相当简单的问题，我想......我刚刚安装了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，然后检查它们是否已添加：测试功能和过程中的列表。
#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
不确定最新情况......任何帮助都会受到赞赏。
下面的工作代码：转换为python 3.x
#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")
答案 0 :(得分：1)
示例中有多个错误，让我们一起来看看。首先，如果你想要一个列表，那么你需要定义它，即：
l = ['joe'] # this is a list
s = {'joe'} # this is a set
现在，由于您使用的是Python 2，因此始终建议使用raw_input代替input。后者将对获取的字符串应用eval，因此它将输入评估为Python代码。出于安全原因，您通常不希望这样（我知道这是一个例子）。所以，让我们将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，实际上将字符串转换为数字。现在您需要执行相同操作，但使用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，则没有为其定义此类方法has_key。如果有问题的对象是dict，那将会有效，但事实并非如此。在给定代码中检查包含的正确方法是something in A。使用set时执行此操作比使用list时效率更高，代码保持不变（除非您需要将append更改为add）。
您现在可以调整代码吗？
答案 1 :(得分：1)
使用firstlist，你可以混合列表的概念和字典的概念。看起来你只想要一个清单......
firstlist = ['Joe']
，而不是使用has_key，写
if printthis in firstlist:
答案 2 :(得分：1)
#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"
答案 3 :(得分：0)
以下是回复中的代码