Analyzing the Annual Republicans vs. Democrats Congressional Baseball Game

Every year, the United States Congress takes a break from blocking each others bills and plays a charity baseball game. The best part, the teams are broken down by party lines, Republicans vs. Democrats. The tradition started in 1909 by Representative John Tener of Pennsylvania, a former professional baseball player. Last week was the annual game and Republicans were able to break a 7 year winning streak by the Democrats.

Below the net wins over the series is shown. The higher on the y axis the more Republican wins and the lower on the y axis the more Democrat wins. From this graph it is fairly obvious that each party has had long winning streaks. The gray dots represent years when the game was not held or I could not find any information about the game. In 1935, 1937, 1938, 1939, and 1941, games were held between members of congress and the press.

The following graph displays the points scored by each team over time. In the early years of the series, the games had much higher total scores than more recent years.

Next the point differentials were explored. The point differential is the difference between the scores of the two teams. Many of the closest games were held in the late seventies through the nineties. This time period also saw few winning streaks because the competition was fairly even between the parties.

A histogram was formed to understand the distribution of point differences. The Democrats have some extremely large wins with three wins over 20 points and the Republicans have none. Another interesting finding is that only one game ended in a tie. This is surprising because the charity event does not have overtime so it is logical to think out of the 81 games played more than one would end in a tie.

Over the years, the annual game has been held at many different locations.  Each party has had different rates of success at each field. The winning percentage at each field was calculated to understand if either party has a home-field-advantage at any park. Langley High School is a bit of an outlier because it was selected as the location after two rain delays and only hosted one game. American League Park II and Georgetown Field were the first two stadiums to host the game and each only hosted one game. Memorial Stadium had the fourth fewest games with only four, but all other locations had nine or more games.

Ironically, RFK Stadium, named after the famous Democratic U.S. Senator, has given Republicans a strong home-field-advantage. Republicans have won 13 out of the 14 games played at the stadium. Democrats have seen similar success at Nationals Park; winning 7 out of the 9 games.

Currently, I am planning to update these graphs each year after the annual game. Please feel free to add ideas for additional graphs or analysis in the comment section.

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Notes:

1. The data came from https://en.wikipedia.org/wiki/Congressional_Baseball_Game#Game_results
2. Some of the stadiums were renamed over the years and the original data set contained both names. For the analysis, the same stadiums were combined with the most recent name.

How Gender and Race Affect Police Interactions

Recently, police violence has become the focus of a lot of media attention. It has formed many protests and organizations around reducing police violence. Many of the organizations are specially focused on reducing violence toward blacks because it is a problem disproportionately effecting the black community. This post seeks to investigate some of these claims and understand the relationship between the violence each ethnic group experiences and their violence against police.

The following graphics come from a conversation about disparities between races when it comes to police killings. The discussion turned to the fact that only some disparities are thought of as problems of the system but others are generally thought to be acceptable. For example, blacks make up about 11% of the population, but 29% of the police killings. This disparity is largely seen as racism in the law enforcement and the overall justice system. Critics of this assumption usually point to the higher rates of crimes committed by blacks compared to whites and other races. However, the use of crime statistics from, what some believe is a racist institution is not a good method for explaining the differences in police kill rates.

Another group that is disproportionately killed, compared to their percentage of the population, is men.  Males make up a little less than half of the population (49.1%), but are 94.2% of the police killing victims. However, no one asserts the justice department to be sexist. The group discussing this matter largely agreed the reason for men to disproportionately be killed by police is because men most likely kill police more than women.

The follow graphic was created to compare the population, the proportion of people killed by police, and the number of police killed, broken down by gender. Men make up 94.2% of police killings, but also were responsible for 97.5% of police murders. This means while only half the population, men are 16 times more likely to be killed by police than compared to women. However, the killer of a police officer is 39 times more likely to be a man compared to a woman.

A similar graphic was created broken down by race (Note: http://killedbypolice.net/ did not use the method of classifying asians as the population data and FBI, so asians was included in "Other" for the people killed by police. Also the FBI defines hispanics as a subset of whites and not their own category so this is why hispanics are not represented in the "Killed Police" section). The chart below shows blacks are much more likely to be killed by police compared to their portion of the population, however while only being 29.5% of the people killed by police and 11% of the population, 43% of police officers are killed by blacks. 

Personally, I do not believe you can say that one race can be expected to be killed more because they kill police more. I believe, unlike gender, there are socio-economic differences between the groups that could lead to a greater likelihood of turning to crime because of lack of economic opportunity. Another factor is the populations are not perfectly comparable. Whites and asians households have fewer children than black and hispanics [3]. This leads to a lower ratio of old people to young people in the black and hispanics populations. Since the vast majority of people committing murders and/or being killed by police are young, populations with fewer old people will look like they commit more murders per capita.

Below is the R code used to generate the plots.

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Sources
[1] http://killedbypolice.net/ (May 2, 2013)
[2] https://www.fbi.gov/about-us/cjis/ucr/leoka/2013/tables/table_44_leos_fk_race_and_sex_of_known_offender_2004-2013.xls (April 10, 2016)
[3] http://www.pewsocialtrends.org/2012/05/17/explaining-why-minority-births-now-outnumber-white-births/ (April 29, 2016)

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This text was recognized by the built-in Ocrad engine. A better transcription may be attained by right clicking on the selection and changing the OCR engine to "Tesseract" (under the "Language" menu). This message can be removed in the future by unchecking "OCR Disclaimer" (under the Options menu). More info: http://projectnaptha.com/ocrad

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########################### R Code ################################ 

######### By Race ###########

# Cop killing graphic
# https://www.fbi.gov/about-us/cjis/ucr/leoka/2013/tables/table_44_leos_fk_race_and_sex_of_known_offender_2004-2013.xls
cop_killer_race <- c("White", "Black", "Asian", "Other")
cop_killer_quanity <- c(289, 243, 9, 24)
variable <- rep("Killed Police", length(cop_killer_race))
percent <- cop_killer_quanity/sum(cop_killer_quanity)
cop_killer <- data.frame(race=cop_killer_race, quanity=cop_killer_quanity, type=variable, percent=percent)
 
# People killed by cops
# Source killedbypolice.net (May 2, 2013)
cop_killed_race <- c("White", "Black", "Hispanic", "Other")
cop_killed_quanity <- c(782, 464, 302, 26)
variable <- rep("Killed by Police", length(cop_killed_race))
percent <- cop_killed_quanity/sum(cop_killed_quanity)
killed_by_cop <- data.frame(race=cop_killed_race, quanity=cop_killed_quanity, type=variable, percent=percent)
 
#Population Data
population_race <- c("White", "Black", "Asian", "Hispanic", "Other")
population_quanity <- c(196817552, 37685848, 14465124, 50477594, 28116441+2932248+540013)
variable <- rep("Population ", length(population_race))
percent <- population_quanity/sum(population_quanity)
population <- data.frame(race=population_race, quanity=population_quanity, type=variable, percent=percent)
 
# Bind the three data frames
data <- rbind(population, killed_by_cop, cop_killer )
 
# Calc the placement of the percent text in the graph
df <- data
df <- transform(df, mid_y = ave(df$percent, df$type, FUN = function(val) cumsum(val) - (0.5 * val)))
 
# Plot
ggplot(data=df, aes(x=type, y=quanity, fill=race, label=paste(round(percent*100,1),"%"))) + 
  geom_bar(stat="identity", position = "fill") + labs(x = "", y = "Percent", fill = "Race") + 
  geom_text(aes(y = mid_y)) + theme_bw() +
  annotate("text", label = "HallwayMathlete.com", x = 2, y = -.03, size = 4, colour = "gray") 
 
######### By Gender ###########
# Gender Women killed
# https://www.fbi.gov/about-us/cjis/ucr/leoka/2013/tables/table_44_leos_fk_race_and_sex_of_known_offender_2004-2013.xls
cop_killer_gender <- c( "Female", "Male", "Not Reported")
cop_killer_quanity <- c(13, 551, 1)
variable <- rep("Killed Police", length(cop_killer_gender))
percent <- cop_killer_quanity/sum(cop_killer_quanity)
cop_killer <- data.frame(race=cop_killer_gender, quanity=cop_killer_quanity, type=variable, percent=percent)
 
# People killed by cops
# Source killedbypolice.net (May 2, 2013)
cop_killed_gender <- c("Female", "Male","Not Reported")
cop_killed_quanity <- c( 177,2916, 2)
variable <- rep("Killed by Police", length(cop_killed_gender))
percent <- cop_killed_quanity/sum(cop_killed_quanity)
killed_by_cop <- data.frame(race=cop_killed_gender, quanity=cop_killed_quanity, type=variable, percent=percent)
 
#Population Data
population_gender <- c("Female","Male")
population_quanity <- c(143368343, 138053563)
variable <- rep("Population ", length(population_gender))
percent <- population_quanity/sum(population_quanity)
population <- data.frame(race=population_gender, quanity=population_quanity, type=variable, percent=percent)
 
# Bind the three data frames
data <- rbind(population, killed_by_cop, cop_killer )
 
# Calc the placement of the percent text in the graph
df <- data
df <- transform(df, mid_y = ave(df$percent, df$type, FUN = function(val) cumsum(val) - (0.5 * val)))
 
# Plot
ggplot(data=df, aes(x=type, y=quanity, fill=race, label=paste(round(percent*100,1),"%"))) + 
  geom_bar(stat="identity", position = "fill") + labs(x = "", y = "Percent", fill = "Race") + 
  geom_text(aes(y = mid_y)) + theme_bw() +
  annotate("text", label = "HallwayMathlete.com", x = 2, y = -.03, size = 4, colour = "gray") 

Salaries of Presidential Primary Voters by Candidate and State

The data used in this post comes from FiveThirtyEight and are put into easy to understand graphics. The first graphic shows the average salary of supports of each candidate by state. The states are ranked by highest average salary, Maryland, to the state with the lowest salary, Mississippi. The red line shows the average salary for each state. The first obvious conclusion is that John Kasich supporters make about $5-10 thousand more than supporters of other candidates. Second, in low salary states Clinton and Sanders supporter's have similar salaries, but when looking at higher salary states, Clinton supporters' salaries are even with Trump and Cruz supporters' salaries.

The following plot shows the distribution of average salaries for each Presidential Candidate. Again we see similar trends as before with Kasich having a high average income and Clinton having a mix of low and high income supporters.

The last plot shows the relationship between the average salary of a state and the average salary of a candidate's supporters. The red line is a perfect 1 to 1 ratio and the closer a candidate is to the red line the closer the candidate's supporters are to having the same salary as the average person of that state. The reason for almost all the dots falling above the line is because people with below average salaries are less likely to vote.

If you have any suggestions for plots using this data, please share in the comment section.

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Note:

[1] All states are not included because all states have not held elections yet.

Introduction to Machine Learning with Random Forest

The purpose of this tutorial is to serve as a introduction to the randomForest package in R and some common analysis in machine learning.

Part 1. Getting Started

First step, we will load the package and iris data set. The data set contains 3 classes of 50 instances each, where each class refers to a type of iris plant. One class is linearly separable from the other 2; the latter are NOT linearly separable from each other.

install.packages('randomForest')
library(randomForest)
data(iris)
head(iris)

Part 2. Fit Model

Now that we know what our data set contains, let fit our first model. We will be fitting 500 trees in our forest and trying to classify the Species of each iris in the data set. For the randomForest() function, "~." means use all the variables in the data frame.

Note: a common mistake, made by beginners, is trying to classify a categorical variable that R sees as a character. To fix this, convert the variable to a factor like this randomForest(as.factor(Species) ~ ., iris, ntree=500)

fit <- randomForest(Species ~ ., iris, ntree=500)

The next step is to use the newly create model in the fit variable and predict the label.

results <- predict(fit, iris)
summary(results)

After you have the predicted labels in a vector (results), the predict and actual labels must be compared. This can be done with a confusion matrix. A confusion matrix is a table of the actual vs the predicted with the diagonal numbers being correctly classified elements while all others are incorrect.

# Confusion Matrix
 table(results, iris$Species) 

Now we can take the diagonal points in the table and sum them, this will give us the total correctly classified instances. Then dividing this number by the total number of instances will calculate the percentage of prediction correctly classified. <- -="" 1="" accuracy="" correctly_classified="" div="" error="" iris="" length="" pecies="" results="" style="overflow: auto;" table="" total_classified="">

# Calculate the accuracy  
correctly_classified <- table(results, iris$Species)[1,1] + table(results, iris$Species)[2,2] + table(results, iris$Species)[3,3] 
total_classified <- length(results)  
# Accuracy   
correctly_classified / total_classified   
# Error  
1 - (correctly_classified / total_classified)

Part 3. Validate Model 

The next step is to validate the prediction model. Validation requires splitting your data into two sections. First, the training set, which will be used to create the model. The second will be the test set and will test the accuracy of the prediction model. The reasoning for splitting the data is to allow a model to be created using one data set and then reserving some data, where the output is already known, to "test" the model accuracy. This more effectively estimates the accuracy of the model by not using the same data used to create the model and predict the accuracy.

# How to split into a training set   
rows <- nrow(iris) col_count <- c(1:rows)  
Row_ID <- sample(col_count, rows, replace = FALSE)   
iris$Row_ID <- Row_ID   
 
# Choose the percent of the data to be used in the training 
data training_set_size = .80   
#Now to split the data into training and test 
 index_percentile <- rows*training_set_size   
# If the Row ID is smaller then the index percentile, it will be assigned into the training set   
train <- iris[iris$Row_ID <= index_percentile,]   
# If the Row ID is larger then the index percentile, it will be assigned into the training set   
test <- iris[iris$Row_ID > index_percentile,]   
train_data_rows <- nrow(train)   
test_data_rows <- nrow(test)   
total_data_rows <- (nrow(train)+nrow(test)) train_data_rows / total_data_rows     
 
# Now we have 80% of the data in the training set  test_data_rows / total_data_rows     
# Now we have 20% of the data in the training set   
# Now lets build the randomforest using the train data set  
fit <- randomForest(Species ~ ., train, ntree=500) 

After the test set is predicted, a confusion matrix and accuracy must be calculated.

# Use the new model to predict the test set   
results <- predict(fit, test, type="response")   
# Confusion Matrix  
table(results, test$Species)  
# Calculate the accuracy   
correctly_classified <- table(results, test$Species)[1,1] + table(results, test$Species)[2,2] + table(results, test$Species)[3,3] total_classified <- length(results)   
# Accuracy   
correctly_classified / total_classified  
# Error  
1 - (correctly_classified / total_classified)

Part 4. Model Analysis 

After the model is created, understanding the relationship between variables and number of trees is important. R makes it easy to plot the errors of the model as the number of trees increase. This allows users to trade off between more trees and accuracy or fewer trees and lower computational time.

fit <- randomForest(Species ~ ., train, ntree=500)   
results <- predict(fit, test, type="response")  
 
# Rank the input variables based on their effectiveness as predictors   
varImpPlot(fit)   
# To understand the error rate lets plot the model's error as the number of trees increases  
plot(fit) 

Part 5. Handling Missing Values 

The last section of this tutorial involves one of the most time consuming and important parts of the data analysis process, missing variables. Very few machine learning algorithms can handle missing data in the data. However the randomForest package contains one of the most useful functions of all time, na.roughfix(). Na.roughfix() takes the most common factor in that column and replaces all the NAs with it. For this section we will first create some NAs in this data set and then replace them and run the prediction algorithm.

# Create some NA in the data.   
iris.na <- iris for (i in 1:4)  
iris.na[sample(150, sample(20)), i] <- NA   
 
# Now we have a dataframe with NAs   
View(iris.na)   
#Adding na.action=na.roughfix   
#For numeric variables, NAs are replaced with column medians.   
#For factor variables, NAs are replaced with the most frequent levels (breaking ties at random) 

iris.narf <- randomForest(Species ~ ., iris.na, na.action=na.roughfix) 

results <- predict(iris.narf, train, type="response")

Congratulations! You now know how to create machine learning models, fit data using those models, test the model’s accuracy and display it in a confusion matrix, how to validate the model, and quickly replace missing variables. All of these are the basic fundamental skills in machine learning!