How can I let the output return True or False based on the input of the list?

Question

So I have a list hand whose data is something like this ['AS', 'AD', 'AC', 'CH', 'CS']. Now I have to find the first character of each element in the list and see if there are 3 identical letters and 2 identical letters or not.

Note: they must all be in order though.. like always x,x,x,y,y or y,y,x,x,x. Any other combination, like x,y,y,x,x or x,y,x,y,x, will return False

So for example

['AS', 'AD', 'AC', 'CH', 'CS'] returns True because there are 3 A and 2 C.

['AS', 'SD', 'SC', 'CH', 'CS'] returns False because the order is not right

['CS', 'CD', 'AC', 'AH', 'AS'] returns True because there are 3 A and 2 C

['AS', 'CD', 'DC', 'AH', 'CS'] returns False because none of the characters appear 3 times and 2 times.

Here is my code so far, which doesn't work...

hand = ['AS', 'AD', 'AC', 'CH', 'CS']
updated_hands = list([x[1] for x in hand])
if ((updated_hands[0] == updated_hands[1] == updated_hands[2]) or (updated_hands[2] == updated_hands[3] == updated_hands[4])) and ((updated_hands[3] == updated_hands[4]) or (updated_hands[0] == updated_hands[1])):
    print('True')
else:
    print('False')

What changes should I make to this?

Why doesn't it work?

defladamouse
– defladamouse

2021-12-08 14:30:31 +00:00
Commented Dec 8, 2021 at 14:30 — defladamouse
– defladamouse, Commented Dec 8, 2021 at 14:30
The first character is x[0], right?

doctorlove
– doctorlove

2021-12-08 14:32:25 +00:00
Commented Dec 8, 2021 at 14:32 — doctorlove
– doctorlove, Commented Dec 8, 2021 at 14:32

TheFlyingObject · Accepted Answer · 2021-12-08 14:48:06Z

3

Here is something that should work for arbitrary lengths and orders:

hands = [['AS', 'AD', 'AC', 'CH', 'CS']]
hands.append(['AS', 'SD', 'SC', 'CH', 'CS'])
hands.append(['CS', 'CD', 'AC', 'AH', 'AS'])
hands.append(['AS', 'CD', 'DC', 'AH', 'CS'])

for hand in hands:

    condition = True

    last_count = 1
    last_item = hand.pop(0)[0]

    while hand:
        item = hand.pop(0)

        if (item[0] == last_item):
            last_count += 1 

        if (last_item != item[0]):
            if (last_count == 2) or (last_count == 3):
                condition = True
            else:
                condition = False
                break

        last_item = item[0]

    print(condition)

Outputs:

True
False
True
False

answered Dec 8, 2021 at 14:48

TheFlyingObject

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Comments

defladamouse · Accepted Answer · 2021-12-08 14:40:57Z

This is the approach I would take, which is effectively the same as what you've tried but I've just expanded it a bit to try and make it more readable. I think probably what you've got is "correct", but the complexity of doing it as a one liner is hiding a small bug.

def test_list(hand):
    test1 = all(let[0] == hand[0][0] for let in hand[:3])
    test2 = all(let[0] == hand[0][0] for let in hand[:2])
    test3 = all(let[0] == hand[2][0] for let in hand[2:])
    test4 = all(let[0] == hand[3][0] for let in hand[3:])
    
    if (test1 and test3) or (test2 and test4):
        print("True")
    else:
        print("False")

hand1 = ['AS', 'AD', 'AC', 'CH', 'CS']
hand2 = ['AS', 'SD', 'SC', 'CH', 'CS']
hand3 = ['CS', 'CD', 'AC', 'AH', 'AS']
hand4 = ['AS', 'CD', 'DC', 'AH', 'CS']

test_list(hand1) #prints True
test_list(hand2) #prints False
test_list(hand3) #prints True
test_list(hand4) #prints True

Akash Nagtilak · Accepted Answer · 2021-12-09 06:46:40Z

0

    Lets Make it Simple.

    hand = ['AS', 'AD', 'AC', 'CH', 'CS']
    a = [sorted(han[0][0]) for han in hand] 
    
    import itertools
    b = itertools.chain(*a)
    c = list(b)
    
    d = [han[0][0] for han in hand] 
    
    if c == d:
    print(True)
    else:
    print(False)
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answered Dec 9, 2021 at 6:46

Akash Nagtilak

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