![\"sample\"](\"http:\/\/jkthinks.synology.me\/wp-content\/uploads\/2015\/02\/sample.jpg\")
<\/a><\/p>\nI will use the sample data, which consists of several variables representing recency, frequency, and monetary. After reading and sorting data, classify customers based on quintile. Actually, there are three approaches to classify customers: sequential, independent, and intuitive. I will use the independent approach, which classifies customers with no consideration of the relation or importance of variables. <\/strong><\/strong><\/p>\n After getting codified segments, you need to get the highly responsive groups who will account for better profitability. To that end, it is required to find a breakeven point, which will divide sample groups into two segments: a target group and a non-target group. The breakeven point is the response rate that makes profits from marketing activities equivalent to costs for those activities.<\/p>\n Number of target customers * Breakeven point * Profit – Number of target customers * Marketing cost = 0<\/strong><\/p>\n From the above formula, the breakeven point is obtained by dividing costs with profits. We need to apply this response rate to R, resulting in the target customers who have higher than the rate. Before this process, you may need to check how customer groups account for response rates. Higher number of groups show higher response rate from marketing activities in a bar chart. i.e. the group of 555 has higher possibility to buy some stuff than the group of 111.<\/p>\n# set a working directory\r\nsetwd(\"D:\/Rstudio\/R files\/Project\/RFM\")\r\n\r\n# read a data file from the working directory\r\nd1 <- read.csv(\"sample.csv\")\r\nhead(d1)\r\n\r\n# make data frame with relevant columns\r\nd2 <- as.data.frame(cbind(d1[,1],d1[,4],d1[,2],d1[,3],d1[,5],d1[,6]))\r\n\r\n# change the column names\r\nnames <- c(\"ID\", \"Recency\", \"Frequency\", \"Monetary\", \"Buy\", \"avgExpense\")\r\nnames(d2) <- names\r\nhead(d2)\r\ndim(d2)\r\n\r\n# 5 quantile for recency\r\nRq <- quantile(d2$Recency, probs = seq(0, 1, 0.2), na.rm = FALSE, names = TRUE)\r\nd2$R_Score[d2$Recency >= Rq[5]] <- \"1\"\r\nd2$R_Score[d2$Recency < Rq[5] & d2$Recency >= Rq[4]] <- \"2\"\r\nd2$R_Score[d2$Recency < Rq[4] & d2$Recency >= Rq[3]] <- \"3\"\r\nd2$R_Score[d2$Recency < Rq[3] & d2$Recency >= Rq[2]] <- \"4\"\r\nd2$R_Score[d2$Recency < Rq[2]] <- \"5\"\r\n\r\n# 5 quantile for frequency\r\nFq <- quantile(d2$Frequency, probs = seq(0, 1, 0.2), na.rm = FALSE, names = TRUE)\r\nd2$F_Score[d2$Frequency >= Fq[5]] <- \"5\"\r\nd2$F_Score[d2$Frequency < Fq[5] & d2$Frequency >= Fq[4]] <- \"4\"\r\nd2$F_Score[d2$Frequency < Fq[4] & d2$Frequency >= Fq[3]] <- \"3\"\r\nd2$F_Score[d2$Frequency < Fq[3] & d2$Frequency >= Fq[2]] <- \"2\"\r\nd2$F_Score[d2$Frequency < Fq[2]] <- \"1\"\r\n\r\n# 5 quantile for monetary\r\nMq <- quantile(d2$Monetary, probs = seq(0, 1, 0.2), na.rm = FALSE, names = TRUE)\r\nd2$M_Score[d2$Monetary >= Mq[5]] <- \"5\"\r\nd2$M_Score[d2$Monetary < Mq[5] & d2$Monetary >= Mq[4]] <- \"4\"\r\nd2$M_Score[d2$Monetary < Mq[4] & d2$Monetary >= Mq[3]] <- \"3\"\r\nd2$M_Score[d2$Monetary < Mq[3] & d2$Monetary >= Mq[2]] <- \"2\"\r\nd2$M_Score[d2$Monetary < Mq[2]] <- \"1\"\r\n\r\n# convert character to numeric\r\nd2$R_Score <- as.numeric(d2$R_Score)\r\nd2$F_Score <- as.numeric(d2$F_Score)\r\nd2$M_Score <- as.numeric(d2$M_Score)\r\n\r\n# calculate the total score\r\nTotal_Score <- c(100*d2$R_Score + 10*d2$F_Score+d2$M_Score)\r\nd3 <- cbind(d2,Total_Score)\r\nhead(d3)\r\n<\/pre>\n