Help to write a function covers that returns a list of courses from a dict of courses?

Question

Let's write some functions to explore set math a bit more. We're going to be using this COURSES dict in all of the examples. Don't change it, though!
So, first, write a function named covers that accepts a single parameter, a set of topics. Have the function return a list of courses from COURSES where the supplied set and the course's value (also a set) overlap.
For example, covers({"Python"}) would return ["Python Basics"].
```sets.py
COURSES = {
    "Python Basics": {"Python", "functions", "variables",
                      "booleans", "integers", "floats",
                      "arrays", "strings", "exceptions",
                      "conditions", "input", "loops"},
    "Java Basics": {"Java", "strings", "variables",
                    "input", "exceptions", "integers",
                    "booleans", "loops"},
    "PHP Basics": {"PHP", "variables", "conditions",
                   "integers", "floats", "strings",
                   "booleans", "HTML"},
    "Ruby Basics": {"Ruby", "strings", "floats",
                    "integers", "conditions",
                    "functions", "input"}
}
def covers(topics):
    courselist = []
    for key in COURSES.keys():
        if topics == COURSES.values():
            courselist.extend(key)
    return course_list
```

mhjp · Answer

2 things of note for this challenge. 
The idea is to use Set Math to come up with the solution.
i.e from the teachers notes
```
| or .union(*others) - all of the items from all of the sets.
& or .intersection(*others) - all of the common items between all of the sets.
- or .difference(*others) - all of the items in the first set that are not in the other sets.

^ or .symmetric_difference(other) - all of the items that are not shared by the two sets.
```
courselist.extend() will iterate over the string and add each letter to the list 
i.e. courselist.extend() will output something like ['P', 'y', 't', 'h', 'o', 'n', ' ', 'B', 'a', 's', 'i', 'c', 's']
instead use course_list.append().
```python
def covers(topic):
    courselist = []
    for key in COURSES.keys():
        if COURSES[key] & topic:
            courselist.append(key)
    return course_list
```

Jeff Muday · Answer

You seem to be on the right track.  Though you cannot really use the COURSES.values() in an equivalence '==' expression. We will want to, instead, use the "in" operator.  This will allow us to see if the topic from the required_topics is contained in one of the key/value pairs of courses.
Note there is no single "right way" to do this challenge, but many different right ways to pass the challenge.
I find it is difficult to do this type of challenge without breaking down the COURSES into its component key/value pairs in a loop.  You should get into the habit of using the pattern below:
```python
this method gets all our key/value pairs into key and value variables from "my_dictionary"
for key, value in my_dictionary.items():
    print(key, value)
```
Finally, lets incorporate this idea into our function...
python
def covers(required_topics):
    matched_courses = set() # the set of our matched courses
    # make sure we check every topic in the required_topics list
    for topic in required_topics:
        # here is where we break down the course/topics key/value pairs from COURSES
        for course, topics in COURSES.items():
            # if the topic is in the course topics, add it to the set
            if topic in topics:
                matched_courses.add(course)
    return list(matched_courses)

Good luck with Python.  It is one of the best languages to learn!

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Kirome Thompson

Kirome Thompson

Help to write a function covers that returns a list of courses from a dict of courses?

2 Answers

Jeff Muday

Jeff Muday

mhjp

mhjp