similar to: Printing a variable in a loop

Hello, I have a dataframe (var_1, var_2, ..., var_n) and I would like to export summary statistics to Latex in the form of a table. I want specific summary statistics by crossing numerous variables 2x2 AT ONCE. In each cell I would like sometimes to have the median (Q1 - Q3), or frequency and proportion, etc. CrossTable, xtab, etc... do not allow for multiple 2 by 2 crosstabulation. The table

Hello, I'm new to R and have a very basic syntax question for the functionglm. I am using the function glm() to do a regression on a data set. I want to run this function in a while loop, so I need to be able to do two things: (1) I need to be able to have my list of variables in the regression be based from a variable vector instead of hard code. (2) I need to be able to collect the

Buenas tardes, Quiero aplicar la función rbind y necesito tener los mismos nombres de columnas. Como tengo unas 195 variables en cada dataframe, necesito hacerlo de una forma rápida. Tengo 9 bases de datos y tengo que fusionar todas. ¿Como puedo comprobar que los nombres de las variables son los mismos? Y de lo contrario, ¿como detecto las diferencias? He probado con

Hi, I know you can access various sip variables via 'Set(sstatus=${SIPPEER(201,status)})' (for example) to get the status of the sip user - but what about variables? I have a user that has setvar=123456 in their users.conf (sip.conf if you prefer). I can read it with a 'sip show peer 201' - but that gives everything and parsing that isn't really an option. Anyone know how

Buenos dias, Tengo la siguiente duda. Desde una hoja de calculo .xls me descargo un fichero en r as.matrix. Time VAR_2 VAR_3 VAR_4........ VAR_N Fechas(n) Precios Precios Precios Precios Fechas (n-1) ¿como puedo hacer para graficar de forma ordenada (fechas) cada uno del vector columnas (son precios de activos)? [[alternative HTML

Hey, I'm trying to create an image file with the results of a regression analysis. In TeX, the line would be something like: $ size = 0.34 + 4.3 var_1 $ Can I create a plot window with just this line in it? I tried playing around with plot.new() or dev.new(), but didn't really find something that worked. Thanks in advance, Alex -- alexx at alexx-fett:~$ vi .emacs

I have the data similar to the following in a data frame: LastName Department Salary 1 Johnson IT 56000 2 James HR 54223 3 Howe Finance 80000 4 Jones Finance 82000 5 Norwood IT 67000 6 Benson Sales 76000 7 Smith Sales 65778 8 Baker HR 56778 9 Dempsey HR 78999 10 Nolan

I am relatively new to R and object oriented programming. I have relied on SAS for most of my data analysis. I teach an introductory undergraduate forecasting course using the Diebold text and I am considering using R in addition to SAS and Eviews in the course. I work primarily with univariate or multivariate time series data. I am having a great deal of difficulty understanding and working with

Dear all I have a work-flow issue with lm(). When I use > lm(y1~x1, anscombe) Call: lm(formula = y1 ~ x1, data = anscombe) Coefficients: (Intercept) x1 3.0001 0.5001 I get as expected the formula, "y1 ~ x1", in the print()ed results or summary(). However, if I pass through a formula object > (form <- formula(y1~x1)) y1 ~ x1 > lm(form, anscombe) Call:

moments::anscombe.test(x) does give errors when x has too few values or if all the values in x are the same > moments::anscombe.test(c(1,2,6)) Error in if (pval > 1) pval <- 2 - pval : missing value where TRUE/FALSE needed > moments::anscombe.test(c(2,2,2,2,2,2,2,2)) Error in if (pval > 1) pval <- 2 - pval : missing value where TRUE/FALSE needed You can use tryCatch() to

Hi all, While running the anscombe.test in R, I'm getting an error of *Error in if (pval > 1) pval <- 2 - pval : missing value where TRUE/FALSE needed* for a few time series columns whereas for most of the series the function is working fine. I have checked for those specific columns for missing values. However, there is no NA/NAN value in the dataset. I have also run kurtosis for

Hi Bill, Thanks for prompt reply and letting me know a way around it. I have more than 1200 observations and not all the values are the same. However, my data points are quite similar, for example, 0.079275, 0.078867, 0.070716 in millions and etc. I have run the data without converting it to millions and I still get the same error message. As I have kurtosis value, it should be fine for the

I have an array with 44800 columns and 24 rows I would like to compute the row average for the array 100 columns at a time, so I would like to end up with an array of 24 rows x 448 columns. I have tried using apply(dataset, 1, function(x) mean(x[])), but I am not sure how to get it to take the average 100 columns at a time. Any ideas would be welcomed. thanks, Spencer [[alternative HTML

Hi, How to mark points on x axis of a graph keeping x axis as constant and changing y from y1 to y2 respectively. I want to highlight the area from y1 to y2. Any suggestions Thank you Jeet [[alternative HTML version deleted]]

I don't have much experience with r. What I am trying to do is to turn regression coefficients (after I run a lm or glm model) into some matrix such that I can do some post-estimation calculation, for example predicted probabilities in glm model, etc.. Or, is there any function in r that I can use to do something along that line? thanks. Jun Xu, Ph.D. Department of Sociology Ball State

useRs, I have a data frame where four of the columns of the data frame represent the values of a two-by-two matrix. I'd like to, row-by-row, go through the data frame and use the four columns, in matrix form, to perform calculations necessary to create new values for variables in the data frame. My first idea was to use apply: apply(as.array(data.frame[,1:4]), 1, matrix, nrow=2) Though

I have a question about the D'Agostino skewness test and the Anscombe-Glynn kurtosis test. agostino.test(x, alternative = c("two.sided", "less", "greater")) anscombe.test(x, alternative = c("two.sided", "less", "greater")) The option "alternative" in those two functions seems to be the null hypothesis. In the output, the

Hi All, I'm using the subset function to select a list of variables, some of which are contiguous in the data frame, and others of which are not. It works fine when I use the form: subset(mydata,select=c(x1,x3:x5,x7) ) In reality, my list is far more complex. So I would like to store it in a variable to substitute in for c(x1,x3:x5,x7) but cannot get it to work. That use of the c function

Another bad case is > moments::anscombe.test(rep(c(1,1.1),length=35)) Error in if (pval > 1) pval <- 2 - pval : missing value where TRUE/FALSE needed I haven't checked the formulas carefully, but I suspect the problem is from taking the cube root of a negative number in z <- (1 - 2/(9 * a) - ((1 - 2/a)/(1 + xx * sqrt(2/(a - 4))))^(1/3))/sqrt(2/(9 * a)) In R, the

Hi, I would like to create a plot of y1,y2,y3,y4 against x for several subjects such that y1 and y2 are plotted against x in one panel and y3 and y4 against x in another panel. Thus if there are 3 subjects I should end up with 6 panels. Is there a simple way of doing so (i.e. without calling xyplot() several times, and then padding the results together)?? Regards S?ren