撲克程式 Python__Python

是Udacity課程的第一個項目。

先從宏觀把握一下思路，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數，它根據傳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張牌進行預先處理，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數，代表同花，順子，以及kind(n, ranks),代表ranks有n張牌的點數。這裡的三個函數實現非常巧妙，利用了set去重的特性。

def straight(ranks):    return (max(ranks) - min(ranks)) == 4 and len(set(ranks)) == 5def flush(hand):    suit = [s, for r, s in hand]    return len(set(suit)) == 1def 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，於是需要一個函數來判斷是否是這種情況，這個函數中調用了kind()函數，由於kind()函數滿足短路特性，只會返回先得到的滿足情況的點數，於是將其翻轉後，在調用一邊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 randommydeck = [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), ranksdef 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), rankscount_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