是Udacity課程的第一個專案。

先從巨集觀把握一下思路，目的是做一個比較德州撲克大小的問題
首先，先抽象出一個處理的函式，它根據返回值的大小給出結果。

之後我們在定義如何比較兩個或者多個手牌的大小，為方便比較大小，我們先對5張牌進行預處理，將其按照降序排序，如下：

def card_ranks(hand):

    ranks = ['--23456789TJQKA'.INDEX(r) for r, s in hand]
    ranks.sort(reverse=True)
    return ranks

然後我們可以枚舉出一共有9種情況，並用數字代表每一種情況的等級，利用Python的比較功能，將等級放在第一位，如果等級相同，那麼再比較後面的。

def hand_rank(hand):
    "Return a value indicating the ranking of a hand."
    ranks = card_ranks(hand) 
    if straight(ranks) and flush(hand):
        return (8, max(ranks))
    elif kind(4, ranks):
        return (7, kind(4, ranks), kind(1, ranks))
    elif kind(3, ranks) and kind(2, ranks):
        return
 
 (6, kind(3, ranks), kind(2, ranks))
    elif flush(hand):
        return (5, ranks)
    elif straight(ranks):
        return (4, max(ranks))
    elif kind(3, ranks):
        return (3, kind(3, ranks), ranks)
    elif two_pair(ranks):
        return (2, two_pair(ranks), ranks)
    elif kind(2, ranks):
        return
 
 (1, kind(2, ranks), ranks)
    else:
        return (0, ranks)

可以看到，如果等級相同，接下來比較的是每套牌中牌的大小了。同時我們需要三個函式，代表同花，順子，以及kind(n, ranks),代表ranks有n張牌的點數。這裡的三個函式實現非常巧妙，利用了set去重的特性。

def straight(ranks):
    return (max(ranks) - min(ranks)) == 4 and len(set(ranks)) == 5

def flush(hand):
    suit = [s, for r, s in hand]
    return len(set(suit)) == 1
def kind(n, ranks):
    for s in ranks:
        if ranks.count(s) == n : return s
    return None

我們發現，有一種情況是含有兩個對，於是需要一個函式來判斷是否是這種情況，這個函式中呼叫了kind()函式，由於kind()函式滿足短路特性，只會返回先得到的滿足情況的點數，於是將其翻轉後，在呼叫一邊kind,若得到的結果相同，那麼就只有一個對（或者沒有），否則就有兩個。

def two_pairs(ranks):
    pair = kind(2, ranks)
    lowpair = kind(2, list(reverse(ranks)))
    if pair != lowpair:
        return (pair, lowpair)
    else:
        return None

好了，整體的骨架算是搭完了，接下來處理會產生bug的情況，首先是A2345，當排序時由於A被算作14，所以針對這個問題需要單獨列一個if

處理A是最低：
def card_ranks(hand):
    ranks = ['--23456789TJQKA'.INDEX(r) for r, s in hand]
    ranks.sort(reverse=True)
    return  [5, 4, 3, 2, 1] if (ranks = [14, 5, 4, 3, 2] else ranks

之後就是進一步的簡化了，思路挺好的

def poker(hands):
    return allmax(hands, key=hand_ranks)


def allmax(iterable, key=None):
    result, maxval = [], None
    ket = key or lambda(x): x
    for x in iterable:
        xval = key(x)
        if not result or xval > maxval:
            result, maxval = [x], xval
        elif:
            result.append(x)
    return result
"""大於就取代，等於就加入，小於不作處理"""

import random

mydeck = [r+s for r in '23456789TJKQA' for s in'SHDC]
def deal(numhands, n=5, deck = [r+s for r in '23456789TJKQA' for s in'SHDC]):
    random.shuffle(deck)
    return [deck[n*i:n*(i + 1)] for i in range(numhands)]



def hand_ranks(hand):
    groups = group['--23456789TJQKA'.index(r) for r, s in hand]
    counts, ranks = unzip(groups)
    if rnaks == (14, 5, 4, 3, 2, 1):
        ransk = (5, 4, 3, 2, 1)

    straight = len(ranks) == 5 and max(ranks) - min(ranks) == 4
    flush = len(set([s for r, s in hand])) ==1
    return(9 if (5,) == count else
          8 if straight and flush else
          7 if (4, 1) == counts else
          6 if (3, 2) == counts else
          5 if flush else
          4 if straight else
          3 if (3, 1, 1) == counts else
          2 if (5, 1, 1) == counts else
          1 if (2, 1, 1, 1) == counts else
          0), ranks
def group(items):
    groups = [(items.count(x), x) for x in set(items)]
    return sorted(groups, reverse = True)
def unzips(pairs):return zip(*pairs)

def hand_ranks(hand):
     groups = group['--23456789TJQKA'.index(r) for r, s in hand]
    counts, ranks = unzip(groups)
    if rnaks == (14, 5, 4, 3, 2, 1):
        ransk = (5, 4, 3, 2, 1)

    straight = len(ranks) == 5 and max(ranks) - min(ranks) == 4
    flush = len(set([s for r, s in hand])) ==1
    return max(count_ranks[counts], 4*straight + 5 * flush), ranks
count_rankings = {(5,):10, (4, 1):7, (3,2):6, (3,1,1):3, (2,2,1):2,
(2,1,1,1): 1,(1,1,1,1,1):0}

總結下，面對一個問題的思維步驟：

started:understand problems look at specification See if it make sense
define the piece of problem reuse the piece you have test! >explore
最後是是的程式在各個方面達到均衡
correctness elegance efficienct featrues