DATA 608 01[46846] : HomeWork1

Question 1

Create a graph that shows the distribution of companies in the dataset by State (ie how many are in each state). There are a lot of States, so consider which axis you should use. This visualization is ultimately going to be consumed on a ‘portrait’ oriented screen (ie taller than wide), which should further guide your layout choices.

# Answer Question 1 here
inc %>% 
  group_by(State) %>%
  summarise(No_Companies = n()) %>%
  arrange(desc(No_Companies)) %>%
  ggplot(aes(x = reorder(State, No_Companies), y = No_Companies)) + 
         geom_bar(stat="identity", fill="firebrick2", width=0.5) + 
         coord_flip() + 
         labs(title="Amount of Fastest Growing Companies by State", subtitle="count vs. State", x="State", y="Count") #+ 

  #theme_grey(base_size = 8) +
  #scale_y_continuous(expand=c(0,0))

# Creating the ggplot for top 10 states
top10 <- inc %>% 
          count(State, sort = TRUE) %>% 
          head(10)
top10$State <- factor(top10$State, levels = top10$State[order(top10$n)])
ggplot(top10, aes(State,n)) + 
  geom_bar(stat="identity", fill='steelblue') + 
  coord_flip() + 
  ylab("Count") + 
  theme_minimal() + 
  ggtitle("Number of Fastest Growing Companies Per State - Top 10") + 
  geom_text(aes(label=n), vjust=0.3, hjust=1.6, color="white", size=3.5)

inc1 <- inc
inc1$State <- factor(inc1$State, rev(levels(inc1$State)))
png(filename = "Figure.png", height = 960, width = 480, units="px")
g <- ggplot(data = inc1, aes(x = State)) + 
      geom_bar(fill="steelblue") +
      coord_flip() + 
      #scale_x_discrete(limits = c(0, 30)) + 
      scale_y_continuous(limits = c(0, 800)) + 
      labs(y = "Count") + 
      ggtitle("Inc. Magazine 5,000 Fastest Growing Companies") + 
      theme(panel.background = element_rect(fill = "white"), 
            panel.grid.major.x = element_line(color = "lightgrey"), 
            axis.text = element_text(size=12, color = "grey55"), 
            axis.title = element_text(size=14, color = "grey55"), 
            plot.title = element_text(size=14, color = "grey55", hjust=0.5))
plot(g)
garbage <- dev.off()

Question 2

Lets dig in on the state with the 3rd most companies in the data set. Imagine you work for the state and are interested in how many people are employed by companies in different industries. Create a plot that shows the average and/or median employment by industry for companies in this state (only use cases with full data, use R’s complete.cases() function.) In addition to this, your graph should show how variable the ranges are, and you should deal with outliers.

# Answer Question 2 here
third_state <- inc %>% 
  group_by(State) %>%
  summarise(No_Companies = n()) %>%
  arrange(desc(No_Companies)) %>%
  summarise(value = nth(State, 3))

paste0("3rd Most Employed State:",third_state[[1, 1]])

## [1] "3rd Most Employed State:NY"

inc[complete.cases(inc), ] %>% 
  filter(State == third_state[[1, 1]]) %>%
  group_by(Industry) %>%
  summarise(No_of_Employees = sum(Employees)) %>%
  arrange(desc(No_of_Employees)) %>%
  ggplot(aes(x = reorder(Industry, No_of_Employees), y = No_of_Employees)) + 
    geom_bar(stat="identity") + 
    coord_flip() + 
    labs(title=paste("Most Employed Companies in", third_state[[1, 1]], "by Industry"))

inc[complete.cases(inc), ] %>% 
  filter(State == third_state[[1, 1]]) %>%
  ggplot(aes(Industry, Employees)) + 
    geom_boxplot() + 
    coord_flip() + 
    labs(title ="New York Employment Overview by Industry") +
    geom_boxplot(outlier.shape=NA) +
    scale_y_continuous(limits = quantile(inc[complete.cases(inc), ]$Employees, c(0.01, 0.99)))

inc[complete.cases(inc), ] %>% 
  filter(State == 'NY') %>%
  group_by(Industry) %>%
  summarize(meanEmployment=mean(Employees),medianEmployment=median(Employees)) %>%
  gather('Metric','Value',2:3) %>%
  ggplot(aes(x=Industry,y=Value)) +
         geom_bar(stat='identity',aes(fill=Metric),position='dodge') + 
         coord_flip() + 
         labs(title='Total employed by Industry in NY')+ylab('Employees')

Business Products & Services appears to be an outlier. Upon checking the average difference between industry means, we may be able to regulate the outlier.

third.state.table <- inc[complete.cases(inc), ] %>% 
                      filter(State == 'NY') %>%
                      group_by(Industry) %>%
                      summarize(meanEmployment=mean(Employees), medianEmployment=median(Employees)) 
pivot_third.state.table <- third.state.table %>%
                            gather(property, count, meanEmployment, medianEmployment)
state.outlier <- pivot_third.state.table %>% filter(property == "meanEmployment")
state.outlier <- state.outlier[-c(2),] # drop the outlier
mean(diff(sort(state.outlier$count))) # calculate the average difference 

## [1] 26.49105

max(diff(sort(state.outlier$count)))

## [1] 191.6224

As can be seen from above, the average difference between consecutive leading industries is 26.49105, with a max of 191.6224. With this in mind, we can cap the outlier at ~200 more than the second-highest.

third.state.edited <- pivot_third.state.table
third.state.edited[2,3] <- 815
ggplot(third.state.edited, aes(x=reorder(Industry, count), y=count)) +
  geom_bar(stat="identity", position="dodge", aes(fill=property)) +
  coord_flip() +
  scale_fill_manual(values=c("lightsteelblue4", "lightsteelblue2"), guide=guide_legend(reverse=T)) +
  scale_y_continuous(expand=c(0,0))

par(mfrow=c(1,2))
# Create a plot showing average employment by industry for companies in the state
ggplot(third.state.table %>% mutate(Industry = fct_reorder(Industry, meanEmployment)) , aes(x=Industry,y=meanEmployment, fill=Industry)) + 
  geom_bar(stat="identity") + 
  ylab('Mean Employment by Industry') + 
  #coord_flip() +
  scale_fill_grey(start=0.8, end=0.1) + 
  theme(legend.position='none') +
  theme(axis.text.x=element_text(angle=90,hjust=1,vjust=0.5))

# Create a plot showing median employment by industry for companies in the state
ggplot(third.state.table %>% mutate(Industry = fct_reorder(Industry, medianEmployment)), aes(x=Industry,y=medianEmployment, fill=Industry)) +
  geom_bar(stat="identity") + 
  ylab('Median Employment by Industry') + 
  #coord_flip() + 
  scale_fill_grey(start=0.8, end=0.1) + 
  theme(legend.position='none') +
  theme(axis.text.x=element_text(angle=90,hjust=1,vjust=0.5))

Question 3

Now imagine you work for an investor and want to see which industries generate the most revenue per employee. Create a chart that makes this information clear. Once again, the distribution per industry should be shown.

inc[is.na(inc)] <- 0 
inc %>% 
  group_by(Industry) %>%
  summarise(Employee_Revenue = sum(Revenue)/sum(Employees)) %>%
  arrange(desc(Employee_Revenue)) %>%
  ggplot(aes(x = reorder(Industry, Employee_Revenue), y = Employee_Revenue, fill = Employee_Revenue)) + 
    geom_bar(stat="identity") + 
    coord_flip() + 
    labs(title ="Revenue Per Employee by Industry", ylab="Industry") +
    scale_fill_gradient(low = "black", high = "darkgreen") + 
    theme(legend.position = "none")

inc3 <- inc[complete.cases(inc),] %>%
          group_by(Industry) %>%
          summarize(total_employees=sum(Employees), total_revenue=sum(Revenue), growth_rate=mean(Growth_Rate))
inc3$avgRev = inc3$total_revenue/inc3$total_employees
compDF = inc3[!inc3$Industry=='Computer Hardware' & !inc3$Industry=='Human Resources',]
compDF$scaledAR = scale(compDF$avgRev)
ggplot(compDF,aes(x=reorder(Industry,scaledAR),y=scaledAR,fill=growth_rate)) + 
  geom_bar(stat="identity") + 
  coord_flip()

# Add Revenue-Per-Employee variable
RPE = inc[complete.cases(inc), ] %>% mutate(RevPerEmp = Revenue/Employees)

# Remove extreme outliers
RPEoutliers = RPE$RevPerEmp %in% boxplot.stats(RPE$RevPerEmp, coef = 1.5)$out
RPEoutclean = RPE[!RPEoutliers, ]

# Revenue/Employee vs Industry
ggplot(RPEoutclean, aes(x = reorder(Industry, RevPerEmp, FUN = mean), y = RevPerEmp)) + 
  geom_boxplot(outlier.shape = NA) +
  ylab("Revenue per Employee") + 
  xlab("Industry")+
  coord_flip()+
  stat_summary(fun.y=mean, colour="blue", geom="point", show_guide = FALSE)

inc3 = inc[inc$Employees > 0, ]
inc3$Revperemp <- with(inc3, Revenue/Employees)
dfSubset <- aggregate(inc3$Revperemp, list(Industry = inc3$Industry), mean)
dfSubset$Legend <- paste(dfSubset$Industry, round(dfSubset$x/1000,0), 'K', sep=' ') 
ggplot(dfSubset, aes(area = x, fill = Industry, label = Legend)) +
  geom_treemap(show.legend = FALSE, na.rm = TRUE, stat="identity", layout="squarified") +
  geom_treemap_text(fontface = "italic", colour = "white", place = "centre",
                    grow = TRUE)

In this case, we have an outlier: Computer Hardware. Thus, computer hardware has the potential for the most revenue per employee, but a safer bet would probably be energy, because of its higher growth rate.

inc3 <- inc %>% 
  mutate(Revenue_Per_Employee = Revenue/Employees) %>% 
  group_by(Industry) %>%
  #summarise(Employee_Revenue = sum(Revenue)/sum(Employees)) %>%
  arrange(desc(Revenue_Per_Employee))
ggplot(inc3, aes(x=Industry, y=Revenue_Per_Employee, fill=Industry)) + 
  geom_boxplot() + 
  coord_flip() + 
  theme(legend.position='none') + 
  ylab("Revenue Per Employee")

ggplot(inc3, aes(x=Industry, y=Revenue_Per_Employee, fill=Industry)) + 
  geom_boxplot() + 
  scale_y_continuous(trans='log10') + 
  coord_flip() + 
  theme(legend.position='none') + 
  ylab("Revenue Per Employee - Logarithmic Transformation Base 10")

ggplot(inc3, aes(x=Industry, y=Revenue_Per_Employee, fill=Industry)) + 
  geom_violin() + 
  scale_y_continuous(trans='log10') + 
  coord_flip() + 
  theme(legend.position='none') + 
  ylab("Revenue Per Employee - Logarithmic Transformation Base 10") + 
  ggtitle("Violin Plot of Revenue Per Employee by Industry")