Sariah Nokes - Intro to the Tidyverse
You are thinking of launching a new digital magazine subscription service, similar to Netflix, but with magazines. To determine the viability of this idea and potential key segments, we will look at this data of magazine subscribers and not subscribers. The dataset we will use in this assignment is the Magazine Subscription Data.
Start by loading in the correct packages and dataset.
install.packages("tidyverse")## Installing package into '/home/rstudio-user/R/x86_64-pc-linux-gnu-library/3.6'
## (as 'lib' is unspecified)install.packages("ggplot2")## Installing package into '/home/rstudio-user/R/x86_64-pc-linux-gnu-library/3.6'
## (as 'lib' is unspecified)library(tidyverse)## ── Attaching packages ───────────────────────────────────────────── tidyverse 1.2.1 ──## ✔ ggplot2 3.2.1 ✔ purrr 0.3.3
## ✔ tibble 2.1.3 ✔ dplyr 0.8.3
## ✔ tidyr 1.0.0 ✔ stringr 1.4.0
## ✔ readr 1.3.1 ✔ forcats 0.4.0## ── Conflicts ──────────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()library(ggplot2)
library(readr)
mag <- read_csv("Magazine Subscription Data.csv")## Parsed with column specification:
## cols(
## age = col_double(),
## gender = col_character(),
## income = col_double(),
## kids = col_double(),
## ownHome = col_character(),
## subscribe = col_character(),
## Segment = col_character()
## )If you recall, we learned about 5 verbs to use when coding in the tidyverse. They were: filter, arrange, mutate, summarize, and group_by. Which of these functions could be used to display only people with 0 children? mutate___________
Use this function to make it so that you only see magazine subscribers with 0 children.
filter(mag, kids=="0")## # A tibble: 121 x 7
## age gender income kids ownHome subscribe Segment
## <dbl> <chr> <dbl> <dbl> <chr> <chr> <chr>
## 1 43.2 Male 44169. 0 ownYes subNo Suburb mix
## 2 28.5 Male 47245. 0 ownNo subNo Suburb mix
## 3 35.2 Female 52568. 0 ownYes subNo Suburb mix
## 4 47.6 Male 47918. 0 ownYes subNo Suburb mix
## 5 37.6 Female 65767. 0 ownNo subNo Suburb mix
## 6 42.0 Female 53127. 0 ownYes subNo Suburb mix
## 7 44.0 Female 41255. 0 ownYes subNo Suburb mix
## 8 44.5 Male 57363. 0 ownNo subNo Suburb mix
## 9 45.3 Female 65170. 0 ownNo subNo Suburb mix
## 10 42.3 Male 49675. 0 ownYes subNo Suburb mix
## # … with 111 more rows- Now let’s arrange the number of people with 0 kids according to their age. What is the youngest age with 0 kids? 19____
arrange(mag,age)## # A tibble: 300 x 7
## age gender income kids ownHome subscribe Segment
## <dbl> <chr> <dbl> <dbl> <chr> <chr> <chr>
## 1 19.3 Female 18593. 0 ownNo subNo Urban hip
## 2 20.7 Male 22517. 3 ownNo subNo Urban hip
## 3 21.0 Female 27244. 1 ownNo subNo Urban hip
## 4 21.2 Male 18419. 1 ownYes subYes Urban hip
## 5 21.4 Male 16646. 3 ownNo subNo Urban hip
## 6 21.5 Female 17083. 2 ownNo subNo Urban hip
## 7 21.8 Male 27807. 2 ownNo subYes Urban hip
## 8 22.1 Male 21107. 0 ownNo subNo Urban hip
## 9 22.2 Female 20222. 2 ownYes subYes Urban hip
## 10 22.3 Female 24541. 1 ownNo subNo Urban hip
## # … with 290 more rows- Let’s now apply the arrange function with descending order. What is the oldest age with 0 kids? 80___
arrange(mag,desc(age))## # A tibble: 300 x 7
## age gender income kids ownHome subscribe Segment
## <dbl> <chr> <dbl> <dbl> <chr> <chr> <chr>
## 1 80.5 Male 82077. 0 ownYes subYes Travelers
## 2 78.2 Female 24604. 0 ownYes subNo Travelers
## 3 75.9 Female 23968. 0 ownYes subNo Travelers
## 4 71.9 Female 60279. 0 ownYes subYes Travelers
## 5 70.6 Male 48697. 0 ownNo subNo Travelers
## 6 68.1 Female 51535. 0 ownNo subNo Travelers
## 7 68.1 Female 25772. 0 ownYes subNo Travelers
## 8 68.1 Male 104312. 0 ownYes subNo Travelers
## 9 68.0 Female 69075. 0 ownNo subNo Travelers
## 10 66.9 Male 54061. 0 ownYes subNo Travelers
## # … with 290 more rows- Let’s take a look at just females. What is the highest income among females? 106430.05
filter(mag, gender=="Female") %>%
arrange(desc(income))## # A tibble: 157 x 7
## age gender income kids ownHome subscribe Segment
## <dbl> <chr> <dbl> <dbl> <chr> <chr> <chr>
## 1 54.9 Female 106430. 0 ownYes subNo Travelers
## 2 57.8 Female 105538. 0 ownYes subNo Travelers
## 3 66.4 Female 101174. 0 ownYes subNo Travelers
## 4 55.2 Female 96509. 0 ownYes subNo Travelers
## 5 47.8 Female 92431. 0 ownYes subNo Travelers
## 6 56.3 Female 91509. 0 ownYes subNo Travelers
## 7 53.7 Female 85770. 0 ownYes subNo Travelers
## 8 62.4 Female 82349. 0 ownYes subNo Travelers
## 9 37.3 Female 81042. 1 ownNo subNo Suburb mix
## 10 47.9 Female 79544. 1 ownYes subNo Suburb mix
## # … with 147 more rows- Let’s start off by grouping according to Segment. How many segments are there?
mag %>% group_by (Segment) %>% summarise(count=n())## # A tibble: 4 x 2
## Segment count
## <chr> <int>
## 1 Moving up 70
## 2 Suburb mix 100
## 3 Travelers 80
## 4 Urban hip 50- Now we will combine a few other principles, including summarize, to calculate the average age for each Segment. Name this output “segment_age”. Name the new average age variable “avg_age”.
segment_age <-
mag %>% group_by (Segment) %>% summarise(avg_age = mean(age))
head(segment_age)## # A tibble: 4 x 2
## Segment avg_age
## <chr> <dbl>
## 1 Moving up 36.3
## 2 Suburb mix 39.9
## 3 Travelers 57.9
## 4 Urban hip 23.9- Let’s take a look at children alongside age. Add average kids named “avg_kids” as another variable. Name this output “segment_age_kids”. What segment has the lowest number of kids on average? __________
segment_age_kids <- mag %>% group_by (Segment) %>% summarise(avg_age = mean(age), avg_kids= mean(kids))
head(segment_age_kids)## # A tibble: 4 x 3
## Segment avg_age avg_kids
## <chr> <dbl> <dbl>
## 1 Moving up 36.3 1.91
## 2 Suburb mix 39.9 1.92
## 3 Travelers 57.9 0
## 4 Urban hip 23.9 1.1- Now let’s make some visualizations. Use ggplot to make a bar chart out of our average age for each segment.
ggplot(data=segment_age_kids, aes(x=Segment, y=avg_age)) +
geom_bar(stat="identity")
10. Let’s look at what the distribution of income is like across our data. Make a histogram showing the distribution of income.
ggplot(mag, aes(x=income)) + geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
- Note the bin-width of the histogram above is pretty bad and distorts the graph. Let’s change the binwidth to 10,500 and re-graph.
ggplot(mag, aes(x=income)) + geom_histogram(binwidth=10500)
- Let’s now take a look at how the distribution of income varies by segment. Make a boxplot of income by segment. What segment has the largest range of income? _____TRAVELERS______
ggplot(mag, aes(x=Segment, y=income)) +
geom_boxplot()
- We also want to know what the distribution of age is like among genders. Make a new boxplot to show this. Are the age distributions similar or different for the genders? Explain Similar because their medians and IQR’s are close in range____________
ggplot(mag, aes(x=gender, y=age)) +
geom_boxplot()
- Let’s look at the relationship between age and income. Plot these two in a scatterplot. Paste your code below.
ggplot(mag, aes(x=age, y=income)) +
geom_point()
- This is a good graph, but it could be more useful. Let’s try adding some color differentiation. Re-graph and make color show gender. Are the trends of the two genders generally similar or different? ______________
ggplot(mag, aes(x=age, y=income, color=gender)) +
geom_point()
- Now let’s see what this graph looks like across Segments. Use the facet_wrap feature to plot the graphs for each segment. Note: This graph should feature all of the previous things we’ve done. We have been building this up with explaining income by age, then adding color for gender, and now we will facet_wrap by segment.
color <- ggplot(mag, aes(x=age, y=income, color=gender)) +
geom_point()
color <- color + facet_wrap(~Segment, ncol=2)
color