similar to: how to compare two datasets in R>?

Hello, Say I have 2 columns, bmi and gender, the first being all the values and the second being male or female. How would I subset this into males only and females only? I have searched these fora and read endlessly about select[] and split() functions but to no avail. Also the table is not ordered. bmi gender -> bmi gender + bmi gender 1 24.78 male

Hello, I need to plot the means of some outcome for two groups (control vs intervention) over time (discrete) on the same plot, for various subsets such as gender and grade level. What I have been doing is creating all possible subsets first, using the aggregate function to create the means over time, then plotting the means over time (as a simple line plot with both control & intervention

Dear R-user, I have a data like this below, age <- c("young","mid","old") married <- c("no","yes") income <- c("low","high","medium") gender <- c("female","male") I want to make some of combination data like these, age.income.dat <- expand.grid(age,

UseRs, I have a dataframe with 2547 rows and several hundred columns in R 3.1.3. I am trying to run a small logistic regression with a subset of the data. know_fin ~ comp_grp2+age+gender+education+employment+income+ideol+home_lot+home+county > str(knowf3) 'data.frame': 2033 obs. of 18 variables: $ userid : Factor w/ 2542 levels

I don't have a logistic regression model and am trying to generate probability curves for all possible combinations of the variables. My logit model has 5+ variables, and I want to draw curves for every scenario. See code below. When home_owner is 0 and 1, I want curves. The same goes for all other variables categories, so that I have permutations for all possible combinations. I've

I have a data set with about 6 million rows and 50 columns. It is a mixture of dates, factors, and numerics. What I am trying to accomplish can be seen with the following simplified data, which is given as dput output below. > head(myData) mydate gender mygroup id 1 2012-03-25 F A 1 2 2005-05-23 F B 2 3 2005-09-08 F B 2 4 2005-12-07 F B 2

Michael, Thank you for the suggestion. I will take your advice and look more critically at the covariates. John On 7/27/2017 8:08 AM, Michael Friendly wrote: > Rather than go to a penalized GLM, you might be better off > investigating the sources of quasi-perfect separation and simplifying > the model to avoid or reduce it. In your data set you have several > factors with large

Dear R users, I have a problem, which I can not find a solution. Probably someone could help me? I have a result from my classification, like this > credit.toy [[1]] age married ownhouse income gender class 1 20-30 no no low male good 2 40-50 no yes medium female good [[2]] age married ownhouse income gender class 1 20-30 yes yes high male

Dear R experts, I'm new in R and a beginner in terms of statistics. It should be simple question, but definitely difficult to solve it by myself. I'd like to see main effect of group(gender: sample size is different(M:F=23:18) and one of condition(cond) and the interaction at each subset from 90 datasets So I perform anova 90 times using a command like below; for(i in 1:90)

Dear List, When I subset a data.frame, the levels are not re-adjusted (see example). Why is this? Am I missing out on some basic stuff here? Thanks Ulrik > m <- data.frame(gender = c("M", "M","F"), ht = c(172, 186.5, 165), wt = c(91,99, 74)) > dim(m) [1] 3 3 > levels(m$gender) [1] "F" "M" > s <- subset(m, m$gender ==

Rather than go to a penalized GLM, you might be better off investigating the sources of quasi-perfect separation and simplifying the model to avoid or reduce it. In your data set you have several factors with large number of levels, making the data sparse for all their combinations. Like multicolinearity, near perfect separation is a data problem, and is often better solved by careful

Hi, Late to the thread here, but I noted that your dependent variable 'know_fin' has 3 levels in the str() output below. Since you did not provide a full c&p of your glm() call, we can only presume that you did specify 'family = binomial' in the call. Is the dataset 'knowf3' the result of a subsetting operation, such that there are only two of the three levels of

Sent from my iPhone > On Jan 19, 2025, at 1:57?PM, David Winsemius <dwinsemius at comcast.net> wrote: > > ?I don?t understand why you don?t include the full text of the error. > > ? > David > Sent from my iPhone > >> On Jan 19, 2025, at 10:00?AM, Sparks, John via R-help <r-help at r-project.org> wrote: >> >> ?Hello R-Helpers, >>

Marc, Sorry for the lack of info on my part. Yes, I did use 'family = binomial' and I did drop the 3rd level before running the model. I think the str(<subset>) that I wrote into my original email might not have been my final step before using glm. Thank you for reminding of the potential problem. I think Michael Friendly's idea is probably the solution I need to consider.

i am using mac OSX 10.7.5, running R version 2.15.2 (2012-10-26) -- "Trick or Treat" when i do: uncountry <- unique(wvsAB[,7]) wvsAB$numcountry <- match(wvsAB$country, uncountry) "unstate" isn't attaching. > library(base) > uncountry <- unique(wvsAB[,7]) > wvsAB$numcountry <- match(wvsAB$country, uncountry) > ls(wvsAB) [1] "age"

Hello R-Helpers, I was looking into how to test whether the beta coefficient from a regression would be the same for two different groups contained in the dataset for the regression. When I put that question into google, AI returned a very nice looking answer (and a couple of variations on it). library(car) data <- data.frame(income = c(30, 45, 50, 25, 60, 55), education =

Hi everyone, I'm new at R (although I'm a Stata user for some time and somehow proficient in it) and I'm trying to use the 'diverse' R package to compute a few diversity measures on a sample of firms for a period of about 10 years. I was wondering if you can give me some hints on how to best proceed on using the 'diverse' package. My sample has the following setup.

Hello, all. I've been working on this for sometime and was almost at the end/ last chunk of code i would need.... When I received an error. Rather than go to bed and think about it in the morning, I messed with my data and now I am not getting anything. I was up until 4am trying to fix this. Zip files of my data are attached (the data which ends in 'a' matches with wvsA and the

I don?t understand why you don?t include the full text of the error. ? David Sent from my iPhone > On Jan 19, 2025, at 10:00?AM, Sparks, John via R-help <r-help at r-project.org> wrote: > > ?Hello R-Helpers, > > I was looking into how to test whether the beta coefficient from a regression would be the same for two different groups contained in the dataset for the

Hi, the standard errors of the coefficients in two regressions that I computed by hand and using lm() differ by about 1%. Can somebody help me to identify the source of this difference? The coefficient estimates are the same, but the standard errors differ. ####Simulate data happiness=0 income=0 gender=(rep(c(0,1,1,0),25)) for(i in 1:100){ happiness[i]=1000+i+rnorm(1,0,40)