问题描述
我想先按字母顺序(俱乐部,钻石,心脏,黑桃)对一手牌进行排序,然后再按等级(从王牌到王牌)排序。我有以下代码(很长且效率很低),但输出似乎错误。我觉得这与我使用的冒泡排序有关。我在这里想念的是什么,什么是这样做的更好方法?对于我的最终输出,我还将将2、11、12和13转换回Ace,Jack,Queen和King,但是我真的不想重做我最初将它们转换为数值的操作。 I / O应该如下所示:
样本输入:[('h','q'),('c','j'),('d','2'),('s','3'),('h ','5'),('d','j'),('d','10')]
样本输出:[['c','j'),('d','2'),('d','10'),('d','j'),('h ','5'),('h','q'),('s','3')]
def sort_hand(hand):
"""
Sorts the a hand of cards by the value of each suit/type of card.
"""
# A=1,2,3,4,5,6,7,8,9,10,J=11,Q=12,K=13
# UPDATING CARDS SO THAT They CAN BE ORDERED BY RANK
new_hand = []
# GIVING NUMERIC VALUES TO JACK,QUEEN,KING,AND ACE
for element in hand:
if element[1] == 'j':
val = element[1].replace('j','11')
new_element = (element[0],val)
new_hand.append(new_element)
elif element[1] == 'q':
val = element[1].replace('q','12')
new_element = (element[0],val)
new_hand.append(new_element)
elif element[1] == 'k':
val = element[1].replace('k','13')
new_element = (element[0],val)
new_hand.append(new_element)
elif element[1] == 'a':
val = element[1].replace('a','1')
new_element = (element[0],val)
new_hand.append(new_element)
else:
pass
new_hand.append(element)
# BUBBLE SORT USED TO ORDER BY RANK
for ix in range(1,len(new_hand)):
value_to_sort = new_hand[ix][1]
while new_hand[ix-1][1] > value_to_sort and ix > 0:
new_hand[ix],new_hand[ix-1] = new_hand[ix-1],new_hand[ix]
ix -= 1
# MAKE SUBLISTS FOR EACH SUIT
c_list = []
d_list = []
h_list = []
s_list = []
for element in new_hand:
if element[0] == 'c':
c_list.append(element)
elif element[0] == 'd':
d_list.append(element)
elif element[0] == 'h':
h_list.append(element)
else:
s_list.append(element)
# COMBINE ORDERED SUIT SUBLISTS TO MAKE FINAL SORTED HAND BY RANK
final_hand = c_list + d_list + h_list + s_list
return final_hand
>>> hand = [('h','q'),('c','j'),('d','2'),('s','3'),('h','5'),'10')]
>>> sort_hand(hand)
[('c','11'),'10'),'12'),'3')]
我觉得我与这段代码很接近,但是不确定从这里开始。希望能得到一个解释清楚的答案,谢谢。
解决方法
使用lambda进行双键排序。
<?php
define('DB_HOST','localhost');
define('DB_USER','mydb_admin');
define('DB_PASS','SomePass');
define('DB_NAME','my_db');
function db_connect()
{
$db = mysqli_connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);
confirm_db_connect();
return $db;
}
function confirm_db_connect()
{
if (mysqli_connect_errno()) {
$msg = 'Database connection failed: ';
$msg .= mysqli_connect_error();
$msg .= ' (' . mysqli_connect_errno() . ')';
exit($msg);
}
}
function db_disconnect()
{
if (isset($db)) {
mysqli_close($db);
}
}
$db = db_connect();
输出:
rank = ['2','3','4','5','6','7','8','9','10','j','q','k','a']
suit = ['c','d','h','s']
inp = [('h','q'),('c','j'),('d','2'),('s','3'),('h','5'),'10')]
outp = sorted(inp,key = lambda x:
(suit.index(x[0]),rank.index(x[1])))
print(outp)
,
您可以使用内置的sort
或sorted
函数来实现此目的
使用上述功能需要的其他功能是比较器逻辑,可以使用key
参数来提供。
def sort_hand(inp):
##### I have created the order of the suites arbitrarily
suite_map = {
'h':1,'c':2,'s':3,'d':4
}
rank_map = {
'a':1
'k':2
'q':3
'j':4
'10':5
'9':6
'8':7
'7':8
'6':9
'5':10
'4':11
'3':12
'2':13
'1':14
}
return suite_map[inp[0]],rank_map[inp[1]]
O / P --->
>>> hand
[('h','10')]
>>> sorted(hand,key=sort_hand)
[('h','10'),'2')]
您可以根据您的订单根据suite_map
和rank_map
控制词典的顺序,并进行相应的排序