[Concept Exercise] Cheese Club

config.json

@@ -213,6 +213,20 @@
         ],
         "status": "wip"
       },
+      {
+        "slug": "cheese-club",
+        "name": "Cheese Club",
+        "uuid": "c9b29ac8-cb91-4808-89ce-81b2358bd586",
+        "concepts": [
+          "dataframes"
+        ],
+        "prerequisites": [
+          "lists",
+          "loops",
+          "functions"
+        ],
+        "status": "wip"
+      },
       {
         "slug": "captains-log",
         "name": "Captains Log",

exercises/concept/cheese-club/.docs/hints.md

@@ -0,0 +1,26 @@
+# Hints
+
+## 1. Classify customers
+
+- Using [`every()`][ref-every] with an anonymous function may be convenient.
+
+## 2. Name customers
+
+- Processing two inputs in parallel: think about [`map2()`][ref-map2] as a possible approach.
+
+## 3. Separate out emphatic customers
+
+- Earlier tasks will help you.
+- Which customers do you want to [`keep`][ref-keep] in the output list?
+
+## 4. Change ratings to binary
+
+- There are several ways to solve this, but [`if_else()`][ref-if_else] from the `dplyr` library may be simplest. This was discussed in the [Switch Concept][concept-switch].
+- Within `purrr`, another possibility is [`map_int()`][ref-map_int].
+
+[ref-every]: https://purrr.tidyverse.org/reference/every.html
+[ref-map2]: https://purrr.tidyverse.org/reference/map2.html
+[ref-keep]: https://purrr.tidyverse.org/reference/keep.html
+[ref-if_else]: https://dplyr.tidyverse.org/reference/if_else.html
+[ref-map_int]: https://purrr.tidyverse.org/reference/map.html
+[concept-switch]: https://exercism.org/tracks/r/concepts/switch

exercises/concept/cheese-club/.docs/instructions.md

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+# Instructions
+
+We are starting up a cheese club, which will use ML to select new cheeses to offer our cheese-loving customers based on their histories and tastes.
+
+New members are required to fill out an initial survey so we can gather some basic data to start with.
+Through this, it's been found that there is a subset of emphatic clients who lack nuance in their critiques.
+As this can end up irrevocably biasing a more nuanced algorithm, there is a separate algorithm set up to handle their needs.
+You are asked to provide some helper functions to wrangle their data.
+
+~~~~exercism/note
+While there may be different ways to solve the following tasks, each can be solved with no more than two higher order functions.
+~~~~
+
+## 1. Classify customers
+
+The rating system has a five star basis, which is simply consists of integers `1:5`.
+The emphatic customers will only give ratings of `1` or `5`, and we want to know if a customer exhibits this behavior.
+
+Implement `all_15()` which takes a vector of ratings and returns `true` if all ratings are either `1` or `5`, and false otherwise.
+
+```R
+all_15(c(2, 3, 4, 4, 1))
+#> [1] FALSE
+
+all_15(c(1, 5, 5, 1, 5))
+#> [1] TRUE
+```
+
+## 2. Name customers
+
+We need to associate the customer name (or ID) with their ratings.
+
+Implement `name_customers()`, which takes a vector of names and a list of ratings, and "zips" them into a list of lists.
+
+Inner list elements should have names `name` and `rating`.
+
+```R
+names <- c("c1", "c2")
+ratings <- list(c(2, 3, 5), c(1, 1, 5))
+name_customers(names, ratings)
+#> [[1]]
+#> [[1]]$name
+#> [1] "c1"
+
+#> [[1]]$rating
+#> [1] 2 3 5
+
+#> [[2]]
+#> [[2]]$name
+#> [1] "c2"
+
+#> [[2]]$rating
+#> [1] 1 1 5
+```
+
+## 3. Separate out emphatic customers
+
+We need to separate the more emphatic customers from the others.
+
+Implement `emphatics()` which takes customers and ratings.
+Returns a list restricted to those who only use only `1` or `5` star ratings.`
+
+```R
+# names and ratings as in the task 2 example
+
+emphatics(names, ratings)
+#> [[1]]
+#> [[1]]$name
+#> [1] "c2"
+
+#> [[1]]$rating
+#> [1] 1 1 5
+```
+
+## 4. Change ratings to binary
+
+Since the emphatic customers only use `1` and `5` ratings, it will more computationally convenient if these are changed these to `0` and `1`.
+
+Implement `to_binary()` which takes vector of emphatic ratings.
+Returns binary ratings, where `1` has been changed to `0` and `5` has been changed to `1`.
+
+```R
+to_binary(c(1, 1, 5, 5, 1))
+#> [1] 0 0 1 1 0
+```