Question

我正在做基本的Python卡技巧，我搜索的内容太复杂了，这是我的代码：

import random
import itertools
deck = ["S A","S 2","S 3","S 4","S 5","S 6","S 7","D A","D 2","D 3","D 4","D 
5","D 6","D 7","C A","C 2","C 3","C 4","C 5","C 6","C 7"]
random.shuffle(deck)
for z in range(3):
    print("Pile ",[z+1])
    for i in range(7):
        print(deck[i])

我想要做的就是得到它，当我得到我的3堆时，我不想要任何重复的值，但我不知道我是如何做到的，或者我是如何得到它们的3个单独的栏目？

Answer 1

使用

piles=[deck[0::3],deck[1::3],deck[2::3]]

此构建3列表分别从位置0,1和2开始，从一个项目跳到另一个项目，长度为3.无需知道初始列表的大小。

首先选择0,3,6，......
第二次选择第1,4,7项......
第三个拿起第2,5,8，......

Answer 2

代码中的轻微修改。

df_noyau_yes

Answer 3

试试这个：

import random

deck = ["S A","S 2","S 3","S 4","S 5","S 6","S 7","D A","D 2","D 3","D 4","D 5","D 6","D 7","C A","C 2","C 3","C 4","C 5","C 6","C 7"]
data = {}
for i in range(3):
    data[i] = []
    for j in range(7):
        data[i].append(deck.pop(random.randint(0,len(deck)-1)))
print(data)

Answer 4

Randex！正如你在问题中提到的那样，我并没有想到制作3列，但这可能是你有用的：

import random

deck = [
  "S A", "S 2", "S 3", "S 4", "S 5", "S 6", "S 7",
  "D A", "D 2", "D 3", "D 4", "D 5", "D 6", "D 7",
  "C A", "C 2", "C 3", "C 4", "C 5", "C 6", "C 7"
  ]

hands = {}

for stack in range(3):
  hands[stack] = []

您也可以稍后为.append（）分配一个随机生成的值，然后从列表中删除以获得更多代码可读性

  for card in range(7):
    hands[stack].append(deck.pop(random.randint(0, len(deck) - 1)))

“牌”将从牌组中分配一张随机牌，直到每个“牌”牌都有7张牌叠加

for stack in hands:
  print(stack, '-', hands[stack])

如何从21个唯一值列表中获取3个唯一值列表？

4 个答案: