尝试根据字典中的值设置变量

Trying to set a variable from a value in a dictionary

我正在学习 python 并且正在制作剪刀石头布游戏。

我卡在了一部分上。

我目前有 4 个变量(虽然我想减少到 2 个)

他们分别在字典中查找 Key 和 Value。

choice = {1:'rock',  2:'paper',  3:'scissors'}

我遇到的问题是使用变量从字典中获取键。

这是给我带来麻烦的代码片段

    print('--- 1 = Rock    2 = Paper     3 = Scissors --- ')
    pKey = input() # this is sets the key to the dictionary called choice
    while turn == 1: # this is the loop to make sure the player makes a valid choice
        if pKey == 1 or 2 or 3:
            pChoice = choice.values(pKey)  # this calls the value from choice dict and pKey variable
            break
        else:
            print('Please only user the numbers 1, 2 or 3 to choose')

    comKey = random.randint(1, 3)  # this sets the computer choices
    comChoice = choice.values(comKey)

具体比较麻烦的地方是

 pChoice = choice.values(pKey)

 comChoice = choice.values(comKey)

我尝试了所有我知道的方法,包括使用括号、尝试不同的方法和使用不同的格式。

很想学这个!谢谢!

你不知道如何从字典中获取元素,你的代码应该是这样的:

import random
choice = {1: 'rock',  2: 'paper',  3: 'scissors'}

print('1 = Rock\t2 = Paper\t3 = Scissors')
pKey = int(input())
if pKey in (1, 2, 3):
    pChoice = choice[pKey]
else:
    print('Please only user the numbers 1, 2 or 3 to choose')
    pChoice = 'No choice'

comKey = random.randint(1, 3)
comChoice = choice[comKey]
print(pChoice, comChoice)

适合我。

听起来你只是想查字典

pKey = 1
pChoice = choices[pKey]  # rock

dict.values 用于创建包含字典所有值的列表(实际上是一个 dict_values 对象)。它不用作查找。


就您的代码结构而言,它可能需要一些工作。 rock/paper/scissors 选择对于 Enum 来说是完美的,但现在你可能有点难以接受。让我们尝试作为顶层模块常量。

ROCK = "rock"
PAPER = "paper"
SCISSORS = "scissors"

def get_choice():
    """get_choice asks the user to choose rock, paper, or scissors and
    returns their selection (or None if the input is wrong).
    """
    selection = input("1. Rock\n2. Paper\n3. Scissors\n>> ")
    return {"1": ROCK, "2": PAPER, "3": SCISSORS}.get(selection)

将它们作为常量寻址可确保它们在您的代码中处处相同,否则您会得到一个非常明确的 NameError(而不是 if 分支未执行,因为您 if comChoice == "scisors"


带有枚举的最小示例如下所示:

from enum import Enum

Choices = Enum("Choices", "rock paper scissors")

def get_choice():
    selection = input(...)  # as above
    try:
        return Choices(int(selection))
    except ValueError:
        # user entered the wrong value
        return None

您可以通过使用更详细的枚举定义来扩展它,并教每个 Choice 实例如何计算获胜者:

class Choices(Enum):
    rock = ("paper", "scissors")
    paper = ("scissors", "rock")
    scissors = ("rock", "paper")

    def __init__(self, loses, beats):
        self._loses = loses
        self._beats = beats

    @property
    def loses(self):
        return self.__class__[self._loses]

    @property
    def beats(self):
        return self.__class__[self._beats]

    def wins_against(self, other):
        return {self: 0, self.beats: 1, self.loses: -1}[other]

s, p, r = Choices["scissors"], Choices["paper"], Choices["rock"]
s.wins_against(p)  # 1
s.wins_against(s)  # 0
s.wins_against(r)  # -1

不幸的是,没有很好的方法来丢失其中的抽象(每次调用时将 "paper" 抽象为 Choices.paper)因为你不知道 Choices["paper"] 是什么时候 Choices.rock 被实例化。