similar to: crosstabs and histograms with flexible binning of dates

Hi R-users, This is a generic question, is there a way to plot a line graph for the output from crosstable function? one of the inputs to the crosstab function is categorical. Taby -- [[alternative HTML version deleted]]

##### BEGIN CODE ###### #!/usr/bin/perl ###### # # --start, -s = The date you would like to start generating regressors #--end, -e = When to stop generating holiday regressros # --scope, -c = D, W for Daily or Weekly respectively (e.g. Does this week have a particular holiday) # --file, -f = Ummm where to write the output silly! # # **NOTE** The EOM holiday is "End of Month" for

Here is the correct version. The old version is the redirect only version of the script. ### BEGIN SCRIPT #### #!/usr/bin/perl ###### # --start, -s = The date you would like to start generating regressors #--end, -e = When to stop generating holiday regressros # --scope, -c = D, W for Daily or Weekly respectively (e.g. Does this week have a particular holiday) # --file, -f = Ummm where to write

Hello all,, Im looking for a simple function to produce a crosstab from a dumped sql query result. Its very hard to produce crosstabs with most databases (Access being the exception), so with the vast array of R packages, Im sure this has to have already been implemented somewhere. Examples are always good: Take a csv dump like name code user1 100 user2 100 user1 200 user2 210 user1 300 user2

Hello, I am wondering if it is possible, or what the correct way to code a three-way crosstab in R using the survey package? I have been using the following code to complete two way crosstabs, but have not seen any three-way code. Two-Way: svyby(~factor(a), ~factor(b), data, svymean) Thanks! Rachel [[alternative HTML version deleted]]

Suppose you have a dataframe with two factors, but not all the factor levels appear together. For instance: stuff<- structure(list(fac1 = structure(c(2, 1, 2, 2), .Label = c("down", "up"), class = "factor"), fac2 = structure(c(1, 2, 1, 2), .Label = c("left", "right"), class = "factor"), x = c(1, 1, 2, 2), y = c(1, 2, 2, 1)),

Hi, I'm a new R user. I noticed that RKWard has a really nice output when it does crosstabs. It even includes the chi-squared values. However, I obviously can't simply copy this into my standalone R script. Does anyone know the code or a way for me to view the code. Secondly, if I am writing an R script I want to be able to view the output as a pdf even if my function is not a graph,

Is there a crosstabs function like the one found in S-Plus available in any of the R addon packages? I can't seem to locate anything other than the chisq.test, which is adequate but does not yield the output customarily required with any journal submission (in my field). (S-Plus, SPSS, SAS all provide examples of the required output). If there isn't such a function already

Hi all! What is the equivalent R function of SPLUS crosstabs? Thank you. Danar. Stat. Inst. Umea Univ. Umea, Sweden -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To:

Hello there, I'm quite new with R. Does anybody know a true variant in R of the Splus function "crosstabs" to create contingency tables, with the possibility of assigning a weight variable? Thanx Tim -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

Is there an equivalent in R to the crosstabs command S-Plus? I've been searching for a while and can't seem to find anything. Thanks. Mark J. Lamias Department of Statistics Department of Political Science Survey Methodology Program/Survey Research Center Institute for Social Research - University of Michigan 426 Thompson Street, Room 315 Ann Arbor, Michigan 48104-2321 (734) 647-5381

Dear Users, I'm a complete novice to R. I need to do a crosstabs in R, but my data is almost completely alphanumeric (with some variables scaled). The Table routine does not seem to accept alphanumeric data. What should I do? Do I need to recode it? How should I do that? Thanks in advance, Wilfred

[Delayed and reposted by list maintainer due to not passing through spam filter.. MM] Hello, I am looking for two R-functions that could generate an Output like SPSS. 1. An equivalent function for (SPSS) FREQUENCIES which generate something like this: Valid Cum Value Frequency Percent Percent Percent 1 1 5.3 5.6 5.6

I've been playing around with various table tools, trying to construct a fairly simple cross-tab. It shouldn't be hard, but for some reason it turning out to be (for me). If I want to see how many men and how many women agree with a agree/disagree question (coded 1,0), I can do this: >attach(mydata) >mytable <- table(male, q1.bin) # gender and a binary response variable

Dear list, How can I calculate the difference in days between the eventdate and basedate in the below dataset? id basedate outcome.3 eventdate daydiff 1 1001 1999-09-28 2 1999-10-01 3 2 1002 1999-09-22 1 3 1003 2000-01-19 1 4 1004 2004-01-25 2 2004-02-03 9 5 1005 2005-08-11 1 6 1006 2000-07-04

Dear list. New to R, I'm looking for a way of using crosstab to output low-dimensional (higher than 2) contingency tables (frequencies, per-cents by rows, % by columns, mean, quantiles....) I'm looking for something of the following sort dataframe: singers, categorical variates: voice category (soprano,mezzo-soprano, ...) , voice type( drammatic, spinto, lirico-spinto, lirico,

Hi, i am playing about mail archive this is what allready works for me require ["variables","date","fileinto","mailbox","envelope","subaddress","regex"]; # Extract date info if currentdate :matches "year" "*" { set "year" "${1}"; } if currentdate :matches "month" "*"

I have the following data: x <- as.factor(c(1,1,1,2,2,2,3,3,3)) y <- as.factor(c(10,10,10,20,20,20,30,30,30)) z <- c(100,100,NA,200,200,200,300,300,300) I could create the cross tab of x and y with Sum of z as its elements using the xtabs function as follows: # X Vs. Y with Sum Z xtabs(z ~ x + y) y x 10 20 30 1 200 0 0 2 0 600 0 3 0 0 900 How do I replace

Perhaps this is a common question but I haven't been able to find the answer. I have data with many factors, each taking many values. However, only relatively few combinations appear in the data, ie have nonzero counts, in other words the resulting table is sparse. Say we have 10 factors each with 10 levels. The result of table() would exceed the memory space (on a 32bit machine). Is there

Karsten, I have a function which might be too simple, but could be modified to what you requested. It is called "xtab" and appears to run okay on my version of R (0.61.2). The tar.gz file includes text coding for "xtab" and "padding" which is called. I set up empty functions and read the text code into the editor, although there is probably a better way. Hope it