Python - 简单的列表函数给出名称错误

Question

非常简单的问题，我想......我从字面上刚刚安装了Python，并正在测试一些初学者教程。

我想创建一个菜单，允许您将项目添加到列表中，然后检查是否添加了它们：测试函数和过程中的列表。

#create empty list and define variables 
firstlist = {'joe'} 
additem = "test" 
printthis = "test" 

#create menu, add or check name 
def menu(): 
    #print what options you have 
    print "Add to list: type '1'" 
    print "Check for name: type '2'" 
    print "To exit program: type '3'" 
    return input ("Choose your option: ") 

def addmenu(): 
    additem = input("Name of list item: ") 
    firstlist.append(additem) 
    print additem, "has been appended" 

def checkmenu(): 
    printthis = input ("What are you looking for?: ") 
    if firstlist.has_key(printthis): 
     print "is in the list" 
    else: 
     print "is not in the list" 

# Perform action 
loop = 1 
choice = 0 
while loop == 1: 
    choice = menu() 
    if choice == 1: 
     addmenu() 
    elif choice == 2: 
     checkmenu() 
    elif choice == 3: 
     loop = 0 
    elif choice > 3: 
     print "You made an incorrect selection" 

继承人我的错误：

Traceback (most recent call last): 
    File "C:\Python27\testing python\tutorials\dictionaryselection", line 32, in <module> 
    addmenu() 
    File "C:\Python27\testing python\tutorials\dictionaryselection", line 15, in addmenu 
    additem = input("Name of list item: ") 
    File "<string>", line 1, in <module> 
NameError: name 'TESTING' is not defined 

不知道怎么回事...任何帮助，将不胜感激。下面

工作代码：转换到Python 3.x的

#create empty list and define variables 
firstlist = ['Joe'] 
additem = "test" 
printthis = "test" 

#create menu, add or check name 
def menu(): 
    #print what options you have 
    print ("") 
    print ("Add to list: type '1'") 
    print ("Check for name: type '2'") 
    print ("To list the whole list '3'") 
    print ("To exit program: type '4'") 
    print ("-------------------------") 
    return input ("Choose your option: ") 

def addmenu(): 
    additem = input("Name of list item: ") 
    firstlist.append(additem) 
    print (additem, "has been appended") 

def checkmenu(): 
    printthis = input("What are you looking for?: ") 
    if printthis in firstlist: 
     print ("is in the list") 
    else: 
     print ("is not in the list") 

def listlist(): 
    print (firstlist[1]) 

# Perform action 
loop = 1 
choice = 0 
while loop == 1: 
    choice = int(menu()) 
    if choice == 1: 
     addmenu() 
    elif choice == 2: 
     checkmenu() 
    elif choice == 3: 
     listlist() 
    elif choice == 4: 
     loop = 0 
    elif (choice > 4): 
     print ("You made an incoorect selection") 

Answer 1

有在本例中的多个错误，让我们通过他们去。首先，如果你想有一个列表，那么你需要将它定义为这样的，即：

l = ['joe'] # this is a list 
s = {'joe'} # this is a set 

现在，因为你使用Python 2，它总是建议代替input使用raw_input。后者将对获得的字符串应用eval，因此它将评估输入为Python代码。你通常不希望出于安全原因（我知道这是一个例子）。所以，我们将每个input更改为raw_input。现在的问题是，在使用eval时输入表示数字的字符串实际上会将字符串转换为数字。现在你需要做同样的事情，但是使用raw_input。由于您的选项仅限于整数值，因此解决方案为int(raw_input())。

第三个问题与has_key有关。如果使用的对象是set或list，那么没有为它们定义的方法has_key。如果有问题的对象是dict，那就行了，但事实并非如此。在给定的代码中检查遏制的正确方法是something in A。在使用set时使用set比使用list时效率更高，且代码保持不变（除非需要将append更改为add）。

你现在可以调整你的代码吗？

Answer 2

有了firstlist，你就可以混合列表的想法和字典的想法。它看起来像你只想要一个列表...

firstlist = ['Joe'] 

也，而不是使用对象的has_key，写

if printthis in firstlist: 

Answer 3

@瑞奇·梅森：我已经修改了代码位。在这里你需要一个字典对象而不是一个集合或一个列表。字典对象#since有一个键值对可以很容易地检索您在checkmenu中查找的值，并且在同一时间您可以从列表中删除值。我希望这有帮助。随意发布任何关于相同的进一步查询。

以下是答复如下

Answer 4

#create empty list and define variables 
firstlist = {} 
additem = "" 
printthis = "" 
removeitem = "" 
#create menu, add, remove or check name 
def menu(): 
    print "Add to list: type '1'" 
    print "Remove from the list: type '2'" 
    print "Check for name: type '3'" 
    print "To exit program: type '4'" 
    return input("Choose your option: ") 
def addmenu(): 
    additem = raw_input("Name of list item: ") 
    firstlist[additem] = additem 
    print additem, "has been appended" 
def removemenu(): 
    removeitem = raw_input("Name of list item: ") 
    firstlist.pop(removeitem) 
    print removeitem, " has been removed" 
def checkmenu(): 
    printthis = raw_input("What are you looking for?: ") 
    if firstlist.has_key(printthis): 
     print printthis, " is in the list" 
    else: 
     print printthis, " is not in the list" 

# Perform action 
loop = 1 
choice = 0 
while loop == 1: 
    choice = menu() 
    if choice == 1: 
     addmenu() 
    elif choice == 2: 
     removemenu() 
    elif choice == 3: 
     checkmenu() 
    elif choice == 4: 
     loop = 0 
    elif choice > 4: 
     print "You made an incorrect selection"