similar to: rfm working in localhost but not in production enviromen

I would like to compute the MC test (rfm.test) available in the package MarkedPointProcess (for the data BITOEK for example) in order to test the dependence between the marks and their locations. Why the syntax of rfm.test is false here? I have the message : ****************************** ML WARNING! Forbidden values! -- if there are too many warnings try narrower lower and upper bounds for the

Hi Hemant, Let's take it one step at a time. Save this code as "qdrfm.R" in your R working directory: It includes the comments I added last time and fixes a bug in the recency scoring. qdrfm<-function(x,rbreaks=3,fbreaks=3,mbreaks=3, date.format="%Y-%m-%d",weights=c(1,1,1),finish=NA) { # if no finish date is specified, use current date if(is.na(finish))

I'm trying to perform a RFM analysis on attached dataset, i'm able to get the results using the auto_rfm function but i want to define my own breaks for RFM, when i tried to define my own breaks i got the identical result i.e 111 for every ID. please help me with this with working R script. Thanks hemantsain.com

hello, I'm confused what you guys are talking about. i just want to set ideal threshold values for my RFM scores which can be done using Quantiles but i don't want to use quantiles because my data is not normally distributed so it will lead to wrong ranges of breaks. to fix this problem I'm looking for an approach which can define the ideal range to breaks to categorize RFM scores into

Hemant's problem is that the indicators are not distributed uniformly. With a uniform distribution, categorization gives a reasonably optimal separation of cases. One approach would be to drop categorization and calculate the overall score as the mean of the standardized indicator scores. Whether this is an option I do not know. I did offer an "eyeball" set of breaks in a previous

Hi Hemant, As I suspected, the code broke when I got to the line: result <- rfm_auto(df, id="user_id", payment ="subtotal_amount", date="created_at") Error in rfm_auto(df, id = "user_id", payment = "subtotal_amount", date = "cr eated_at") : could not find function "rfm_auto" It looks like you are using the hoxo-m/easyRFM

I'm trying to perform an RFM analysis on the attached dataset, I'm able to get the results using the auto_rfm function but i want to define my own breaks for RFM. as follow r <-c(30,60,90) f <-c(2,5,8) m <-c(10,20,30) but when i tried to define my own breaks i got the identical result for RFM i.e 111 for every ID. please help me with this with working R script so that i can get

> On Oct 13, 2017, at 2:51 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > You expect us to solve your problem but you ignore advice already recieved. > > Your data are unreadable, use dput(yourdata) instead. see ?dput > >> test<-read.table("clipboard", heade=T) > Error in scan(file = file, what = what, sep = sep, quote = quote,

One possibility is to look at predict.nnet(), and - Write an R function that write out parts of an nnet object that are needed by predict.nnet() to an external file. - Re-write predict.nnet() in C, reading the model information from the external file. Obviously you'll also need the C source for the code that predict.nnet() calls, and modify those as needed to strip out dependency on R, if

Hi You expect us to solve your problem but you ignore advice already recieved. Your data are unreadable, use dput(yourdata) instead. see ?dput > test<-read.table("clipboard", heade=T) Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec, : line 115 did not have 6 elements What is ?ideal interval? can you define it? Should it be such to provide eqal

Six Fried Rice Contact: Geoff Coffey Phone: 602-384-1804 Email: geoff-5v4lRIDW0s4umAx9uhQu3g@public.gmane.org Ruby On Rails, Today''s Hottest Web Framework, Now Supports FileMaker Server Six Fried Rice, a FileMaker Pro Services Provider, Releases Rfm, a Free Open Source FileMaker API for Ruby and Ruby on Rails. The buzz about Ruby on Rails can be heard all over the web. This world

I have a matrix of, say, M and N dimensions: my_matrix=matrix(c(1:60),nrow=6,ncol=10) I have two "id" vectors corresponding to the rows and columns, e.g.: id_m=seq(10,60,by=10) id_n=seq(100,1000,by=100) I would like to create a "proper" database (let's say a data.frame for this example -- i'm going to be loading these into an SQLite database, but we'll leave

Hi Your statement about attaching data is problematic. We cannot do much with it. Instead use output from dput(yourdata) to show us what exactly your data look like. We also do not know how do you want to split your data. It would be nice if you can show also what should be the bins with respective data. Unless you provide this information you probably would not get any sensible answer. Cheers

Dear wise ones, I have a problem assigning different colors to bars in a barplot. The data I'm using is the following dataframe (truncated) : > L0 r n p t [...] 18 19 1 1 RFM 19 20 1 1 RFM 20 21 2 1 RFM 21 23 6 1 RIH 22 24 2 1 ROC 23 25 1 1 ROC 24 26 1 1 ROC 25 27 2 1 ROC 26 28 2 1 RFT 27 29 1 1 RFT 28 30 2 1 RFT 29 31 1 1 ROH [...] My barplot should

I'm getting all the rows as NA in Cscore and almost most of the observation in R and F and M are also NA. what can be the reason for this. also suggest me the appropriate solution. On 9 October 2017 at 15:51, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Hemant, > Here is an example that might answer your questions. Please don't run > previous code as it might not work.

At a snail's pace I keep on translating an introduction to R into italian; I have reached the section describing the glm() function, in which some example code is presented. The very last line of code, before the beginning of the section on Poisson models is: ldp <- ld50(coef(fmp)); ldl <- ld50(coef(fmp)); c(ldp, ldl) which of course gives results 43.663 and 43.663; the correct code

At a snail's pace I keep on translating an introduction to R into italian; I have reached the section describing the glm() function, in which some example code is presented. The very last line of code, before the beginning of the section on Poisson models is: ldp <- ld50(coef(fmp)); ldl <- ld50(coef(fmp)); c(ldp, ldl) which of course gives results 43.663 and 43.663; the correct code

I have always avoided missing data by keeping my distance from the real world. But I have a student who is doing a study of real patients. We're trying to test regression models using multiple imputation. We did the following (roughly): f <- aregImpute(~ [list of 32 variables, separated by + signs], n.impute=20, defaultLinear=T, data=t1) # I read that 20 is better than the default of

Can someone please explain these smb.log entries??? [2003/08/26 18:58:31, 0] smbd/service.c:make_connection(381) fmouse logged in as admin user (root privileges) [2003/08/26 18:58:35, 0] lib/util_sock.c:read_socket_with_timeout(298) read_socket_with_timeout: timeout read. read error = Connection reset by peer. [2003/08/26 18:58:35, 0] smbd/oplock.c:oplock_break(758) oplock_break:

I just updated my web server box, running Debian testing, to samba v3, and am having some problems with a share. Can anyone advise me on this? I have a share called "allsrvr" which is defined in my smb.conf file as follows. Please note that security isn't an issue here. The server is connected to my home system via a VPN and all smb/cifs ports are closed to the world at the