similar to: Proportions of a vector

Hi all, I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6) I want to return a vector with the corresponding probabilities based on the amount of times the numbers occurred. For example, I should get the following vector for xm: prob.xm = c(1/9, 1/9, 1/9, 1/9, 3/9, 3/9, 3/9, 2/9, 2/9) Any help greatly appreciated. -- Thanks, Jim. [[alternative HTML version deleted]]

Hello, I would like to be able to update an existing column in a dataframe, like this... data$score[data$type=="1" & data$year=="2001"]<-data$score * 0.111 data$score[data$type=="1" & data$year=="2002"]<-data$score * 0.222 data$score[data$type=="1" & data$year=="2003"]<-data$score * 0.333 ...but, if possible,

Hello there, I have two groups (men and women) and I know per group how many of them smoke or don't smoke (women 40 of 200; men 100 of 300). I would like to know how I can compare in R if men and women differ significantly in their smoking. However, because there are more men in the sample than women I cannot just compare the number of smokers and non-smokers in both groups, right?! (I would

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Does anybody knows what this error message means: Error in object$terms : $ operator not defined for this S4 class I have peformed the following steps: > library(lattice) > library(Matrix) > library(lme4) > inkm inkm$Gamie glm.incm drop1(glm.incm,test="Ch") Error in object$terms : $ operator not defined for this S4 class Your suggestin would be of a greatl help to

Dear R-Helpers, I've looked high and low for a function that provides frequencies, proportions and cumulative proportions side-by-side. Below is the table I need. Is there a function that already does it? Thanks, Bob > # Generate some test scores > myValues <- c(70:95) > Score <- ( sample( myValues, size=1000, replace=TRUE) ) > head(Score) [1] 77 71 81 88 83 93 > >

Hi r-users,   I would like to use prop.test code and I also calculate manually to test the proportions for 2 groups.  The problem is the answer for the p-value calculated manually are different from prop.test.  Here are the results:   ## Manually   z value: z= (phat-p)/sqrt(pq/n) = (.084-.081)/sqrt(.081(1-.081)/691)=0.289, pvalue=0.7718   ## Using prop.test code > low <- c(56,58) > tot

Dear list members, I want to perform in R the analysis "simultaneous confidence interval for multiple proportions", as illustrated in the article of Agresti et al. (2008) "Simultaneous confidence intervals for comparing binomial parameter", Biometrics 64, 1270-1275. If I am not wrong the R function implementing the Agresti et al. method is prop.test(). I ask an help because I

Hi, I'm try to compute the minimum sample size needed to have at least an 80% of power, with alpha=0.05. The problem is that empirical proportions are really small: 0.00154 in one case and 0.00234. These are the estimated failure proportion of two medical treatments. Thomas and Conlon (1992) suggested Fisher's exact test and proposed a computational method, which according to their table

Hello, Does anyone know which method from Newcombe (1998)* is implemented in prop.test for comparing two proportions? I would guess it is the method based on the Wilson score (for single proportion), with and without continuity correction for prop.test(..., correct=FALSE) and prop.test(..., correct=TRUE). These methods would correspond to no. 10 and 11 tested in Newcombe, respectively. Can

When one has raw data it is easy to create a table of one variable against another and then calculate proportions For example a.nice.table<-table(a,b) prop.table(a.nice.table,1) However, I looked at several papers and created a data frame of the aggregate data. That means I acually created a table except it is a data frame. The first column lists the name of the first author and the year. I

Hi, I have a seemingly simple proportional test. ?here is the question I am trying to answer: ? There is a test running each day in the lab, the test comes out as either positive or negative. So at the end of each month, we can calculate a positive rate in that month as the proportion of positive test results. The data look like: ? Month??? ??# positive?????? # total tests??? positive rate

Hello. I asked before and it was great, cause as a beginner I learned a lot. But, if I have this in R (1 and 2 are codes for sex): > sex<-c(1,2,2,1,1,2,2,2) > sex [1] 1 2 2 1 1 2 2 2 I´d like to obtain the proportion according to sex.So I type: > prop.table(sex) [1] 0.07692308 0.15384615 0.15384615 0.07692308 0.07692308 0.15384615 0.15384615 [8] 0.15384615 The result is OK, but I

Dear Josh, Thanks for your help! Does your answer mean, that you agree the two methods should do the same, and what I was guessing, despite the small differences? What I prefer about ci.pd is, that the help clearly says which method is implemented, which is not the case for prop.test. But I do not know who has programmed the function. Best wishes Steffi Stefanie von Felten, PhD Statistician

Hi, I have a dataset that has 2 groups of samples. For each sample, then response measured is the number of success (no.success) obatined with the number of trials (no.trials). So a porportion of success (prpop.success) can be computed as no.success/no.trials. Now the objective is to test if there is a statistical significant difference in the proportion of success between the 2 groups of

Hello, I'm looking for some guidance with the following problem: I've 2 samples A (111 items) and B (10 items) drawn from the same unknown population. Witihn A I find 9 "positives" and in B 0 positives. I'd like to know if the 2 samples A and B are different, ie is there a way to find out whether the number of "positives" is significantly different in A and B?

Hi, I am wondering if anyone can suggest how to test the equality of 2 proportions. The caveat here is that the 2 proportions were calculated from the same number of samples using 2 different tests. So essentially we are comparing 2 accuracy rates from same, say 100, samples. I think this is like a paired test, but don't know if really we need to consider the "paired" nature of the

Hi All, I have a table based on ordial data and i want to compare proportions and i've seen in the pwr package i can use power.prop.test however i want to find out what the sig. value is based on n1,n2,p1,p2 and this package doesn't contain this.. Does anyone know of a package that does or is it a case of writting a function specifically for this? Many thanks in advance [[alternative

Dear R-users, I have output files having a variable number of tables in the following format: ------------- 1 Pietje I1 I2 Value 1 1 0.11 1 2 0.12 2 1 0.21 2 Jantje I1 I2 I3 Value 1 1 1 0.111 3 3 3 0.333 ... ------------- Would there be an easy way of turning this into (a list of) data.frames with names Pietje, Jantje and variables I1,I2,...Value? (I1,I2 are string or categorical,

Hi, xtabs() creates a table of counts. I want a table of proportions -- that is, I want to divide every vector (along a particular dimension) by its sum. The tiny example below does that. The call to xtabs() creates a matrix "A" with dimensions ("x1","x2","y"). I transform "A" using aperm() and aaply() to get the matrix "B". The