In this post, I conduct a simple risk analysis of employee turnover using the Human Resources Analytics data set from Kaggle.

I describe this analysis as an example of simple risk segmenting because I would like to have a general idea of which combination of employee characteristics can provide evidence towards higher employee turnover.

To accomplish this, I developed a function in R that will take a data frame and two characteristics of interest in order to generate a matrix whose entries represent the probability of employee turnover given the two characteristics. I call these values, turnover rates.

## Human Resources Analytics Data

Firstly, let us go over the details of the human resources analytics data set.

hr_data <- read.csv("HR_comma_sep.csv", header = TRUE) str(hr_data)

The variables are described as follows:

**satisfaction_level**represents the employee’s level of satisfaction on a 0 – 100% scale**last_evaluation**represents the employee’s numeric score on their last evaluation**number_project**is the number of projects accomplished by an employee to date**average_montly_hours**is the average monthly hours an employee spends at work**time_spend_company**is the amount of years an employee worked at this company**work_accident**is a binary variable where 1 the employee experienced an accident, and 0 otherwise**left**variable represents the binary class where 1 means the employee left, and 0 otherwise.**promotion_last_5years**is a binary variable where 1 means the employee was promoted in the last 5 years, and 0 otherwise**sales**is a categorical variable representing the employee’s main job function**salary**is a categorical variable representing an employee’s salary level

## The Rate Function

The following R code presents the function used to conduct this analysis.

# To use rate_matrix, a data frame df must be supplied and two column names from df must be known. The data frame must contain a numeric binary class feature y. # If any of the characteristics are numeric on a continuous scale, a cut must be specified to place the values into categorical ranges or buckets. rate_matrix <- function(df, y, c1 = NA, c2 = NA, cut = 10, avg = TRUE) { # If y is not a binary integer, then stop the function. if (is.integer(df[[y]]) != TRUE) { stop("Please ensure y is a binary class integer.") } df_col_names <- colnames(df) # If c1 and c2 are not available if (is.na(c1) & is.na(c2)) { stop("Please recall function with a c1 and/or c2 value.") } # If only c1 is provided else if (is.na(c2)) { if (is.integer(df[[c1]])) { var1 <- as.character(df[[c1]]) var1 <- unique(var1) var1 <- as.numeric(var1) var1 <- sort(var1, decreasing = FALSE) } else if (is.numeric(df[[c1]])) { var1 <- cut(df[[c1]], cut) df[[c1]] <- var1 var1 <- levels(var1) } else { var1 <- df[[c1]] var1 <- as.character(var1) var1 <- unique(var1) var1 <- sort(var1, decreasing = FALSE) } c1_pos <- which(df_col_names == c1) # Number of column of characteristic c1 var1_len <- length(var1) m <- matrix(NA, nrow = var1_len, ncol = 1) rownames(m) <- var1 colnames(m) <- c1 for (i in 1:var1_len) { bad <- df[,1][which(df[,c1_pos] == var1[i] & df[[y]] == 1)] bad_count <- length(bad) good <- df[,1][which(df[,c1_pos] == var1[i] & df[[y]] == 0)] good_count <- length(good) m[i,1] <- round(bad_count / (bad_count + good_count), 2) } } # If c1 and c2 are provided else { if (is.integer(df[[c1]])) { var1 <- as.character(df[[c1]]) var1 <- unique(var1) var1 <- as.numeric(var1) var1 <- sort(var1, decreasing = FALSE) } else if (is.numeric(df[[c1]])) { var1 <- cut(df[[c1]], cut) df[[c1]] <- var1 var1 <- levels(var1) } else { var1 <- df[[c1]] var1 <- as.character(var1) var1 <- unique(var1) var1 <- sort(var1, decreasing = FALSE) } if (is.integer(df[[c2]])) { var2 <- as.character(df[[c2]]) var2 <- unique(var2) var2 <- as.numeric(var2) var2 <- sort(var2, decreasing = FALSE) } else if (is.numeric(df[[c2]])) { var2 <- cut(df[[c2]], cut) df[[c2]] <- var2 var2 <- levels(var2) } else { var2 <- df[[c2]] var2 <- as.character(var2) var2 <- unique(var2) var2 <- sort(var2, decreasing = FALSE) } c1_pos <- which(df_col_names == c1) # Number of column of characteristic c1 c2_pos <- which(df_col_names == c2) # Number of column of characteristic c2 var1_len <- length(var1) var2_len <- length(var2) m <- matrix(NA, nrow = var1_len, ncol = var2_len) rownames(m) <- var1 colnames(m) <- var2 class_1 <- max(df[[y]]) class_0 <- min(df[[y]]) for (i in 1:var1_len) { for (j in 1:var2_len) { bad <- df[,1][which(df[,c1_pos] == var1[i] & df[,c2_pos] == var2[j] & df[[y]] == class_1)] bad_count <- length(bad) good <- df[,1][which(df[,c1_pos] == var1[i] & df[,c2_pos] == var2[j] & df[[y]] == class_0)] good_count <- length(good) m[i,j] <- round(bad_count / (bad_count + good_count), 2) } } } # Create class 1 matrix report that includes averages if (avg == TRUE) { ColumnAverage <- apply(m, 2, mean, na.rm = TRUE) ColumnAverage <- round(ColumnAverage, 2) RowAverage <- apply(m, 1, mean, na.rm = TRUE) RowAverage <- round(RowAverage, 2) RowAverage <- c(RowAverage, NA) m <- rbind(m, ColumnAverage) m <- cbind(m, RowAverage) return(m) } else { return(m) } }

## Employee Turnover Data Investigation

To begin this data investigation, I use the assumption that I have gained significant amounts of experience and field knowledge within Human Resources. I begin this heuristic analysis with the thought that employee turnover is greatly affected by how an employee feels about their job and about the company.

### Are employees with small satisfaction levels more likely to leave?

The first thing I would like to confirm is that employees with small satisfaction levels are more likely to leave.

satisfaction <- rate_matrix(df = hr_data, y = "left", c1 = "satisfaction_level", cut = 20, avg = TRUE) View(satisfaction)

The function call here uses a cut value of 20 with no particular reason. I want a large enough cut value to provide evidence of my claim.

As seen in the matrix, satisfaction levels between 0.0891 and 0.136 shows that 92% of employees categorized in this range will leave. This provides evidence that low satisfaction levels among employees are at highest risk of leaving the company.

As we would expect, the highest levels of satisfaction of 0.954 to 1 experience 0% employee turnover.

For simplicity and ease of understanding, I define 0.5 as the average satisfaction level. By taking a look at below average satisfaction levels between 0.363 to 0.408 and 0.408 to 0.454, there is an odd significant increase to the risk of employees leaving. This particular area of employee satisfaction requires more investigation because it goes against intuition.

### Are employees with below average satisfaction levels more likely to leave across different job functions?

To alleviate this concern of odd satisfaction levels defying our intuition, I continue the investigation by seeing whether satisfaction levels vary across other characteristics from the data. It is likely possible that these below average satisfaction levels are tied to their job function.

satisfaction_salary <- rate_matrix(df = hr_data, y = "left", c1 = "satisfaction_level", c2 = "sales", cut = 20, avg = TRUE) View(satisfaction_salary)

Here, the same ranges of 0.363 to 0.408 and 0.408 to 0.454 satisfaction levels are generally at high risk to leave even across all job functions. There is evidence to suggest that somewhat unhappy workers are willing to leave regardless of their job function.

### Is an unhappy employee’s likelihood of leaving related to average monthly hours worked?

To continue answering why below average satisfaction levels ranges experience higher employee turnover than we expect, I take a look at the relationship between satisfaction levels and average monthly hours worked. It could be that below average satisfaction levels at this company are tied to employees being overworked.

# First, convert the integer variable average_montly_hours into a numeric variable to take advantage of the function's ability to breakdown numeric variables into ranges. average_montly_hours <- hr_data["average_montly_hours"] average_montly_hours <- unlist(average_montly_hours) average_montly_hours <- as.numeric(average_montly_hours) hr_data["average_montly_hours"] <- average_montly_hours satisfaction_avghours <- rate_matrix(df = hr_data, y = "left", c1 = "satisfaction_level", c2 = "average_montly_hours", cut = 20, avg = TRUE) View(satisfaction_avghours)

To reiterate, the row ranges represent the satisfaction levels and the column ranges represent the average monthly hours worked. Here, there is strong evidence to suggest that employees within the below average satisfaction level range of 0.363 to 0.408 and 0.408 to 0.454 work between 117 to 160 hours a month.

Using domain knowledge, typically, a full-time employee will work at least 160 hours a month, given that a full-time position merits 40 hours a week for 4 weeks in any given month. The data suggests here that we have a higher probability of workers leaving given they work less than a regular full-time employee! This was different from my initial train of thought that the employees were potentially overworked.

Given this finding, I come to one particular conclusion: **employees with highest risk of leaving are those that are on contract, seasonal employees, or are part-time employees**.

By considering other variables such as the number of projects worked on by an employee, it is possible to further support this conclusion.

satisfaction_projects <- rate_matrix(df = hr_data, y = "left", c1 = "satisfaction_level", c2 = "number_project", cut = 20, avg = TRUE) View(satisfaction_projects)

Here, it is evident to see that the below average satisfaction levels of 0.363 to 0.408 and 0.408 to 0.454 may in fact correspond to contract or part-time employees as the probability of turnover sharply decreases after 2 projects completed.

### Are contract, part-time or seasonal employees more likely to be unhappy if the job is accident-prone?

Now that we identified the high risk groups of employee turnover within this data set, this question comes to mind because we would like to address the fact that an employee’s enjoyment in their role should be tied to their satisfaction levels. It could be that these part-time employees are experiencing hardships during their time at work, thereby contributing to their risk of leaving.

To answer this question, I take a look at the satisfaction level and number of projects completed given that an employee experienced a workplace accident.

# I use the package dplyr in order to filter the hr_data dataframe to only include observations that experienced a workplace accident require(dplyr) accident_obs <- filter(hr_data, Work_accident == 1) satisfaction_accident <- rate_matrix(df = accident_obs, y = "left", c1 = "satisfaction_level", c2 = "number_project", cut = 20, avg = TRUE) View(satisfaction_accident)

Here, given the below average satisfaction levels of 0.363 to 0.408 and 0.408 to 0.454 for number of projects equal to 2 and given that employees experienced a workplace accident, there is evidence to suggest that there is a higher chance of turnover.

## Further Work

The purpose of this analysis was to apply a risk segmenting method on human resources analytics data to identify potential reasons for employee turnover. I used probabilities or turnover rates to help identify some groups of employees that were at risk of leaving the company.

I found that there were higher chances of turnover given the employee had an extremely low satisfaction level, but also discovered that the type of employee (contract, part-time, seasonal) could be identified as groups of high risk of turnover. I addressed a possible fact that the likelihood of unhappiness for part-time employees was attributed to them working on jobs that were accident-prone.

With the example presented in this post, Human Resources can use this information to put more efforts into ensuring contract, part-time, or seasonal employees experience lower turnover rates. This analysis allowed us to identify which groups of employees are at risk and allowed us to identify potential causes.

This risk analysis approach can be applied to any other field of practice other than Human Resources, including Health and Finance. It is useful to be able to come up with quick generic risk segments within your population so that further risk management solutions can be implemented for specific problems at hand.

Lastly, this post only provides a simple way to segment and analyze risk groups but it is not the only way! More advanced methods such as clustering and decision trees can help identify risk groups more thoroughly and informatively to provide an even bigger picture. For quick checks to domain expertise in any particular field of practice, the rate function I present here can be sufficient enough in identifying risk groups.