Question

我试图在python中选择2D列表中的随机元素。我正在创造一个二十一点游戏。

我知道这段代码有多长时间没有优化，但我是编程新手，我想犯错误去学习。

这是我设置的初始代码。我创造了西装（黑桃，俱乐部，心形，钻石）。我将它附加到名为list_of_cards的列表中，该列表已初始化为空数组。

然后我有另一个名为player_card的列表供玩家查看。这里是参考。

list_of_suits= []       #Creating list of cards
player_card  = []

king_of_spades  =  10   #Creating face cards for the spades deck.
queen_of_spades =  10   
jack_of_spades  =  10

king_of_clubs  =  10   
queen_of_clubs =  10      
jack_of_clubs  =  10

king_of_hearts  =   10   
queen_of_hearts =   10    
jack_of_hearts  =   10

king_of_diamonds  =  10   
queen_of_diamonds =  10 
jack_of_diamonds  =  10

ace_of_spades =   [1,11]     # Aces are either 1 or 11
ace_of_clubs =    [1,11]
ace_of_hearts =   [1,11]
ace_of_diamonds = [1,11] 


spades = [ace_of_spades,2,3,4,5,6,7,8,9,10, jack_of_spades, queen_of_spades, king_of_spades]
clubs =  [ace_of_clubs,2,3,4,5,6,7,8,9,10, jack_of_clubs, queen_of_clubs, king_of_clubs]
hearts =  [ace_of_hearts,2,3,4,5,6,7,8,9,10, jack_of_hearts, queen_of_hearts, king_of_hearts]
diamonds =  [ace_of_diamonds,2,3,4,5,6,7,8,9,10, jack_of_diamonds, queen_of_diamonds, king_of_diamonds]

list_of_suits.append(spades)
list_of_suits.append(clubs)
list_of_suits.append(hearts)
list_of_suits.append(diamonds)

这是随机卡的选择器。它将遍历数组以随机选择四张卡中的一张。接下来，它将进入该阵列并选择随机卡。

random_list_of_suits = random.choice(list_of_suits)

random_card_number = random.choice(random_list_of_suits)    
random_list_of_suits.remove(random_card_number)
player_card.append(random_card_number)

print player_card
print list_of_suits

以下是我想弄清楚的问题：如何每次创建一个新的随机数？

我有点卡住，我尝试通过for循环，但是如果我把random_card_number放在一个循环中，它将选择它最初做了四次的相同随机卡。

Answer 1

import random

list2d = [range(0, 5), range(5, 10), range(10, 15), range(15, 20)]

for i in range(0, 10):
    random_suite = random.choice(list2d)
    random_card = random.choice(random_suite)
    print random_suite, random_card

输出：

[5, 6, 7, 8, 9] 7
[5, 6, 7, 8, 9] 5
[5, 6, 7, 8, 9] 5
[10, 11, 12, 13, 14] 11
[10, 11, 12, 13, 14] 13
[10, 11, 12, 13, 14] 12
[10, 11, 12, 13, 14] 14
[10, 11, 12, 13, 14] 10
[10, 11, 12, 13, 14] 10
[0, 1, 2, 3, 4] 2

Answer 2

您在random.choice（）中输入的列表必须是一维数组。所以你可以做到这一点，

random_list_of_suits = np.random.choice(list_of_suits) # you can get spades/clubs/hearts/diamonds
index_of_random_card_number = random.choice(len(random_list_of_suits))
random_card = random_list_of_suits[index_of_random_card_number] # You get a random card

Answer 3

#include <stdafx.h>
#include <iostream>
#include <cstdlib> // for rand() and srand()
#include <ctime> // for time()
using namespace std;

int main()
{

    srand(time(0)); // set initial seed value to system clock
    for (int nCount=0; nCount < 100; ++nCount)
    {
        cout << rand() << "\t";

        if ((nCount+1) % 5 == 0)
            cout << endl;
    }

srand使用时间作为第一个初始种子来创建随机数，因此每次运行程序时，您将获得不同的随机数，因为进样种子将随时间变化。

如何从2d列表中选择随机元素

3 个答案: