初学者的问题：从Python中的函数返回一个布尔值 [英] Beginner question: returning a boolean value from a function in Python

问题描述

我试图让这个剪刀石头布的游戏要么返回一个布尔值，如集 player_wins 来真或假，这取决于如果玩家获胜，或完全重构这个code，以便它不使用whil​​e循环。
我是从世界的系统管理员那边传来，如果这是写在错误的风格，所以请温柔。
我已经尝试了一些事情，我理解TIMTOWTDI，而只是想一些投入。

感谢。

 进口随机全球player_wins
player_wins =无高清RPS（）：    player_score = 0
    cpu_score = 0    而player_score＆LT; 3 cpu_score＆LT; 3：        武器='摇滚'，'纸'，'剪刀'        对于i在范围（0，3）：
          打印％D％的％（I + 1，武器[I]）        玩家= INT（输入（选择1-3）） -  1
        CPU = random.choice（范围（0，3））        打印％s和％的％（武器[播放]武器[CPU]）
        如果CPU =玩家！
          如果（球员 -  CPU）％，第3版; （CPU  - 玩家）％3：
            player_score + = 1
            打印玩家赢得％d个游戏\\ n％player_score
          其他：
            cpu_score + = 1
            打印CPU赢得％d个游戏\\ n％cpu_score
        其他：
          打印扎！\\ n
RPS（）
 

我试图做这样的事情：

 打印％s和％的％（武器[播放]武器[CPU]）
    如果CPU =玩家！
      如果（球员 -  CPU）％，第3版; （CPU  - 玩家）％3：
        player_score + = 1
        打印玩家赢得％d个游戏\\ n％player_score
        如果player_score == 3：
            返回player_wins ==真
      其他：
        cpu_score + = 1
        打印CPU赢得％d个游戏\\ n％cpu_score
        如果cpu_score == 3：
            返回player_wins ==假
    其他：
      打印扎！\\ n
 

解决方案

忽略的重构问题，你需要了解的功能和返回值。你并不需要一个全球性的。永远。你可以这样做：

  DEF RPS（）：
    ＃code，以确定是否玩家获胜
    如果player_wins：
        返回True    返回False
 

然后，只需赋值给变量这一功能外，像这样：

  player_wins = RPS（）
 

这将您刚才调用的函数指定返回值（True或False）。

---

的意见后，我决定要补充的是习惯用法，这将是更好的EX从而pressed：

  DEF RPS（）：
     ＃code，以确定是否玩家赢，分配一个布尔值（TRUE或FALSE）
     ＃给变量player_wins。     返回player_wins PW = RPS（）
 

这赋予 player_wins 的布尔值（在函数内部）的 PW 变量之外的功能。

I'm trying to get this rock paper scissors game to either return a Boolean value, as in set player_wins to True or False, depending on if the player wins, or to refactor this code entirely so that it doesn't use a while loop. I'm coming from the sysadmin side of the world, so please be gentle if this is written in the wrong style. I have tried a few things, and I understand TIMTOWTDI, and would just like some input.

Thanks.

import random

global player_wins
player_wins=None

def rps():

    player_score = 0
    cpu_score = 0

    while player_score < 3 and cpu_score < 3:

        WEAPONS = 'Rock', 'Paper', 'Scissors'

        for i in range(0, 3):
          print "%d %s" % (i + 1, WEAPONS[i])

        player = int(input ("Choose from 1-3: ")) - 1
        cpu = random.choice(range(0, 3))

        print "%s vs %s" % (WEAPONS[player], WEAPONS[cpu])
        if cpu != player:
          if (player - cpu) % 3 < (cpu - player) % 3:
            player_score += 1
            print "Player wins %d games\n" % player_score
          else:
            cpu_score += 1
            print "CPU wins %d games\n" % cpu_score
        else:
          print "tie!\n"
rps()

I'm trying to do something like this:

   print "%s vs %s" % (WEAPONS[player], WEAPONS[cpu])
    if cpu != player:
      if (player - cpu) % 3 < (cpu - player) % 3:
        player_score += 1
        print "Player wins %d games\n" % player_score
        if player_score == 3:
            return player_wins==True
      else:
        cpu_score += 1
        print "CPU wins %d games\n" % cpu_score
        if cpu_score == 3:
            return player_wins==False
    else:
      print "tie!\n"

解决方案

Ignoring the refactoring issues, you need to understand functions and return values. You don't need a global at all. Ever. You can do this:

def rps():
    # Code to determine if player wins
    if player_wins:
        return True

    return False

Then, just assign a value to the variable outside this function like so:

player_wins = rps()

It will be assigned the return value (either True or False) of the function you just called.

---

After the comments, I decided to add that idiomatically, this would be better expressed thus:

 def rps(): 
     # Code to determine if player wins, assigning a boolean value (True or False)
     # to the variable player_wins.

     return player_wins

 pw = rps()

This assigns the boolean value of player_wins (inside the function) to the pw variable outside the function.

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