G. Games Customer LTV Model

Nir Regev
Principal Data Scientist
Sisense Ltd.

April 16, 2016

G Games Use-Case

G Games use daily customer revenues per country and fit a linear regression model to predict the life time value of each customers segmentaion (per country in this case)

Agenda

Gram Games Customer Revenue Life Time Model
Specific Implementation for input in fixed length
Generic Implementation for varying input length

Generating data for demo

args <- list(c(2, 4), c(3, 5), c(2, 3), c(4, 5), c(4, 1))
args

## [[1]]
## [1] 2 4
## 
## [[2]]
## [1] 3 5
## 
## [[3]]
## [1] 2 3
## 
## [[4]]
## [1] 4 5
## 
## [[5]]
## [1] 4 1

Defining parsing and A wrapper function

tlv <- function (args) {
  # data generation (for simulation purposes)
  #args <- list(c(2, 4), c(3, 5), c(2, 3), c(4, 5), c(4, 1))
  arg1=args[[1]];
  arg2=args[[2]];
  arg3=args[[3]];
  arg4=args[[4]];
  arg5=args[[5]];
  sisense_result <-c()
  for (i in 1:length(arg1)) {
    retention <- c(arg1[i],arg2[i],arg3[i],arg4[i],arg5[i])
    result <- tlv_model(retention)
    sisense_result <-c(sisense_result,result)
  }
  #retention <- data.frame(t(mapply( FUN = c,args)))
  #sisense_result <- lapply(X = retention, FUN = tlv_model)
  return (sisense_result)
  #sisense_result # result vector after "tlv function" calculation.
}

defining a model (linear regression) function

tlv_model <- function(res){
  retentiondf = data.frame(day = c(1,7,14,21,28), retention=res);
  modell = lm(log((retentiondf$retention), base = exp(1)) ~ log(retentiondf$day, base = exp(1)));
  a = exp(modell$coefficients[1]);
  m = modell$coefficients[2];predictedRetentiondf = data.frame(day = 1:360);
  predictedRetentiondf = data.frame(day = predictedRetentiondf$day, pRetention =    a*predictedRetentiondf$day^m);
  lt = sum(predictedRetentiondf$pRetention)*0.55;
  return(lt)
}

simulation of entire model execution

first_results <- tlv(args)
first_results

## [1] 936.4313 349.0783

simulation with different args length

args <- list(c(2, 4), c(3, 5), c(2, 3), c(4, 5), c(4, 1),c(2,7))
second_results <- tlv(args)
first_results; second_results

## [1] 936.4313 349.0783

## [1] 936.4313 349.0783

same results as before - how is that possible ? The new data point (2,7) is not being parsed (parsing is hard coded)

Make the code more generic and robust

Let’s adjust the funciton to parse vaying args length (no assumption on args length)
give up loops (slow) and use apply functions (fast)
Day vector should be dynamically created or be receieved as additional argument

tlv <- function (args) {
  
  retention <- data.frame(t(mapply( FUN = c,args)))
  sisense_result <- lapply(X = retention, FUN = tlv_model)
  return (sisense_result)
}

tlv_model <- function(res){
  retentiondf = data.frame(day = seq(1,30, by=round(30/length(args))), retention=res);
  modell = lm(log((retentiondf$retention), base = exp(1)) ~ log(retentiondf$day, base = exp(1)));
  a = exp(modell$coefficients[1]);
  m = modell$coefficients[2];predictedRetentiondf = data.frame(day = 1:360);
  predictedRetentiondf = data.frame(day = predictedRetentiondf$day, pRetention = a*predictedRetentiondf$day^m);
  lt = sum(predictedRetentiondf$pRetention)*0.55;
  return(lt)
}

Let’s run the new code

args <- list(c(2, 4), c(3, 5), c(2, 3), c(4, 5), c(4, 1), c(4,5))
tlv(args)

## $X1
## [1] 1077.17
## 
## $X2
## [1] 475.286