CODEWITHSUNDEEP
pythonlistdictionaryindexing
Student J
Student J

Reputation: 203

Python 3: List inside dictionary, list index out of range

I have created a dictionary where the entries are lists like this:

new_dict
{0: ['p1', 'R', 'p2', 'S'], 1: ['p3', 'R', 'p4', 'P'], 2: ['p5', 'R', 'p6', 'S'], 3:   ['p7', 'R', 'p8', 'R'], 4: ['p9', 'P', 'p10', 'S'], 5: ['p11', 'R', 'p12', 'S'], 6: ['p13', 'S', 'p14', 'S']}

From here I am trying to check that the elements inside the lists are in my list below Moves. For example, in new_dict[0], I want to check that the 1st element and the third element in Moves, if not, raise the class exception. (This is a snippet of the code.)

class NoStrategyError(Exception): pass

j=0
while j < len(new_dict):
    i = 0
    while i < 4:
        # Write the code for the NoSuchStratgyError 
        Moves = ['R', 'S', 'P', 'r', 's', 'p']
        if new_dict[j][1+4*i] not in Moves or new_dict[j][3+4*i] not in Moves:
            raise NoStrategyError("No such strategy exists!")
        i+=1
    j+=1

Now here is the problem, when I run this, I get the following error:

if new_dict[j][1+4*i] not in Moves or new_dict[j][3+4*i] not in Moves: IndexError: list index out of range

What does this mean?

Is there a better way to write the inner while loop? And change it instead to a for loop? Something like, for elem in new_dict[j]?

Upvotes: 0

Views: 9146

Answers (1)

blehman
blehman

Reputation: 1970

Notice that in the first interation of your nested loop, we see the following values that are both in Moves:

>>>new_dict[0][1], new_dict[0][3] 
('R', 'S')

However, on your second iteration in the nested loop, you are trying to evaluate terms that are not included in the dictionary:

>>>new_dict[1][5], new_dict[1][7]
IndexError: list index out of range

Notice that new_dict[1] only has 4 elements: 

>>>new_dict[1]
['p3', 'R', 'p4', 'P']

So you can only reference new_dict[1][0],new_dict[1][1],new_dict[1][2],new_dict[1][3]:

>>>new_dict[1][0],new_dict[1][1],new_dict[1][2],new_dict[1][3]
('p3', 'R', 'p4', 'P')

Upvotes: 1

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