human
I found an interesting problem on https://www.codewars.com/kata/5739174624fc28e188000465:
A famous casino is suddenly faced with a sharp decline of their revenues. They decide to offer Texas hold'em also online. Can you help them by writing an algorithm that can rank poker hands? 

## Task

Create a poker hand that has a method to compare itself to another poker hand:

```python
compare_with(self, other_hand)
```

A poker hand has a constructor that accepts a string containing 5 cards:
```python
PokerHand("KS 2H 5C JD TD")
```

The characteristics of the string of cards are:
* Each card consists of two characters, where
 * The first character is the value of the card: `2, 3, 4, 5, 6, 7, 8, 9, T(en), J(ack), Q(ueen), K(ing), A(ce)`
 * The second character represents the suit: `S(pades), H(earts), D(iamonds), C(lubs)`
* A space is used as card separator between cards

The result of your poker hand compare can be one of these 3 options:

```python
[ "Win", "Tie", "Loss" ]
```

## Notes
* Apply the [Texas Hold'em](https://en.wikipedia.org/wiki/Texas_hold_%27em) rules for ranking the cards.
* Low aces are **NOT** valid in this kata.
* There is no ranking for the suits.

If you finished this kata, you might want to continue with [Sortable Poker Hands](https://www.codewars.com/kata/sortable-poker-hands)

I tried it in Python, but could not do it. Can you solve it?

 ... Give me a Python code to compare itself to another poker hand:?
assistant
#Here is Python code to compare itself to another poker hand:
.

class PokerHand(object):

    def __init__(self, hand):
        self.st = ['2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A']
        self.h = sorted(hand.split(), key=lambda x: self.st.index(x[0]))
        self.v=lambda i,j :"Win" if self.st.index(i) > self.st.index(j) else 'Loss'
    
    def compare_with(self, o):
        o = sorted(o.h, key=lambda x: self.st.index(x[0]))
        if o == self.h : return 'Tie'
        s,s1 = [i[0] for i in o],[i[0] for i in self.h]
        a = [self.straight_flush, self.four_of_kind, self.full_house, self.flush, self.straight, self.three_of_kind,self.two_pair, self.one_pair]
        b = [0, 1, 2, self.flush_, self.straight_,5,6, 7]
        for u, j in enumerate(zip(a, b)):
            k,l = j[0],j[1]
            i,j = k(self.h),k(o)
            if i != j : return 'Win' if i else 'Loss'
            elif i == j == 1 : 
                if u in [3,4] : return l(s1,s) 
                return self.common(s1, s)
        for i, j in zip(s1[::-1], s[::-1]):
            if i != j : return self.v(i,j)
        return 'Tie'
    
    def straight_flush(self, p):
        return len(set([i[1] for i in p])) == 1 and all(abs(self.st.index(p[i][0]) - self.st.index(p[i + 1][0])) == 1 for i in range(len(p) - 1))

    def four_of_kind(self, p):
        p = [i[0] for i in p]
        return any(p.count(i) == 4 for i in p)

    def full_house(self, p):
        p = [i[0] for i in p]
        return any(p.count(i) == 3 for i in p) and any(p.count(i) == 2 for i in p)

    def flush(self, p):
        return len(set([i[1] for i in p])) == 1

    def straight(self, p):
        return all(abs(self.st.index(p[i][0]) - self.st.index(p[i + 1][0])) == 1 for i in range(len(p) - 1))

    def three_of_kind(self, p):
        p = [i[0] for i in p]
        return any(p.count(i) == 3 for i in p)

    def two_pair(self, p):
        p = [i[0] for i in p]
        return [p.count(i) for i in set(p)].count(2) == 2

    def one_pair(self, p):
        p = [i[0] for i in p]
        return [p.count(i) for i in...