Federico C. F. Calboli Department of Epidemiology and Biostatistics Imperial College, St. Mary's Campus Norfolk Place, London W2...

Dear All, I'm new to use dovecot for the Mail service... So, how to enable the mail service ( pop3 / pop2 / imap ) ? Edward.

...human adult lifespan. For example, if I wanted to plot the hazard (i.e., mortality) functions: pop1 <- function (t) { x=c(0.03286343, 0.04271132) a3<-x[1] b3<-x[2] shift<-15 # only considering mortality after 15 years h.t<-a3*exp(b3*(t-shift)) return<-h.t } pop2 <- function (t) { x=c(0.02207778, 0.04580059) a3<-x[1] b3<-x[2] shift<-15 # only considering mortality after 15 years h.t<-a3*exp(b3*(t-shift)) return<-h.t } ylab.name <- expression(paste(italic(h),"(",italic(a),")")) plot(seq(15,80,1),p...

..., n10, type='b', xlim=range(c(1:t),c (1:t)), ylim=range(n10, n20), xlab='tempo', ylab='tamanho populacional') points(t, n20, type='b', col="red") points(t,n10,type="b", col="black") legend("topleft", c("Pop1","Pop2"), cex=0.8, col=c ("black","red"), pch=21:21, lty=1:1); } if(t>0){ for (i in 1:t){ n1[i==1]<-n1 n2[i==1]<-n2 n1[i+1]<-k1-alfa*n2[i] n2[i+1]<-k2-beta*n1[i] if(n1[i]==0){n1[i:t]==0} if(n2[i]==0){n2[i:t]==0} } plot(c(1:t), n1[1:t], type='b', xlim=ran...

Dear R user: I am studying the allele data of two populations. the following is the data: a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16 a17 pop1 0.0217 0.0000 0.0109 0.0435 0.0435 0.0000 0.0109 0.0543 0.1739 0.0761 0.1413 0.1522 0.1087 0.0870 0.0435 0.0217 0.0109 pop2 0.0213 0.0213 0.0000 0.0000 0.0000 0.0426 0.1702 0.2128 0.1596 0.1809 0.0957 0.0745 0.0106 0.0106 0.0000 0.0000 0.0000 a1,a2,a3 ...... a17 is the frequency of 17 alleles , the sum is 1. I want to test the significance of the distribution of 17 alleles between two populations. How...

...eles = 10 Therefore AD = (2 * n) / 10 = 0.2 What I want to do is compare two populations of 200 organisms each but sampling for only 20 at a time. So there are 200!/((200-20)! * 20!) possible combinations of samples in each population. For all possible combinations of sample pop1 and sample pop2 I want to measure AD ie (200!/((200-20)! * 20!) * 200!/((200-20)! * 20!) ) calculations. As well as the unique allele problem, can someone suggest how I can do the sampling loops? Thanks, Phil. -- Philip Rhoades Pricom Pty Limited (ACN 003 252 275 ABN 91 003 252 275) GPO Box 3411 Sydney...

...of each combination through which the line should be fit for subsequent plotting. This hopefully shows the idea: ---<---------------cut here---------------start-------------->--- toydf <- expand.grid(1:100, c("A", "B"), c("pop1", "pop2", "pop3", "pop4", "pop5")) toydf <- data.frame(facA = toydf[[3]], facB = toydf[[2]], x = toydf[[1]], y = rnorm(1000)) xyplot(y ~ x | facA, groups = facB, data = toydf, panel.groups = function(x, y, subscripts, ...) { panel.x...

...the per bin count, and the factors identifying each record. My problem is including the latter: ---<---------------cut here---------------start-------------->--- toydf <- expand.grid(sample(0:50, 50, TRUE), c("A", "B"), c("pop1", "pop2", "pop3", "pop4", "pop5")) "getFreq" <- function(x, bks) { fhist <- hist(x, breaks = bks, plot = FALSE, include.lowest = TRUE) matrix(c(fhist$breaks[-1], fhist$counts), nrow = length(fhist$breaks[-1]), ncol = 2) } f...

Dear All, I am trying to build a program which will take repeated samples without replacement from a population of values. The interesting catch is that I would like the sample values to be removed from the population, after each sample is taken. For example: pop<-c(1,5,14,7,9,12,18,19,65,54) sample(pop, 2) = lets say, (5,54) ## This is where I would like values (5, 54) to be removed from

greeting. a short while ago, i may have gone to a site i should not have. maybe. after visiting, i decided i would check for rpm updates. when yumex opened to available packages, it showed that; openssl.x86_64 0:1.0.1e-42.el6_7.4 was available, so i checked it, then clicked install button. during log display, i got error message, tried again. still got error. seems that there is a

The text below is a part of, some work I have to do, which is due in 2 days and I am strung up with a lot of other stuff, so I was hoping someone would take 5 mins and help me ?? Here is a part of my data.frame: year country1 country2 contig comlang pop1 gdp1 pop2 gdp2 rta dist avgflow 1 1992 AUS AUT 0 0 17.4950008 321708.281 7.7825189 194684.078 0 15608.4 1.075999e+02 2 1992 AUS BEL 0 0 17.4950008 321708.281 10.0450001 231762.094 0 16319.2 4.767162e+02 3 1992 AUS CAN...

...#elapsed ~ 12s... by far the slowest approach!!!! # Suggestions #Peter Langfelder values = c(500, 2169, 1121, 2500, 300, 625) system.time( elevation.PL <- values[as.numeric(factor(Population))] ) # ~ 0.85s # Values need to be in the order in which the levels of the factor are sorted #i.e. Pop2 <- rep(c("Ga", "CO", "CN", "Ng", "KO", "Mw"), 10) # levels(factor(Pop2)) would not work. #or codeToElev = data.frame(codes = c("CO", "CN","Ga","KO", "Mw", "Ng"),...

Hi people, I have a text file like this one posted: snp_id gene chromosome distance_from_gene_center position pop1 pop2 pop3 pop4 pop5 pop6 pop7 rs2129081 RAPT2 3 -129993 "upstream" 0.439009 1.169210 NA 0.233020 0.093042 NA -0.902596 rs1202698 RAPT2 3 -128695 "upstream" NA 1.815000 NA 0.399079 1.8142...

....Dallas1.Level3.net (64.159.3.41) 42.432 ms 42.468 ms 42.399 ms 14 pop1-dal-P0-3.atdn.net (66.185.141.49) 44.038 ms 43.801 ms 43.575 ms 15 bb1-dal-P0-0.atdn.net (66.185.146.144) 44.262 ms 43.778 ms 43.730 ms 16 bb1-hou-P6-0.atdn.net (66.185.152.133) 48.333 ms 48.300 ms 48.188 ms 17 pop2-hou-P1-0.atdn.net (66.185.146.113) 48.127 ms 48.091 ms 48.131 ms 18 rr-Houston.atdn.net (66.185.146.38) 49.331 ms 49.778 ms 49.286 ms 19 srp2-0.hstntxtid-rtr1.texas.rr.com (24.93.33.145) 48.566 ms 48.574 ms 48.577 ms 20 pos3-0.hstntxeas-rtr1.houston.rr.com (24.28.96.198) 49.660 ms 49...