Working thru the Gradebook coding challenge and struggling with #8 (You decided to switch from a numerical grade value to a Pass/Fail option for your poetry class. Find the grade value in your gradebook for your poetry class and use the .remove() method to delete it).
Thought I had a grasp on the use of “remove”, but turns out that is wrong.
My current code for gradebook is gradebook is gradebook = [[“physics”, 98], [“calculus”, 97], [“poetry”, 85], [“history”, 88]].
Output is “[[‘physics’, 98], [‘calculus’, 97], [‘poetry’, 85], [‘history’, 88], [‘computer science’, 100], [‘visual arts’, 98]]” as computer science and Visual arts have both been added.
To remove the Poetry score, shouldn’t the following work: “gradebook.remove[2][1]”?
Certainly possible I am way off though. As of two weeks I had never coded, so please have mercy…
Thanks, Ted
If someone answers your question, please mark their response as a solution
.remove does not work that way (and you would need the argument for the function between ( and ) instead of between [ and ]
If you have this list: listA = ['a', 'b', 'c', 'd']
and you want to remove 'c' (which is at index 2)
you would do listA.remove('c')
It’s more complicated with a list of lists (a 2D list) listB = [ ['a', 'b', 'c', 'd'], ['x', 'y', 'z'] ]
If I want to remove 'c', I first have to access the list that contains it;
the list that contains 'c' is an element of listB ;
the list that contains 'c' is the first element of listB, so its at an index of 0 for listB
Therefore the list that contains 'c' is listB[0] .
I’d remove 'c' from this list by doing listB.remove('c') .
Similarly, the list containing ‘poetry’ is gradebook[2]
which is ["poetry", 85]
so to remove the 85, you would do gradebook[2].remove(85)
