🔒 full house - how can I determine high card in 3 of a kind?

[spare bear]

as in title how can I determine high card in 3 of a kind
for example 22288, 44433

[neon arch]

@spare bear

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[spare bear]

determining just high card its easy

but I need restriction here

so would be same logic for rank of 3 of a kind

I thought also about sum cards in three of a kind

2+2+2, 4+4+4

but maybe high card in 3 of a kind better option

of course poker game texas holdem

I think I found solution it depends on position in hand so I can just hardcode highcard from index 0

so if 22288
then
value = hand[0]
in pseudocode

and elif 88222
then
value = hand[2]

I think ✅ solved

[vapid ember]

your question unfortunately is no good to someone who doesn't know the terminology you're using... for me, I can't help because I don't know what a 'high card' is

I thought it was the highest of the spare cards (not the 3 from the fuller part of the house)

[spare bear]

so high card is with higher value

for example A is higher than K

[vapid ember]

so all you're doing is looking for the highest value of all 5 cards?

[spare bear]

no not in 5 cards but subset of it

not like in case of two pair when you look at 5 ards to determine for example last card is high card

example 55JJA

and A is high card

[vapid ember]

I get the 'full house' - 3 cards of one value and 2 of another.
but when do you look at 'high card'? does a full house have a 'high card' or is this something entirely separate?

[spare bear]

full house so 3 of a kind and pair

[vapid ember]

yep, 3 of one value and 2 of another, that's the only part I understand. sorry 🙂

[spare bear]

yes I need to determine high card in 3 of a kind so for example 22288, 44433
so it will be respectively 2 and 4

[vapid ember]

from what you said here:

value = hand[0]
in pseudocode
and elif 88222
then
value = hand[2]
it looks like you are just trying to get the value of the 3 of a kind card?

[spare bear]

no I grab high card maybe in this way

its ok

ah yes right

[vapid ember]

I'm sure regex would have a good solution, otherwise perhaps:

from collections import Counter
print(Counter("44433").most_common())
# output: [('4', 3), ('3', 2)]

[woeful shard]

ranking hands may become easier if you rearrange the cards first

[spare bear]

so sort them

yes because I could have 28822 etc

[vapid ember]

I grouped them using a Counter, but yeah sorting on the value alone doesn't help? your 28822 could be 22288 or 88222 depending on sorting order, which isn't helpful because you still haven't identified the 3 of a kind

[spare bear]

so it must some custom sorting if I want 3 of a kind and pair

but in this case I have 2 cases

like I above typed

if and elif

22288 and 88222

but its same

2's over 8's

[vapid ember]

yeah, it's the same set of cards but you still haven't found the value of the 3 of a kind. My Counter does that.

[spare bear]

yes I will use it thanks

[vapid ember]

I imagine if you have a bunch of different possibilities (ie more than just full house) that regex might be the better option. Several regex patterns that act on the sorted card string and produce what you want.

[spare bear]

hmm but I dont have string I have list of tuples

[vapid ember]

You gave a string as your example... but still, a string could be easily gathered. A string is an iterable just like a list or tuple.

[spare bear]

yes sorry

Counter works on iterable not strings only?

[vapid ember]

class Counter(builtins.dict)
 |  Counter(iterable=None, /, **kwds)
 |
 |  Dict subclass for counting hashable items.  Sometimes called a bag
 |  or multiset.  Elements are stored as dictionary keys and their counts
 |  are stored as dictionary values.

[spare bear]

ok

[vapid ember]

In [8]: Counter((1,2,3,3))
Out[8]: Counter({1: 1, 2: 1, 3: 2})

[spare bear]

ah its bag, multiset

[vapid ember]

eh?

[spare bear]

I mean result or output

yes so shows frequency

[vapid ember]

That's the representation of it, but yeah it is a subclass of Dict.
It has methods that will output other formats, like .most_common() does

[spare bear]

ok I start with high card and back to full house

it helped

[vapid ember]

Otherwise you might be looking at something like this which would return None if not full house else the value of the 3 card set

import re

def get_three_of_a_kind_value(hand_string):
    # Sort the string to ensure it's in order
    sorted_hand = ''.join(sorted(hand_string))
    
    # Define the regex pattern for a full house
    pattern = r'^(.)\1{2}(.)\2$|^(.)\3(.)\4{2}$'
    
    # Match the pattern against the sorted hand
    match = re.match(pattern, sorted_hand)
    
    # If there's a match, return the three-of-a-kind value
    if match:
        # Check which part of the regex matched
        if match.group(1):
            # The first group is the three-of-a-kind when the pattern matches the first option
            return match.group(1)
        else:
            # The fourth group is the three-of-a-kind when the pattern matches the second option
            return match.group(4)
    
    return None

# Example usage
hand_string = "AKAKK"  # Sorting this gives "AAKKK"
three_of_a_kind_value = get_three_of_a_kind_value(hand_string)
print(f"The three of a kind is: {three_of_a_kind_value}")

I've got to shuffle on to something else, good luck with it.

[neon arch]

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