相关疑难解决方法(0)

我正在尝试创建一个函数,将多个变量与一个整数进行比较,并输出一个由三个字母组成的字符串.我想知道是否有办法将其翻译成Python.所以说:

x = 0
y = 1
z = 3
mylist = []

if x or y or z == 0 :
    mylist.append("c")
if x or y or z == 1 :
    mylist.append("d")
if x or y or z == 2 :
    mylist.append("e")
if x or y or z == 3 : 
    mylist.append("f")

这将返回一个列表

["c", "d", "f"]

这样的事情可能吗？

我正在编写一个拒绝访问未授权用户的安全系统.

import sys

print("Hello. Please enter your name:")
name = sys.stdin.readline().strip()
if name == "Kevin" or "Jon" or "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

它按预期授予对授权用户的访问权限,但它也允许未经授权的用户访问!

Hello. Please enter your name:
Bob
Access granted.

为什么会这样？我明确表示,只有在与nameKevin,Jon或Inbar相同时才授予访问权限.我也尝试过相反的逻辑if "Kevin" or "Jon" or "Inbar" == name,但结果是一样的.

def Forest(Health,Hunger):
    print'You wake up in the middle of the forest'
    Inventory = 'Inventory: '
    Squirrel =  'Squirrel'
    while True:
        Choice1 = raw_input('You...\n')
        if Choice1 == 'Life' or 'life':
            print('Health: '+str(Health))
            print('Hunger: '+str(Hunger))
        elif Choice1 == 'Look' or 'look':
            print 'You see many trees, and what looks like an edible dead Squirrel, \na waterfall to the north and a village to the south.'
        elif Choice1 == 'Pickup' or 'pickup':
            p1 = raw_input('Pickup what?\n')
            if p1 == Squirrel:
                if Inventory == …