similar to: Help RFM analysis in R (i want a code where i can define my own breaks instead of system defined breaks used in auto_RFM package)

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

Hi Hemant, Here is an example that might answer your questions. Please don't run previous code as it might not work. I define the break values as arguments to the function (rbreaks,fbreaks,mbreaks) If you want the breaks to work, make sure that they cover the range of the input values, otherwise you get NAs. # expects a three (or more) column data frame where # column 1 is customer ID,

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.

I seriously doubt that you are running the code I sent. What you have probably done is to run your data, which has a different date format, without changing the breaks or the date format arguments. As you haven't provided any example that shows what you are doing, I can't guess what the problem is. Jim On Mon, Oct 9, 2017 at 9:40 PM, Hemant Sain <hemantsain55 at gmail.com> wrote:

Hello Jim, i have converted all my variable data type according to your attached example including date, and my dataset looks like this. ID purchase date 1234 10.2 2017-02-18 3453 18.9 2017-03-22 7689 8 2017-03-24 but when I'm passing the data

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

Hey, i want to define 3 ideal breaks (bin) for each variable one of those variables is attached in the previous email, i don't want to consider quartile method because quartile is not working ideally for that data set because data distribution is non normal. so i want you to suggest another method so that i can define 3 breaks with the ideal interval for Recency, frequency and monetary to

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))

Full_Name: Heather Turner Version: 2.2.0 OS: Windows XP Submission from: (NULL) (137.205.240.44) Standardized residuals as calculated by rstandard.lm, rstandard.glm and plot.lm are Inf/NaN rather than zero when the un-standardized residuals are zero. This causes plot.lm to break when calculating 'ylim' for any of the plots of standardized residuals. Example:

Dear all, can somebody give me some pointer how I can fit a "linear-by-linear association model" (i.e. loglinear model for the ordinal variables) in R? A brief description can be found here 'https://onlinecourses.science.psu.edu/stat504/node/141'. Thanks for your help

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

Hello, I'm working on RFM analysis and i wanted to define my own breaks but my frequency distribution is not normally distributed so when I'm using quartile its not giving the optimal results. so I'm looking for a better approach where i can define breaks dynamically because after visualization i can do it easily but i want to apply this model so that it can automatically define the

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

Hi guys, I want to transfer the following code from R into .Call compatible form. How can i do that? Thanks!!! INT sim; for(i in 1:sim){ if(i>2) genemat <- genemat[,sample(1:ncol(genemat))] ranklist[,1] <- apply(genemat, 1, function(x){ (mean(x[cols]) - mean(x[-cols]))/sd(x)}) ranklist <- ranklist[order(ranklist[,1]),]

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

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

> 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,

I tried toy problems and there doesn't seem to be a basic problem between lm() and ts objects: X = data.frame(x=c(1,2,7,9), y=c(7,2,3,1)) lm(y ~ x, X) X <- lapply(X, function(x) ts(x, frequency=12, start=c(1994,7))) lm(y ~ x, X) and this works fine - whether you do an lm() before or after making ts objects, it's okay. But I have a situation where things aren't okay.

In their paper, "Lexical Scope and Statistical Computing", the authors ( Gentleman and Ihaka ) go to great length explaining why R's use of lexical scoping creates advantages when doing statistical computations. If anyone has or is familiar with this paper, could they provide the main program code for how the "newton" function would be called in their example on page 500

Dear List, How do I extract the approximate Wald test for the frailty (in the following example 17.89 value)? What about the P-values, other Chisq, DF, se(coef) and se2? How can they be extracted? ######################################################> kfitm1 Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id, dist = "gauss"), data = kidney)