Class Activity 6

# load the necessary libraries
library(dplyr)
library(ggplot2)
library(babynames)

We will work with the babynames dataset again in this class activity. The header of the dataset looks like this:

knitr::kable(head(babynames))

year	sex	name	n	prop
1880	F	Mary	7065	0.0723836
1880	F	Anna	2604	0.0266790
1880	F	Emma	2003	0.0205215
1880	F	Elizabeth	1939	0.0198658
1880	F	Minnie	1746	0.0178884
1880	F	Margaret	1578	0.0161672

In this tutorial, we will learn about the five main verbs of dplyr and how to use them to manipulate data:

select(): Choose columns from a data frame
filter(): Choose rows based on a condition
arrange(): Sort the rows of a data frame
mutate(): Add new columns based on existing columns
summarise(): Aggregate data and compute summary statistics

Problem 1: `select()`

Which of these is NOT a way to select the name and n columns together?

select(babynames, -c(year, sex, prop))  #1
select(babynames, name:n)   #2
select(babynames, starts_with("n")) #3
select(babynames, ends_with("n"))  #4

Click for answer

Answer: 4 is not the way to select the name and n columns together

Problem 2: `filter()`

Use filter() with the logical operators to extract:

a. All of the names where prop is greater than or equal to 0.08

Click for answer

filter(babynames, prop >= 0.08)

# A tibble: 3 × 5
   year sex   name        n   prop
  <dbl> <chr> <chr>   <int>  <dbl>
1  1880 M     John     9655 0.0815
2  1880 M     William  9532 0.0805
3  1881 M     John     8769 0.0810

# alternate
babynames %>% filter(prop >= 0.08)

# A tibble: 3 × 5
   year sex   name        n   prop
  <dbl> <chr> <chr>   <int>  <dbl>
1  1880 M     John     9655 0.0815
2  1880 M     William  9532 0.0805
3  1881 M     John     8769 0.0810

b. All of the babies named “Rose”

Click for answer

babynames %>% filter(name == "Rose")

# A tibble: 247 × 5
    year sex   name      n      prop
   <dbl> <chr> <chr> <int>     <dbl>
 1  1880 F     Rose    700 0.00717  
 2  1880 M     Rose      7 0.0000591
 3  1881 F     Rose    734 0.00743  
 4  1882 F     Rose    886 0.00766  
 5  1883 F     Rose    877 0.00730  
 6  1883 M     Rose      5 0.0000445
 7  1884 F     Rose   1060 0.00770  
 8  1884 M     Rose      5 0.0000407
 9  1885 F     Rose   1164 0.00820  
10  1885 M     Rose      9 0.0000776
# ℹ 237 more rows

c. Use filter() to choose all rows where name is “John” and sex is “M”.

Click for answer

babynames %>% filter(name == "John", sex == "M")

# A tibble: 138 × 5
    year sex   name      n   prop
   <dbl> <chr> <chr> <int>  <dbl>
 1  1880 M     John   9655 0.0815
 2  1881 M     John   8769 0.0810
 3  1882 M     John   9557 0.0783
 4  1883 M     John   8894 0.0791
 5  1884 M     John   9388 0.0765
 6  1885 M     John   8756 0.0755
 7  1886 M     John   9026 0.0758
 8  1887 M     John   8110 0.0742
 9  1888 M     John   9247 0.0712
10  1889 M     John   8548 0.0718
# ℹ 128 more rows

Problem 3: `arrange()`

a. Use `arrange()` to sort the babynames dataset by the prop column in descending order.

Click for answer

babynames %>% arrange(desc(prop))

# A tibble: 1,924,665 × 5
    year sex   name        n   prop
   <dbl> <chr> <chr>   <int>  <dbl>
 1  1880 M     John     9655 0.0815
 2  1881 M     John     8769 0.0810
 3  1880 M     William  9532 0.0805
 4  1883 M     John     8894 0.0791
 5  1881 M     William  8524 0.0787
 6  1882 M     John     9557 0.0783
 7  1884 M     John     9388 0.0765
 8  1882 M     William  9298 0.0762
 9  1886 M     John     9026 0.0758
10  1885 M     John     8756 0.0755
# ℹ 1,924,655 more rows

b. Use `arrange()` to sort the babynames dataset by year (ascending) and then by prop (descending).

Click for answer

babynames %>% arrange(year, desc(prop))

# A tibble: 1,924,665 × 5
    year sex   name        n   prop
   <dbl> <chr> <chr>   <int>  <dbl>
 1  1880 M     John     9655 0.0815
 2  1880 M     William  9532 0.0805
 3  1880 F     Mary     7065 0.0724
 4  1880 M     James    5927 0.0501
 5  1880 M     Charles  5348 0.0452
 6  1880 M     George   5126 0.0433
 7  1880 M     Frank    3242 0.0274
 8  1880 F     Anna     2604 0.0267
 9  1880 M     Joseph   2632 0.0222
10  1880 M     Thomas   2534 0.0214
# ℹ 1,924,655 more rows

Problem 4: `mutate()`

a. Use `mutate()` to create a new column called decade which contains the decade the record is in (e.g., 1990 for the years 1990-1999).

Click for answer

babynames %>% mutate(decade = (year %/% 10) * 10)

# A tibble: 1,924,665 × 6
    year sex   name          n   prop decade
   <dbl> <chr> <chr>     <int>  <dbl>  <dbl>
 1  1880 F     Mary       7065 0.0724   1880
 2  1880 F     Anna       2604 0.0267   1880
 3  1880 F     Emma       2003 0.0205   1880
 4  1880 F     Elizabeth  1939 0.0199   1880
 5  1880 F     Minnie     1746 0.0179   1880
 6  1880 F     Margaret   1578 0.0162   1880
 7  1880 F     Ida        1472 0.0151   1880
 8  1880 F     Alice      1414 0.0145   1880
 9  1880 F     Bertha     1320 0.0135   1880
10  1880 F     Sarah      1288 0.0132   1880
# ℹ 1,924,655 more rows

Problem 5: `summarize()` or `summarise()`

Use the codes mentioned so far to compute three statistics:

the total number of children who ever had your name
the maximum number of children given your name in a single year
the mean number of children given your name per year/decade (optional)

Click for answer

babynames %>% 
  filter(name == "Dee", sex == "M")

# A tibble: 136 × 5
    year sex   name      n     prop
   <dbl> <chr> <chr> <int>    <dbl>
 1  1880 M     Dee      20 0.000169
 2  1881 M     Dee      32 0.000296
 3  1882 M     Dee      23 0.000188
 4  1883 M     Dee      22 0.000196
 5  1884 M     Dee      27 0.000220
 6  1885 M     Dee      28 0.000241
 7  1886 M     Dee      26 0.000218
 8  1887 M     Dee      39 0.000357
 9  1888 M     Dee      35 0.000269
10  1889 M     Dee      24 0.000202
# ℹ 126 more rows

babynames %>% 
  filter(name == "Dee", sex == "M") %>% 
  summarise(max_number = max(n))

# A tibble: 1 × 1
  max_number
       <int>
1        125

babynames %>% 
  filter(name == "Dee", sex == "M") %>% 
  mutate(decade = (year %/% 10) * 10) %>% 
  group_by(decade) %>%
  summarise(total = sum(n),
            max = max(n), 
            mean = mean(n))

# A tibble: 14 × 4
   decade total   max   mean
    <dbl> <int> <int>  <dbl>
 1   1880   276    39  27.6 
 2   1890   271    43  27.1 
 3   1900   302    38  30.2 
 4   1910   818   125  81.8 
 5   1920  1090   125 109   
 6   1930  1010   118 101   
 7   1940   967   120  96.7 
 8   1950   957   118  95.7 
 9   1960   683   102  68.3 
10   1970   380    57  38   
11   1980   217    30  21.7 
12   1990   130    17  13   
13   2000    87    13   9.67
14   2010    52    12   7.43

Problem 6

a. Use `min_rank()` and `mutate()` to rank each row in babynames from largest prop to smallest prop.

Click for answer

babynames %>% mutate(rank = min_rank(desc(prop))) %>% arrange(rank)

# A tibble: 1,924,665 × 6
    year sex   name        n   prop  rank
   <dbl> <chr> <chr>   <int>  <dbl> <int>
 1  1880 M     John     9655 0.0815     1
 2  1881 M     John     8769 0.0810     2
 3  1880 M     William  9532 0.0805     3
 4  1883 M     John     8894 0.0791     4
 5  1881 M     William  8524 0.0787     5
 6  1882 M     John     9557 0.0783     6
 7  1884 M     John     9388 0.0765     7
 8  1882 M     William  9298 0.0762     8
 9  1886 M     John     9026 0.0758     9
10  1885 M     John     8756 0.0755    10
# ℹ 1,924,655 more rows

b. Compute each name’s rank within its year and sex.

Click for answer

babynames %>% group_by(year, sex) %>% mutate(rank = min_rank(desc(prop)))

# A tibble: 1,924,665 × 6
# Groups:   year, sex [276]
    year sex   name          n   prop  rank
   <dbl> <chr> <chr>     <int>  <dbl> <int>
 1  1880 F     Mary       7065 0.0724     1
 2  1880 F     Anna       2604 0.0267     2
 3  1880 F     Emma       2003 0.0205     3
 4  1880 F     Elizabeth  1939 0.0199     4
 5  1880 F     Minnie     1746 0.0179     5
 6  1880 F     Margaret   1578 0.0162     6
 7  1880 F     Ida        1472 0.0151     7
 8  1880 F     Alice      1414 0.0145     8
 9  1880 F     Bertha     1320 0.0135     9
10  1880 F     Sarah      1288 0.0132    10
# ℹ 1,924,655 more rows

c. (Extra) Then compute the median rank for each combination of name and sex, and arrange the results from highest median rank to lowest.

Click for answer

babynames %>% 
  group_by(year, sex) %>% 
  mutate(rank = min_rank(desc(prop))) %>%
  group_by(name, sex) %>%
  summarize(score = median(rank)) %>%
  arrange(score)

# A tibble: 107,973 × 3
# Groups:   name [97,310]
   name      sex   score
   <chr>     <chr> <dbl>
 1 Mary      F       1  
 2 James     M       3  
 3 John      M       3  
 4 William   M       4  
 5 Robert    M       6  
 6 Michael   M       7.5
 7 Charles   M       9  
 8 Elizabeth F      10  
 9 Joseph    M      10  
10 Thomas    M      11  
# ℹ 107,963 more rows

Class Activity 6

Problem 1: select()

Problem 2: filter()

a. All of the names where prop is greater than or equal to 0.08

b. All of the babies named “Rose”

c. Use filter() to choose all rows where name is “John” and sex is “M”.

Problem 3: arrange()

a. Use arrange() to sort the babynames dataset by the prop column in descending order.

b. Use arrange() to sort the babynames dataset by year (ascending) and then by prop (descending).

Problem 4: mutate()

a. Use mutate() to create a new column called decade which contains the decade the record is in (e.g., 1990 for the years 1990-1999).

Problem 5: summarize() or summarise()

Problem 6

a. Use min_rank() and mutate() to rank each row in babynames from largest prop to smallest prop.

b. Compute each name’s rank within its year and sex.

c. (Extra) Then compute the median rank for each combination of name and sex, and arrange the results from highest median rank to lowest.

Problem 1: `select()`

Problem 2: `filter()`

Problem 3: `arrange()`

a. Use `arrange()` to sort the babynames dataset by the prop column in descending order.

b. Use `arrange()` to sort the babynames dataset by year (ascending) and then by prop (descending).

Problem 4: `mutate()`

a. Use `mutate()` to create a new column called decade which contains the decade the record is in (e.g., 1990 for the years 1990-1999).

Problem 5: `summarize()` or `summarise()`

a. Use `min_rank()` and `mutate()` to rank each row in babynames from largest prop to smallest prop.