模拟分发扑克牌（python实现）

最新推荐文章于 2023-10-27 14:07:37 发布

原创最新推荐文章于 2023-10-27 14:07:37 发布 · 2.4w 阅读

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本文介绍了一种使用Python实现的自动发牌程序，该程序能够自动生成一副扑克牌，并将其随机分配给四位玩家，每位玩家各得13张牌。发牌后，程序还会将每位玩家手中的牌按花色及大小顺序排列。

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52张扑克牌发个4个玩家，每人13张。
要求：
自动生成一幅扑克牌组；洗牌；发牌到玩家手中；将玩家手中扑克牌按花色大小整理好。
思路一

import random
import operator
def auto():
    pokers=[]
    poker=[]
    for i in ['♥','♠','♦','♣']:
        for j in ['A','2','3','4','5','6','7','8','9','10','J','Q','K']:
            poker.append(i)
            poker.append(j)
            pokers.append(poker)
            poker=[]
    return pokers
poker=auto()
random.shuffle(poker)
li={}
for k in ['player1','player2','player3','player4']:
    b=random.sample(poker,13)    
    for s in b:
        poker.remove(s)
    li.setdefault(k,b)        
print('player1：',sorted(li['player1'],key=operator.itemgetter(0,1)))
print('player2：',sorted(li['player2'],key=operator.itemgetter(0,1)))   
print('player3：',sorted(li['player3'],key=operator.itemgetter(0,1)))
print('player4：',sorted(li['player4'],key=operator.itemgetter(0,1)))

思路二

import random
import time
A=['♥','♠','♦','♣']
B=['A','2','3','4','5','6','7','8','9','10','J','Q','K']
poker=[]
pokers=[]
n=1
for i in A:
    for j in B:
         pokers.append((n,(i+j)))
         n=n+1
print("开始洗牌....")
random.shuffle(pokers)
def xipai(x):
    for i in x:
        pokers.remove(i)
    return pokers
def fapai(y):
    for i in y:
        print(i[1],',',end=" ")
def paixu(z):
    for i in z:
        print(i[1],',',end=" ")
time.sleep(3)
a=random.sample(pokers,13) 
pokers=xipai(a)   
print("开始给player1发牌：\n")
print(fapai(a))
b=random.sample(pokers,13) 
pokers=xipai(b)   
print("开始给player2发牌：\n")
print(fapai(b))
c=random.sample(pokers,13) 
pokers=xipai(c)   
print("开始给player3发牌：\n")
print(fapai(c))
d=random.sample(pokers,13) 
pokers=xipai(d)   
print("开始给player4发牌：\n")
print(fapai(d))

a.sort()
b.sort()
c.sort()
d.sort()

time.sleep(3)
print("player1的牌：\n")
print(paixu(a))
print("player2的牌：\n")
print(paixu(b))
print("player3的牌：\n")
print(paixu(c))
print("player4的牌：\n")
print(paixu(d))