# Coefficient of Variation

Author

Derek Sollberger

Published

September 2, 2022

For a future lecture in my Sports Analytics course, I want an example of a baseball statistic where the averages for two players are similar, but their variances in that same statistic are quite different. It is still early in the semester, so I am looking for an easy-to-understand statistic. Therefore, I will explore home runs per season.

## A Walk Through the Data

The Lahman database contains a lot of baseball statistics, and today I will focus on the Batting data frame.

library("Lahman")
library("tidyverse")

For interest, I will filter the observations to retain the players from the past 18 seasons (since my students are about 18 years old, haha) and player-seasons that included at least 100 at-bats.

df <- Batting |>
filter(yearID >= (2021 - 18)) |>
filter(AB >= 100)

To be illustrative, permit me to select mainly the seasons and home run columns.

df <- df |>
select(playerID, yearID, AB, HR)

At the moment, here is what our data looks like.

head(df)
   playerID yearID  AB HR
1 abreubo01   2003 577 20
2 alfoned01   2003 514 13
3 almoner01   2003 100  1
4 alomaro01   2003 263  2
5 alomaro01   2003 253  3
6 alomasa02   2003 194  5

Since I am concerned with season averages, I am going to group_by the player name. From there, let us then compute the averages and standard deviations for home runs.

df <- df |>
group_by(playerID) |>
mutate(xbar = mean(HR, na.rm = TRUE),
s    = sd(HR, na.rm = TRUE)) |>
ungroup()
head(df)
# A tibble: 6 × 6
playerID  yearID    AB    HR  xbar      s
<chr>      <int> <int> <int> <dbl>  <dbl>
1 abreubo01   2003   577    20 14.3   8.94
2 alfoned01   2003   514    13  8.67  5.86
3 almoner01   2003   100     1  1    NA
4 alomaro01   2003   263     2  2.67  0.577
5 alomaro01   2003   253     3  2.67  0.577
6 alomasa02   2003   194     5  2.33  2.52 

For various reasons, some players only appear once in this subset of data, so their variance is effectively zero (missing in the computation). To answer my original inquiry mathematically, I will now compute the coefficient of variation (and avoid a divide-by-zero error from those players who hit zero home runs).

$CoV = \frac{s}{\bar{x}}$

df <- df |>
# filter(!is.na(s)) |>
filter(xbar > 0) |>
mutate(CoV = s/xbar) |>
arrange(desc(CoV))

In this metric, the top scores are

head(df)
# A tibble: 6 × 7
playerID  yearID    AB    HR  xbar     s   CoV
<chr>      <int> <int> <int> <dbl> <dbl> <dbl>
1 gathrjo01   2005   203     0  0.2  0.447  2.24
2 gathrjo01   2006   154     0  0.2  0.447  2.24
3 gathrjo01   2006   229     1  0.2  0.447  2.24
4 gathrjo01   2007   228     0  0.2  0.447  2.24
5 gathrjo01   2008   279     0  0.2  0.447  2.24
6 burriem01   2008   240     1  0.25 0.5    2   

and the bottom scores are

tail(df)
# A tibble: 6 × 7
playerID  yearID    AB    HR  xbar     s   CoV
<chr>      <int> <int> <int> <dbl> <dbl> <dbl>
1 wadela01    2021   336    18    18    NA    NA
2 wallsta01   2021   152     1     1    NA    NA
3 walshja01   2021   530    29    29    NA    NA
4 whiteel04   2021   198     6     6    NA    NA
5 williju02   2021   119     4     4    NA    NA
6 wisdopa01   2021   338    28    28    NA    NA

## Heavy Hitters

So far, these results seem to be fine mathematically, but they might be uninteresting to the casual baseball fan. To further prune down to recognizable players, let me further filter the data down to players with at least 5 of these 100+ at-bat seasons in the past 18 seasons—and had an average of over 10 home runs per season.

df <- df |>
group_by(playerID) |>
mutate(seasons = n()) |>
ungroup() |>
filter(seasons >= 5) |>
filter(xbar >= 15) |>
arrange(desc(CoV))

Now, the top scores in using the coefficient of variation as my metric are

head(df)
# A tibble: 6 × 8
playerID  yearID    AB    HR  xbar     s   CoV seasons
<chr>      <int> <int> <int> <dbl> <dbl> <dbl>   <int>
1 solerjo01   2015   366    10  16.3  14.2 0.869       7
2 solerjo01   2016   227    12  16.3  14.2 0.869       7
3 solerjo01   2018   223     9  16.3  14.2 0.869       7
4 solerjo01   2019   589    48  16.3  14.2 0.869       7
5 solerjo01   2020   149     8  16.3  14.2 0.869       7
6 solerjo01   2021   308    13  16.3  14.2 0.869       7

and the lowest $$Cov$$ are

tail(df)
# A tibble: 6 × 8
playerID  yearID    AB    HR  xbar     s   CoV seasons
<chr>      <int> <int> <int> <dbl> <dbl> <dbl>   <int>
1 delgaca01   2003   570    42  34.5  6.32 0.183       6
2 delgaca01   2004   458    32  34.5  6.32 0.183       6
3 delgaca01   2005   521    33  34.5  6.32 0.183       6
4 delgaca01   2006   524    38  34.5  6.32 0.183       6
5 delgaca01   2007   538    24  34.5  6.32 0.183       6
6 delgaca01   2008   598    38  34.5  6.32 0.183       6

Well, I do recognize more of the player names, but I realize that the $$Cov$$ alone does not completely solve my question since I still wanted similar averages between two players.

## A Metric on top of a Metric

Now, hear me out. If there are two players $$A$$ and $$B$$, then what I am looking for is a rate of change of the form

$Y = \bigg|\frac{s_{A} - s_{B}}{\bar{x}_{A} - \bar{x}_{B}}\bigg|$

First, let me simplify the data frame down to just the player names, home run averages, and their standard deviations.

df2 <- df |>
select(playerID, xbar, s) |>
distinct()
head(df2)
# A tibble: 6 × 3
playerID   xbar     s
<chr>     <dbl> <dbl>
1 solerjo01  16.3  14.2
2 valenjo03  16.2  13.3
3 muncyma01  20.5  16.6
4 yelicch01  17.8  13.8
5 moralke01  16.1  11.7
6 carpema01  15.5  11.1

Presently, I have about 221 observations, so a pair-wise calculations would be computed over about 4.8841^{4} pairs (hopefully manageable by a computer).

N <- nrow(df2)
df3 <- data.frame(player1 = rep(NA, N^2),
player2 = rep(NA, N^2),
Y       = rep(NA, N^2))

for(i in 1:N){
for(j in 1:N){
row_number = j*(i-1) + j
df3$player1[row_number] <- df2$playerID[i]
df3$player2[row_number] <- df2$playerID[j]

if(i == j){
this_Y_value <- 0
} else {
denominator <- df2$xbar[i] - df2$xbar[j]
if(denominator == 0){ denominator <- 0.1 }

this_Y_value <- abs((df2$s[i] - df2$s[j]) / denominator)
}

df3\$Y[row_number] <- this_Y_value
}

# if((row_number %% 10000) == 0){
#   print(paste("Currently computing row number", row_number))
# }
}

In this improvised metric, the top 10 scores were

df3 |>
arrange(desc(Y)) |>
top_n(10)
Selecting by Y
     player1   player2        Y
1  mccutan01 ramirjo01 692.3109
2  polloaj01 utleych01 348.8826
3  mccutan01 ensbemo01 291.9631
4  hosmeer01 quentca01 244.8742
5  mccanbr01  belljo02 209.0920
6  jacobmi02   bayja01 198.8301
7   kentje01 giambja01 190.9222
8  jonesga02 kinslia01 182.2589
9  contrwi01  shawtr01 180.9609
10 polloaj01 morrilo01 179.5927

Using the playerInfo() helper function in the Lahman package, we can verify the names of the players.

playerInfo("mccutan01")
       playerID nameFirst  nameLast
11832 mccutan01    Andrew McCutchen
playerInfo("ramirjo01")
       playerID nameFirst nameLast
14966 ramirjo01      Jose  Ramirez

## Data Visualization

This whole time, I was hoping for a neat boxplot.

df |>
filter(playerID == "mccutan01" | playerID == "ramirjo01") |>
ggplot(aes(x = playerID, y = HR)) +
geom_boxplot(color = c("#27251F", "#00385D"),
fill = c("#FDB827", "#E50022")) +
stat_summary(fun=mean, geom="point", shape=20, size=14, color="blue", fill="blue") +
scale_x_discrete(labels = c("Andrew McCutchen", "Jose Ramirez")) +
labs(title = "Similar Means, Different Variances",
subtitle = stringr::str_wrap("Andrew McCutchen and Jose Ramirez have averaged about 20.4 home runs per season, but in different ways (Qualifiers: after 2002 season, 100+ AB seasons, at least 5 100+ AB seasons, home run average over 15 HR/season)"),
caption = "Derek Sollberger, 2022-09-02",
x = "MLB Player",
y = "home runs in a season") +
theme(axis.text.x = element_text(size = 15),
legend.position = "none",
panel.background = element_blank(),
plot.background = element_rect(
fill = "#FFFFFF",
color = "#27251F"
),
plot.title = element_text(color = "#E50022", size = 20, hjust = 0.5),
plot.title.position = "plot",
plot.subtitle = element_text(color = "blue", size = 12, hjust = 0.0),
plot.caption = element_text(color = "#092C5C", size = 10, hjust = 1.0),
plot.caption.position = "plot",
plot.margin = margin(20, 20, 20, 20))