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Python Python Collections (2016, retired 2019) Sets Set Math

Posted by thomas2017
thomas2017
1,342 Points

Python, Collection, Set Math task 1 of 2 *Alternative way using for loop and if statements

I was wondering my thinking process would work? I know I was suppose to use intersection to solve this but this was the best that I could muster up before I looked up for the answer. How the heck were we suppose to solve this using the material from the video? When I ran it I got an error of 'dict' object has no attribute 'append'. Can someone please let me know why I got that error.

This was the answer someone else posted.

def covers(arg):
    hold = []
    for key, value in COURSES.items():
        if value & arg:
            hold.append(key)
    return hold
covers({'Ruby'})
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)
#x,y,z
    set1 = {}
    for item in topics:
        if item in "Python Basics":
            if "Python Basics" not in set1:
                set1.append(item)
            else: 
                continue
        elif item in "Java Basics":
            if "Java Basics" not in set1:
                set1.append(item)
            else:
                continue

        elif item in "PHP Basics":
            if "PHP Basics" not in set1:
                set1.append(item)
            else: 
                continue
        else:
            if "Ruby Basics" not in set1:
                set1.append(item)
            else:
                continue
    return set1

1 Answer

Steven Parker
Steven Parker
229,644 Points
Steven Parker
Steven Parker
229,644 Points

The other person's answer appends to a variable named "hold" which was initialized with "hold = []" :point_left: notice brackets. This makes it an (empty) list, and a list is what the challenge asked for as a result anyway.

In your code you are attempting to append to "set1" which was initialized with "set1 = {}" :point_left: notice braces. That makes it a dict instead of a list, and as you discovered, a dict has no "append" method.

So that explains that particular error. But also, tests like this probably don't do what you were expecting:

        if item in "Python Basics":

This does not check the item against the values associated with the dictionary key "Python Basics". What it actually does instead is check to see if the item is contained in the literal string "Python Basics". So if item contains "sic" the result would be True, but if item contains "arrays" the result would be False.

thomas2017
thomas2017
1,342 Points

Hi Steven thanks for the explanation. I was wondering what do you mean "sic" ? " So if item contains "sic" the result would be True, but if item contains "arrays" the result would be False."

Steven Parker
Steven Parker
229,644 Points
Steven Parker
Steven Parker
229,644 Points

The string "sic" would match part of "Python Basics" (note underlined substring). My point was that the code line is looking for matches in the words "Python Basics", not in the dictionary values associated with that key.