similar to: Nested loop R code

Hello Ulrik, Can you please explain this code means how and what this code is doing because I'm not able to understand it, if you can explain it i can use it in future by doing some Lil bit manipulation. Thanks data_help <- data_help %>% mutate(Purchase_ID = 1:n()) %>% group_by(Purchase_ID) %>% do(split_items(.)) cat_help %>% gather("Foo",

Hi Hemant, the solution is really quite similar, and the logic is identical: library(readr) library(dplyr) library(stringr) library(tidyr) data_help <- read_csv("data_help.csv") cat_help <- read_csv("cat_help.csv") # Helper function to split the Items and create a data_frame split_items <- function(items){ x <- items$Items_purchased_on_Receipts %>%

i have a vector containing values ranging from 0 to 24 i want to create another variable which can categorize those values like this please help me with an R code Thanks *Value New_Var*1 0 -5 3 0 -5 5 0 -5 9 6-10 7 6-10 5 6-10 4 0-5 11 11-15 12 11-15 18 16-20 23 21 -25 --

by using these two tables we have to create third table in this format where categories will be on the top and transaction will be in the rows, On 30 August 2017 at 16:42, Hemant Sain <hemantsain55 at gmail.com> wrote: > Hello Ulrik, > Can you please once check this code again on the following data set > because it doesn't giving same output to me due to absence of quantity,a

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 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))

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

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:

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

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

> 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,

Hey PIKAL, It's not a homework neithe that is the real dataset i have signer NDA for my company so that i can share the original data file, Actually I'm working on a market basket analysis task but not able to convert my existing data table to appropriate format so that i can apply Apriori algorithm using R, and this is very important me to get it done because I'm an intern and if i

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 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

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.

Hi It seems to me like homework, there is no homework policy on this help list. What do you want to do with your table 3? It seems to me futile. Anyway, some combination of melt, merge, cast and regular expressions could be employed in such task, but it could be rather tricky. But be aware that Suger does not match sugar (I wonder that sugar is dairy product) and you mix uppercase and

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,

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

Thanks, I think I've found the most succinct expression of differences in two data.frames... length(which( rowSums( x1 != x2 ) > 0)) gives a count of the # of records in two data.frames that do not match. // ________________________________________ From: Henrik Bengtsson [henrik.bengtsson at gmail.com] Sent: Sunday, January 28, 2018 11:12 AM To: Ulrik Stervbo Cc: Marsh Hardy ARA/RISK;

The diffobj package (https://cran.r-project.org/package=diffobj) is really helpful here. It provides "diff" functions diffPrint(), diffStr(), and diffChr() to compare two object 'x' and 'y' and provide neat colorized summary output. Example: > iris2 <- iris > iris2[122:125,4] <- iris2[122:125,4] + 0.1 > diffobj::diffPrint(iris2, iris) < iris2 >