similar to: Naming rows oe columns in a data frame.

Hello! I would like to add names to a table. The table presents median values, numbers of values (n) and percentage(%) for a given agegroup. Thus I would like to add a label above these three variables for each category. But if I try to use colnames I get a message telling me that the number of labels must be equal to the number of columns in the table. I hope someone could understand what I mean!

Hello! I've generated multiple data frames that I wish to export to excel using the function write.table. When I do so all the data is merged into a single column in excel. I Would like the columns of my data frame in separate columns in excel as well but it doesn't work. I tried to download the package WriteXLS and installed Perl on my computer just like it was recommended. I also tried

With R 1.8.1 running in Fedora Core 1 Linux, I am having some trouble recoding and ploting some factor variables. First, can I give you some example data? Here is a column of names for age groups: agegroups <- c( "15-19", "20-24", "25-29","30-34", "35-39",

sorry - hope this question is not tooo silly, but i needed to "autologin" to a remote machine found out that this works fine for me: sftp -opassword=PASSWORD USER at 192.168.1.1 << EOF cd ANYDIRECTORY get ANYFILE bye EOF why isn't this (sftp -opassword=SECRET USER at 192.168.1.1), setting the password with -opassword=PASSWORD, documentated anywhere? bug or feature? kind

I'm creating a stacked bar chart using the likert command in the HH package. My data are in a data frame, with two numeric variables and a categorical variable, I can't get likert to use the column containing the categorical variable as a my y axis label. Here is a quick example: library(HH) #my data are:

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

I am trying to import a SPSS.sav file into R. The attached file is not technically the file I am trying to import, but does replicate my problem. The actual file is much too large to attach. No matter what I do, I can not get R (base or Hmisc) to apply the value labels in the .sav file to the dataframe created in R. Here's the code that I am using. maine <- spss.get("test.sav") #

Hello. About year ago I tested OE and dovecot imap and met this problem (http://wiki.dovecot.org/Clients#head-129bcbfc238b5c34362b8216db87c05e7a34af85) - Using "Headers only" synchronization is buggy and can cause "Message is no longer available on this server" error when opening a mail. I configured OE to sync All messages, and it worked, but then it's acts like pop3 and

v 2.0.1 (sooooh old!) on Win2k I think I know the answer to this but I can hope ... I have data for continuous variables (measures of residents) by a categorical variable in range (1,22), the units in which they live. I want to plot these data with a pair of boxplots one above another with same x-axis (1,22) using par(mfrow=c(2,1)) and then plotting first for the women then for the men.

Greetings: Recently I converted my mailserver to Postfix with some local accounts being accessed via Dovecot and POP3S. The clients are OE. The users are reporting that they aren't getting their messages. In the past with my old (IW?) POP server, I just used mail -f /var/spool/user and deleted the "DO NOT DELETE THIS MESSAGE...." message #1 and this cleared any problems with

Hello. This isn't actually a dovecot question, but I'm running in clienys servers dovecot mostly as pop3 server and now started to migrate some servers to imap. But as clients uses Outlook Express how to transfer their pop3 settings and mailboxes to imap?. Making a new mail account for imap is easy, and drag & drop the default folders (inbox, outbox etc) messages to imap accounts

Is there an easy way to make graphs for the following data. I have pretest and posttest scores for men and women. I would like to form a 'titlted segment' plot for the data. That is, make segments joining the scores, with different types of segments for men and women. Example data: menpre <- c(43,42,26,39,60,60,46) menpost <- c(40,41,36,42,54,58,43) womenpre <-

Hi all, I got curious about something, so in proper scientific fashion I obtained some data and analyzed it. Question: what is the female participation in the R-help email list? Data: the most recent list postings, obtained from the website. I took my best shot at classifying the names given in the email header as male/female, but ended with a fair number of unknowns. This dataset had 2797

Didn''t know what else to call this subject, sorry. Anyway this is more a general discovery question then a coding one. Is it possible (and I already know it should be) to create forms where depending on a particular choice in one select tag, one or another select tags would be presented in the form. As an example, a shopping site , where a dropdown has options of: 1-Men''s shoes

I'm trying to obtain within-group correlations on a subset of variables. I first selected my variables using the following command: mydata$x<-mydata[c("iq","education","achievement")] I'd like to look at correlations among those variables separately for men and women. My gender variable in mydata is coded 1 (women) and 0 (men). I have successfully used

On 12/24/18 7:21 AM, vsnsdualce at memeware.net wrote: > Debian is not ruled by the men who actually write the software, but > instead women. *snip* Can we please ban the person who sent that disgusting rant to the list?

Any thoughts on the following I'd be most grateful - I'm sure there is an easy and quick way to do this but I'm having a mental block this evening. Essentially, I'm trying to replace missing data in my dataset with reference values based on age and sex. So an example dataset is set.seed(1) X = data.frame(age=rnorm(10, 50, 10), sex=rbinom(10, 1, 0.5), A=rnorm(10), B=rnorm(10))

Dear R-helpers, I'm not a speciallist in writing complex functions, and the function still very rusty (any kind of suggestions are very welcome). I want to implement Gale-Shapley algorithm for R Language. It is based on http://www.jstor.org/stable/10.2307/2312726 Gale and Shapley (1962) , and it has evolved to

Using the weight argument with a variance function in lme (nlme), you can allow for heteroscedasticity of the within-group error. Is there a way to do this for the other variance components? For example, suppose you had subjects, days nested within subjects, and visits nested within days within subjects (a fully nested two-way design) and you had, say men and women subjects. Could you allow for

Hello samba, I would like an advice on following: is there any way how to tell samba what transfer mode to use? I think I read that samba does only binary and to use text you have to use something else. We have windows machines using text files on linux samba server and i need to transfer those files in text mode. Can anyone help me please? -- Best regards, Martin