Bechdel analysis using the tidyverse

Chester Ismay

2018-02-11

This vignette is based on tidyverse-ifying the R code here and reproducing some of the plots and analysis done in the 538 story entitled “The Dollar-And-Cents Case Against Hollywood’s Exclusion of Women” by Walt Hickey available here.

Load required packages to reproduce analysis. Also load the bechdel dataset for analysis.

library(fivethirtyeight)
# tidyverse includes ggplot2, tibble, tidyr, readr, purrr, dplyr:
library(tidyverse)
library(knitr)
library(magrittr)
library(broom)
library(stringr)
library(ggthemes)
library(scales)
# Turn off scientific notation
options(scipen = 99)

Filter to only 1990 - 2013

Focus only on films from 1990 to 2013

bechdel90_13 <- bechdel %>% filter(between(year, 1990, 2013))

Calculate variables

Create international gross only and return on investment (ROI) columns and add to bechdel_90_13 data frame

bechdel90_13 %<>% 
  mutate(int_only = intgross_2013 - domgross_2013,
         roi_total = intgross_2013 / budget_2013,
         roi_dom = domgross_2013 / budget_2013,
         roi_int = int_only / budget_2013)

Create generous variable

bechdel90_13 %<>%
  mutate(generous = ifelse(test = clean_test %in% c("ok", "dubious"),
                           yes = TRUE,
                           no = FALSE))

Determine median ROI and budget based on categories

ROI_by_binary <- bechdel90_13 %>% 
  group_by(binary) %>% 
  summarize(median_ROI = median(roi_total, na.rm = TRUE))
ROI_by_binary
binary median_ROI
FAIL 2.454209
PASS 2.696553 
bechdel90_13 %>% 
  summarize(
    `Median Overall Return on Investment` = median(roi_total, na.rm = TRUE))
Median Overall Return on Investment
2.569013 
budget_by_binary <- bechdel90_13 %>% 
  group_by(binary) %>% 
  summarize(median_budget = median(budget_2013, na.rm = TRUE))
budget_by_binary
binary median_budget
FAIL 48385984
PASS 31070724 
bechdel90_13 %>% 
  summarize(`Median Overall Budget` = median(budget_2013, na.rm = TRUE))
Median Overall Budget
37878971

View Distributions

Look at the distributions of budget, international gross, ROI, and their logarithms

ggplot(data = bechdel90_13, mapping = aes(x = budget)) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of budget")

ggplot(data = bechdel90_13, mapping = aes(x = log(budget))) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of Logarithm of Budget")

ggplot(data = bechdel90_13, mapping = aes(x = intgross_2013)) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of International Gross")

ggplot(data = bechdel90_13, mapping = aes(x = log(intgross_2013))) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of Logarithm of International Gross")

ggplot(data = bechdel90_13, mapping = aes(x = roi_total)) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of ROI")

The previous distributions were skewed, but ROI is so skewed that purposefully limiting the x-axis may reveal a bit more information about the distribution: (Suggested by Mustafa Ascha.)

ggplot(data = bechdel90_13, mapping = aes(x = roi_total)) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of ROI") +
  xlim(0, 25)

ggplot(data = bechdel90_13, mapping = aes(x = log(roi_total))) +
  geom_histogram(color = "white", bins = 20) +
  labs(title = "Histogram of Logarithm of ROI")

Linear Regression Models

Movies with higher budgets make more international gross revenues using logarithms on both variables

ggplot(data = bechdel90_13, 
       mapping = aes(x = log(budget_2013), y = log(intgross_2013))) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE)

gross_vs_budget <- lm(log(intgross_2013) ~ log(budget_2013), 
                      data = bechdel90_13)
tidy(gross_vs_budget)
term estimate std.error statistic p.value
(Intercept) 2.4300342 0.3898688 6.232953 0
log(budget_2013) 0.9073902 0.0225334 40.268661 0

Bechdel dummy is not a significant predictor of log(intgross_2013) assuming log(budget_2013) is in the model

Note that the regression lines nearly completely overlap.

ggplot(data = bechdel90_13, 
       mapping = aes(x = log(budget_2013), y = log(intgross_2013), 
                     color = binary)) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE)

gross_vs_budget_binary <- lm(log(intgross_2013) ~ log(budget_2013) + factor(binary), 
                      data = bechdel90_13)
tidy(gross_vs_budget_binary)
term estimate std.error statistic p.value
(Intercept) 2.3585962 0.3988812 5.9130285 0.0000000
log(budget_2013) 0.9100772 0.0227566 39.9918064 0.0000000
factor(binary)PASS 0.0539207 0.0635194 0.8488861 0.3960713

Note the \(p\)-value on factor(binary)PASS here that is around 0.40.

Movies with higher budgets have lower ROI

ggplot(data = bechdel90_13, 
       mapping = aes(x = log(budget_2013), y = log(roi_total))) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE)

roi_vs_budget <- lm(log(roi_total) ~ log(budget_2013), 
                      data = bechdel90_13)
tidy(roi_vs_budget)
term estimate std.error statistic p.value
(Intercept) 2.4300342 0.3898688 6.232953 0.0000000
log(budget_2013) -0.0926098 0.0225334 -4.109890 0.0000416

Note the negative coefficient here on log(budget_2013) and its corresponding small \(p\)-value.

Bechdel dummy is not a significant predictor of log(roi_total) assuming log(budget_2013) is in the model

Note that the regression lines nearly completely overlap.

ggplot(data = bechdel90_13, 
       mapping = aes(x = log(budget_2013), y = log(roi_total), 
                     color = binary)) +
  geom_point() +
  geom_smooth(method = "lm", se = FALSE)

roi_vs_budget_binary <- lm(log(roi_total) ~ log(budget_2013) + factor(binary), 
                      data = bechdel90_13)
tidy(roi_vs_budget_binary)
term estimate std.error statistic p.value
(Intercept) 2.3585962 0.3988812 5.9130285 0.0000000
log(budget_2013) -0.0899228 0.0227566 -3.9515046 0.0000810
factor(binary)PASS 0.0539207 0.0635194 0.8488861 0.3960713

Note the \(p\)-value on factor(binary)PASS here that is around 0.40.

Dollars Earned for Every Dollar Spent graphic

Calculating the values and creating a tidy data frame

passes_bechtel_rom <- bechdel90_13 %>% 
  filter(generous == TRUE) %>% 
  summarize(median_roi = median(roi_dom, na.rm = TRUE))
median_groups_dom <- bechdel90_13 %>% 
  filter(clean_test %in% c("men", "notalk", "nowomen")) %>% 
  group_by(clean_test) %>% 
  summarize(median_roi = median(roi_dom, na.rm = TRUE))
pass_bech_rom <- data_frame(clean_test = "pass", 
                  median_roi = passes_bechtel_rom$median_roi)
med_groups_dom_full <- bind_rows(pass_bech_rom, median_groups_dom) %>% 
  mutate(group = "U.S. and Canada")
## Warning in bind_rows_(x, .id): binding character and factor vector,
## coercing into character vector
passes_bechtel_int <- bechdel90_13 %>% 
  filter(generous == TRUE) %>% 
  summarize(median_roi = median(roi_int, na.rm = TRUE))
median_groups_int <- bechdel90_13 %>% 
  filter(clean_test %in% c("men", "notalk", "nowomen")) %>% 
  group_by(clean_test) %>% 
  summarize(median_roi = median(roi_int, na.rm = TRUE))
pass_bech_int <- data_frame(clean_test = "pass", 
                  median_roi = passes_bechtel_int$median_roi)
med_groups_int_full <- bind_rows(pass_bech_int, median_groups_int) %>% 
  mutate(group = "International")
## Warning in bind_rows_(x, .id): binding character and factor vector,
## coercing into character vector
med_groups <- bind_rows(med_groups_dom_full, med_groups_int_full) %>% 
  mutate(clean_test = str_replace_all(clean_test, 
                                      "pass",
                                      "Passes Bechdel Test"),
         clean_test = str_replace_all(clean_test, "men",
                                      "Women only talk about men"),
         clean_test = str_replace_all(clean_test, "notalk",
                                      "Women don't talk to each other"),
         clean_test = str_replace_all(clean_test, "nowoWomen only talk about men",
                                      "Fewer than two women"))
med_groups %<>% mutate(clean_test = factor(clean_test, 
                                 levels = c("Fewer than two women", 
                                            "Women don't talk to each other",
                                            "Women only talk about men",
                                            "Passes Bechdel Test"))) %>% 
  mutate(group = factor(group, levels = c("U.S. and Canada", "International"))) %>% 
  mutate(median_roi_dol = dollar(median_roi))

Using only a few functions to plot

ggplot(data = med_groups, mapping = aes(x = clean_test, y = median_roi, 
                                        fill = group)) +
  geom_bar(stat = "identity") +
  facet_wrap(~ group) +
  coord_flip() +
  labs(title = "Dollars Earned for Every Dollar Spent", subtitle = "2013 dollars") +
  scale_fill_fivethirtyeight() +
  theme_fivethirtyeight()

Attempt to fully reproduce Dollars Earned for Every Dollar Spent plot using ggplot

ggplot(data = med_groups, mapping = aes(x = clean_test, y = median_roi, 
                                        fill = group)) +
  geom_bar(stat = "identity") +
  geom_text(aes(label = median_roi_dol), hjust = -0.1) +
  scale_y_continuous(expand = c(.25, 0)) +
  coord_flip() +
  facet_wrap(~ group) +
  scale_fill_manual(values = c("royalblue", "goldenrod")) +
  labs(title = "Dollars Earned for Every Dollar Spent", subtitle = "2013 dollars") +
  theme_fivethirtyeight() +
  theme(plot.title = element_text(hjust = -1.6), 
        plot.subtitle = element_text(hjust = -0.4),
        strip.text.x = element_text(face = "bold", size = 16),
        panel.grid.major = element_blank(), 
        panel.grid.minor = element_blank(),
        axis.title.x = element_blank(),
        axis.text.x = element_blank(),
        axis.ticks.x = element_blank()) +
  guides(fill = FALSE)