Resolved: Import counters (Part II), Exercises, Question 6

Hey Dmitriy,

Thank you for reaching out and for learning and practicing new skills with us!

To answer your question, let me draw your attention to the second for-loop in your code more specifically, the k iterator. The last thing that happens in the for-loop is that k becomes equal to three and that aux_list is appended to cards[mast[k]]. After this operation is completed, the following line is executed:

cards[mast[k]] = aux_list.append(value[v]).

While executing it, the value of k is still equal to three. Therefore, the code above corresponds to

cards[mast[3]] = aux_list.append(value[v])

which, in turn, corresponds to

cards['Hearts'] = aux_list.append(value[v]).

Now, note what you have on the right-hand side of this expression. Try and modify your code in the following way:

for v in range(len(value)):
    for k in range(len(mast))
        cards[mast[k]] = aux_list
    print(type(aux_list.append(value[v])))
    cards[mast[k]] = aux_list.append(value[v])
print(cards.items())

where we have added an additional print function. As you can confirm by executing the code, aux_list.append(value[v]) is of type NoneType. The reason for that is it is an operation – it doesn't bear any value. Therefore, every time you exit the second for-loop, you are assigning a NoneType to cards['Hearts']. This doesn't affect the
rest of the suits (Clubs, Diamonds and Spades) as they are handled inside the for-loop.

To obtain the desired result, modify your code in the following way:
for v in range(len(value)):
    for k in range(len(mast)):
        cards[mast[k]] = aux_list
    aux_list.append(value[v])
    cards[mast[k]] = aux_list
print(cards.items())

In order to truly excel in programming, practicing and studying the code are the way to go. I suggest you examine the original solution presented by Giles and compare to your own to see what the advantages and disadvantages of both are.

Good luck!

Kind regards,
365 Hristina