similar to: FW: how to use by() ?

Hello, All! How might one accomplish this using the by() function? m1 is a data frame. # populate column "m1$major_allele" for ( i in 1:length(m1$major_allele)) { if ( m1$Freq1[i] == m1$MAF[i]){ m1$major_allele[i] = m1$Al1[i] } else{ m1$major_allele[i] = m1$Al2[i] } } Jim [[alternative HTML version deleted]]

I am new in R and i am having trouble here. I?ve already searched in the list but hasn?t helped When i run this script above i get the message "Error in gen[j, i] : incorrect number of dimensions". However gen is 1000x200 (ind x loc) and so is g could anybody help me for (i in 1 : loc) { #loc=200 for (j in 1 : ind) { #ind=1000 g1 <= function ( gen ) matrix ( if (gen[j, i]

Hey, I have a dataset and I want to identify the records by groups for further use in ggplot. Here is a sample data: ID Value AL1 1 AL2 2 CA1 3 CA4 4 I want to identify all the records that in the same state (AL1 AND A2), group them as "AL", and do the same for CA1 and CA4. How can I have an output like: ID Value State AL1 1 AL AL2 2 AL CA1 3 CA CA4 4

Is there a way to combine two columns within a data frame? Example data: id snp AL1 AL2 1500 30 A B 1510 30 A A 1520 30 A B This is what I would like: indv snp AL1AL2 1500 30 AB 1510 30 AA 1520 30 AB Any help is greatly appreciated. Alysta

Greetings: I would like comments on this example and after fixing it up, I need help from someone who has access to insert this in R's help page for plotmath. I uploaded a drawing http://pj.freefaculty.org/R/Normal-2009.pdf that is created by the following code http://pj.freefaculty.org/R/Normal1_2009_plotmathExample.R This will be a good addition to the plotmath help page/example.

Hello All, I have a toy dataframe like this. It has 8 columns separated by tab. Name SampleID Al1 Al2 X Y R Th rs191191 A1 A B 0.999 0.09 0.78 0.090 abc928291 A1 B J 0.3838 0.3839 0.028 0.888 abcnab A1 H K 0.3939 0.939 0.3939 0.77 rx82922 B1 J K 0.3838 0.393

Hi John, Thank-you very much for replying, - in answer to your questions: *"As I understand what you wrote, all factors but linelabel are crossed, with linelabel nested within all the other factors?" That's exactly it, *" What is in the highest order cells? Single observations, or multiple scores?" I would say "single observations" - but i have to

I want to shade the area under the curve of the standard normal density. Specifically color to the left of -2 and on. How might i go about doing this? Thanks -- View this message in context: http://n4.nabble.com/Shade-area-under-curve-tp1586439p1586439.html Sent from the R help mailing list archive at Nabble.com.

Hello R useRs, I have a function which returns a list of functions : freq1 <- function(x) { lev <- unique(x[!is.na(x)]) nlev <- length(lev) args <- alist(x=) if (nlev == 1) { body <- c("{", "sum(!is.na(x))", "}") f <- function() {} formals(f) <- as.pairlist(args) body(f) <- parse(text = body) namef <-

Good Mourning, I have a function to generate a matrix as I show part of it; g[j,i]<-if (gen[j,i]==0) al1[i,1]+al1[i,1] else ... However i would like that this function occurred with a probability P and that another function (another formula to generate g matrix) with probability P-1 That?s it, if P is .7, i would like that in 70% of the times (for random i and j) the matrix g was generated

Hi, I have a grouped data set and would like to calculate weighted proportions for a large number of factor variables within each group member. Rather than using dplyr::count() on each of these factors individually, the idea would be to do it for all factors at once. Does anyone know how this would work? Here is a reproducible example: ############################################################

Hi, I have loaded a dataset in R : data = label freq1 freq2 news 54 35 fun 37 21 milk 19 7 food 3 3 .... etc And I have a vector flist<-c("fun","food") Now I want to use the vector 'flist' for selecting these values from 'data' so that I get the following dataset : label freq1 freq2 fun 37 21 food

Dear masters, I have data that looks like this: #val Freq1 Freq2 0.000 178 202 0.001 4611 5300 0.002 99 112 0.003 26 30 0.004 17 20 0.005 15 20 0.006 11 14 0.007 11 13 0.008 13 13 ...many more lines.. Full data can be found here: http://dpaste.com/173536/plain/ What I intend to do is to have a cumulative graph with "val" as x-axis with "Freq1" & "Freq2" as

Good day, All, Is there any way to maintain the number of decimal places in the type of situation below? I would like to maintain the number of decimal places in 0.667, despite the fact that its column-mates have a fourth decimal place. Thank you for your time. Jim dat.txt contents: MARKER ALLELES FREQ1 RSQR EFFECT2 STDERR CHISQ PVALUE rs6599753 C,T

Dear all, I am fitting an nonlinear glm() using optim() by first minimising glm(resp~ var1 + var2, family=binomial, data=data)$deviance where var1= exp(-a1*dist1), and var2= exp(-a2*dist2), where a1 and a2 are parameters and dist1 and dist2 are independent variables. Next, I calculate the value of var1 (and var2) by plugging in the value of al1 (and al2) that minimises deviance, and fit

I am trying to work my way through the book "Singer, JD and Willett, JB, Applied Longitudinal Data Analysis. Oxford University Press, 2003" using R. I have the SAS code and S-Plus code from the UCLA site (doesn't include chapter 8 or later problems). In chapter 8, there is a structural equation/path model which can be specified for the sem package as follows S <- cov(al2)

Hello all, I've been having trouble with assessing a breakpoint in a logistic GLM with two explanatory variables. For this analysis I've been using the 'segmented' package version 0.2-9.1. But I keep getting an error and I don't see where I would be going awry. The situation is the following: Two explanatory variables: bedekking - a variable with possible values between 0 and

Dea'R' helpers I have following data - prob = c(0.1, 0.2, 0.3, 0.4, 0.5) frequency = c(100, 75, 45, 30, 25) no_trials = c(10, 8, 6, 4, 2) freq1 = rbinom(frequency[1], no_trials[1], prob[1]) freq2 = rbinom(frequency[2], no_trials[2], prob[2]) freq3 = rbinom(frequency[3], no_trials[3], prob[3]) freq4 = rbinom(frequency[4], no_trials[4], prob[4]) freq5 = rbinom(frequency[5],

Hi all. 1) If I have a dataframe with variable names as follow: PC1 PC2 ... PCn and I want to pass only some of them to a function, e.g. glm(resp~from PC1 to PC10, PC15, etc.,...) is there a faster way than simply writing each variable name in the formula? 2) Again, I have a dataframe, say ali.df, with tha following variables: ali1, ali2, ...ali78 I want to sum, for example, ali1+al2+ali7+f rom

Dear R experts, I am trying to analyze data from an article, the data looks like this Patient Age Sex Aura preCSM preFreq preIntensity postFreq postIntensity postOutcome 1 47 F A 4 6 9 2 8 SD 2 40 F A/N 5 8 9 0 0 E 3 49 M N 5 8 9 2 6 SD 4 40 F A 5 3 10 0 0 E 5 42 F N 5 4 9 0 0 E 6 35 F N 5 8 9 12 7 NR 7 38 F A 5 NA 10 2 9 SD 8 44 M A 4 4 10 0 0 E 9 47 M A 4 5 8 2 7 SD 10 53 F A 5 3 10 0 0 E 11