R4DSchap12

Using prose, describe how the variables and observations are organised in each of the sample tables.

• table2: 二つの変数が一つの行にまとめられている
• table3: 二つの観測が一つにまとめられている
• table4: 一つのデータセットが二つのtibbleに分けられ、観測が列になっている

Compute the rate for table2, and table4a + table4b. You will need to perform four operations:

• Extract the number of TB cases per country per year.
• Extract the matching population per country per year.
• Divide cases by population, and multiply by 10000.
• Store back in the appropriate place.
• Which representation is easiest to work with? Which is hardest? Why?

table2のほうがマシ table2はyearでgroup_byしてしまえばいいが、table4だとそうはいかない

Recreate the plot showing change in cases over time using table2 instead of table1. What do you need to do first?

countryとyearでgroup_byする

Why are gather() and spread() not perfectly symmetrical?

Carefully consider the following example:

stocks <- tibble(
  year   = c(2015, 2015, 2016, 2016),
  half  = c(   1,    2,     1,    2),
  return = c(1.88, 0.59, 0.92, 0.17)
)
stocks %>% 
  spread(year, return) %>% 
  gather("year", "return", `2015`:`2016`)

## # A tibble: 4 x 3
##    half year  return
##   <dbl> <chr>  <dbl>
## 1     1 2015    1.88
## 2     2 2015    0.59
## 3     1 2016    0.92
## 4     2 2016    0.17

chr以外の型の変数をspreadすると、自動的にchr型になってしまう

Why does this code fail?

table4a %>% 
  gather(1999, 2000, key = "year", value = "cases")
> Error in inds_combine(.vars, ind_list) : Position must be between 0 and n

table4a %>%
  gather(`1999`, `2000`, key = "year", value = "cases")

## # A tibble: 6 x 3
##   country     year   cases
##   <chr>       <chr>  <int>
## 1 Afghanistan 1999     745
## 2 Brazil      1999   37737
## 3 China       1999  212258
## 4 Afghanistan 2000    2666
## 5 Brazil      2000   80488
## 6 China       2000  213766

Why does spreading this tibble fail? How could you add a new column to fix the problem?

people <- tribble(
  ~name,             ~key,    ~value,
  #-----------------|--------|------
  "Phillip Woods",   "age",       45,
  "Phillip Woods",   "height",   186,
  "Phillip Woods",   "age",       50,
  "Jessica Cordero", "age",       37,
  "Jessica Cordero", "height",   156
)

people %>%
  spread(key = key, value = value)
  
> Error: Each row of output must be identified by a unique combination of keys. Keys are shared for 2 rows: * 1, 3 Do you need to create unique ID with tibble::rowid_to_column()?

Phillip Woodsは何歳かわからない

Tidy the simple tibble below. Do you need to spread or gather it? What are the variables?

preg <- tribble(
  ~pregnant, ~male, ~female,
  "yes",     NA,    10,
  "no",      20,    12
)

preg %>%
  gather(male, female, key = sex, value = population) %>%
  spread(key = pregnant, value = population) %>%
  rename(
    not_preg = no,
    preg = yes)

## # A tibble: 2 x 3
##   sex    not_preg  preg
##   <chr>     <dbl> <dbl>
## 1 female       12    10
## 2 male         20    NA

What do the extra and fill arguments do in separate()? Experiment with the various options for the following two toy datasets.

tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
  separate(x, c("one", "two", "three"), extra = "merge")

## # A tibble: 3 x 3
##   one   two   three
##   <chr> <chr> <chr>
## 1 a     b     c    
## 2 d     e     f,g  
## 3 h     i     j

tibble(x = c("a,b,c", "d,e,f,g", "h,i,j")) %>% 
  separate(x, c("one", "two", "three"), extra = "drop")

## # A tibble: 3 x 3
##   one   two   three
##   <chr> <chr> <chr>
## 1 a     b     c    
## 2 d     e     f    
## 3 h     i     j

tibble(x = c("a,b,c", "d,e", "f,g,i")) %>% 
  separate(x, c("one", "two", "three"), fill = "right")

## # A tibble: 3 x 3
##   one   two   three
##   <chr> <chr> <chr>
## 1 a     b     c    
## 2 d     e     <NA> 
## 3 f     g     i

tibble(x = c("a,b,c", "d,e", "f,g,i")) %>% 
  separate(x, c("one", "two", "three"), fill = "left")

## # A tibble: 3 x 3
##   one   two   three
##   <chr> <chr> <chr>
## 1 a     b     c    
## 2 <NA>  d     e    
## 3 f     g     i

Both unite() and separate() have a remove argument. What does it do? Why would you set it to FALSE?

tibble(x = c("a,b,c", "d,e,f", "g,h,i")) %>% 
  separate(x, c("one", "two", "three"), remove = F)

## # A tibble: 3 x 4
##   x     one   two   three
##   <chr> <chr> <chr> <chr>
## 1 a,b,c a     b     c    
## 2 d,e,f d     e     f    
## 3 g,h,i g     h     i

Compare and contrast separate() and extract(). Why are there three variations of separation (by position, by separator, and with groups), but only one unite?

extract() は文字列の区切りに正規表現を使う。 一つの文字列を区切る方法は複数あるが、複数の列をまとめる方法は一つしかないから。

Compare and contrast the fill arguments to spread() and complete().

complete()は非明示的な欠損値を推測してNAで埋めるが、fill()は明示的な欠損値を近くのセルの値で埋める

What does the direction argument to fill() do?

どの方向に接しているセルの値で欠損値を埋めるかを指定する。

In this case study I set na.rm = TRUE just to make it easier to check that we had the correct values. Is this reasonable? Think about how missing values are represented in this dataset. Are there implicit missing values? What’s the difference between an NA and zero?

na.rm = TRUE にしてしまうとexplicitだった欠損値がimplicitになってしまう。

What happens if you neglect the mutate() step? (mutate(key = stringr::str_replace(key, “newrel”, “new_rel”)))

who4 <- who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = T) %>%
  separate(key, c("new", "type", "sexage"), sep = "_")

## Warning: Expected 3 pieces. Missing pieces filled with `NA` in 2580 rows
## [73467, 73468, 73469, 73470, 73471, 73472, 73473, 73474, 73475, 73476,
## 73477, 73478, 73479, 73480, 73481, 73482, 73483, 73484, 73485, 73486, ...].

who[73467,]

## # A tibble: 1 x 60
##   country iso2  iso3   year new_sp_m014 new_sp_m1524 new_sp_m2534
##   <chr>   <chr> <chr> <int>       <int>        <int>        <int>
## 1 <NA>    <NA>  <NA>     NA          NA           NA           NA
## # ... with 53 more variables: new_sp_m3544 <int>, new_sp_m4554 <int>,
## #   new_sp_m5564 <int>, new_sp_m65 <int>, new_sp_f014 <int>,
## #   new_sp_f1524 <int>, new_sp_f2534 <int>, new_sp_f3544 <int>,
## #   new_sp_f4554 <int>, new_sp_f5564 <int>, new_sp_f65 <int>,
## #   new_sn_m014 <int>, new_sn_m1524 <int>, new_sn_m2534 <int>,
## #   new_sn_m3544 <int>, new_sn_m4554 <int>, new_sn_m5564 <int>,
## #   new_sn_m65 <int>, new_sn_f014 <int>, new_sn_f1524 <int>,
## #   new_sn_f2534 <int>, new_sn_f3544 <int>, new_sn_f4554 <int>,
## #   new_sn_f5564 <int>, new_sn_f65 <int>, new_ep_m014 <int>,
## #   new_ep_m1524 <int>, new_ep_m2534 <int>, new_ep_m3544 <int>,
## #   new_ep_m4554 <int>, new_ep_m5564 <int>, new_ep_m65 <int>,
## #   new_ep_f014 <int>, new_ep_f1524 <int>, new_ep_f2534 <int>,
## #   new_ep_f3544 <int>, new_ep_f4554 <int>, new_ep_f5564 <int>,
## #   new_ep_f65 <int>, newrel_m014 <int>, newrel_m1524 <int>,
## #   newrel_m2534 <int>, newrel_m3544 <int>, newrel_m4554 <int>,
## #   newrel_m5564 <int>, newrel_m65 <int>, newrel_f014 <int>,
## #   newrel_f1524 <int>, newrel_f2534 <int>, newrel_f3544 <int>,
## #   newrel_f4554 <int>, newrel_f5564 <int>, newrel_f65 <int>

newrelを含む行ではsexageにNAが挿入されてしまっている

I claimed that iso2 and iso3 were redundant with country. Confirm this claim.

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = T) %>%
  select(country, iso2, iso3) %>%
  distinct_all() %>%
  nrow()

## [1] 219

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = T) %>%
  select(country, iso2, iso3) %>%
  distinct(country) %>%
  nrow()

## [1] 219

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = T) %>%
  select(country, iso2, iso3) %>%
  distinct(iso2) %>%
  nrow()

## [1] 219

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = T) %>%
  select(country, iso2, iso3) %>%
  distinct(iso3) %>%
  nrow()

## [1] 219

country, iso2, iso3は全て対応している

For each country, year, and sex compute the total number of cases of TB. Make an informative visualisation of the data.

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = T) %>%
  mutate(key = stringr::str_replace(key, "newrel", "new_rel")) %>%
  separate(key, c("new", "type", "sexage"), sep = "_") %>%
  select(-new, -iso2) %>%
  separate(sexage, c("sex", "age"), sep = 1) %>%
  group_by(iso3) %>%
  summarise(sum(cases)) %>%
  joinCountryData2Map(joinCode="ISO3", nameJoinColumn="iso3") %>%
  mapCountryData(nameColumnToPlot="sum(cases)", catMethod="fixedWidth", mapTitle = "TB cases of the World", addLegend = TRUE)

## 215 codes from your data successfully matched countries in the map
## 4 codes from your data failed to match with a country code in the map
## 28 codes from the map weren't represented in your data

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = TRUE) %>%
  mutate(key = stringr::str_replace(key, "newrel", "new_rel")) %>%
  separate(key, c("new", "type", "sexage"), sep = "_") %>%
  select(-new, -iso2) %>%
  separate(sexage, c("sex", "age"), sep = 1) %>%
  group_by(year) %>%
  summarise(sum_cases = sum(cases)) %>%
  ggplot(mapping = aes(x = year, y = sum_cases)) +
           geom_line()

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = TRUE) %>%
  mutate(key = stringr::str_replace(key, "newrel", "new_rel")) %>%
  separate(key, c("new", "type", "sexage"), sep = "_") %>%
  select(-new, -iso2) %>%
  separate(sexage, c("sex", "age"), sep = 1) %>%
  group_by(year) %>%
  summarise(countries = n_distinct(country)) %>%
  ggplot(mapping = aes(x = year, y = countries)) +
           geom_line()

who %>% 
  gather(new_sp_m014:newrel_f65, key = "key", value = "cases", na.rm = TRUE) %>%
  mutate(key = stringr::str_replace(key, "newrel", "new_rel")) %>%
  separate(key, c("new", "type", "sexage"), sep = "_") %>%
  select(-new, -iso2) %>%
  separate(sexage, c("sex", "age"), sep = 1) %>%
  group_by(sex) %>%
  summarise(sum_cases = sum(cases)) %>%
  ggplot(mapping = aes(x = sex, y = sum_cases)) +
           geom_point()