Python实现比较扑克牌大小程序代码示例

时间:2022-11-07 07:51:19

是Udacity课程的第一个项目。

先从宏观把握一下思路,目的是做一个比较德州扑克大小的问题
首先,先抽象出一个处理的函数,它根据返回值的大小给出结果。

之后我们在定义如何比较两个或者多个手牌的大小,为方便比较大小,我们先对5张牌进行预处理,将其按照降序排序,如下:

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

然后我们可以枚举出一共有9种情况,并用数字代表每一种情况的等级,利用Python的比较功能,将等级放在第一位,如果等级相同,那么再比较后面的。

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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,如果等级相同,接下来比较的是每套牌中牌的大小了。同时我们需要三个函数,代表同花,顺子,以及kind(n, ranks),代表ranks有n张牌的点数。这里的三个函数实现非常巧妙,利用了set去重的特性。

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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

我们发现,有一种情况是含有两个对,于是需要一个函数来判断是否是这种情况,这个函数中调用了kind()函数,由于kind()函数满足短路特性,只会返回先得到的满足情况的点数,于是将其翻转后,在调用一边kind,若得到的结果相同,那么就只有一个对(或者没有),否则就有两个。

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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,整体的骨架算是搭完了,接下来处理会产生bug的情况,首先是A2345,当排序时由于A被算作14,所以针对这个问题需要单独列一个if

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处理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

之后就是进一步的简化了,思路挺好的

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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
"""大于就取代,等于就加入,小于不作处理"""
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}

总结下,面对一个问题的思维步骤:

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started:understand problems look at specification See if it make sense
define the piece of problem reuse the piece you have test! >explore
最后是是的程序在各个方面达到均衡
correctness elegance efficienct featrues

总结

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原文链接:http://blog.csdn.net/u013795429/article/details/49823715