Question

我试图创建一个菜单驱动的程序，它将生成0到9之间的5个随机整数，并将它们存储在List中。我想让它然后允许用户输入一个整数，然后搜索List，报告List中整数的位置（如果找到），否则返回-1。然后将搜索结果显示在屏幕上并重新显示菜单。

def main():
    choice = displayMenu()
    while choice != '4':
        if choice == '1':
            createList()
        elif choice == '2':
            print(createList)
        elif choice == '3':
            searchList()
        choice = displayMenu()

    print("Thanks for playing!")

def displayMenu():
    myChoice = '0'
    while myChoice != '1' and myChoice != '2' \
                  and myChoice != '3' and myChoice != '4':
         print ("""Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit
                        """)
         myChoice = input("Enter option-->")

         if myChoice != '1' and myChoice != '2' and \
            myChoice != '3' and myChoice != '4':
             print("Invalid option. Please select again.")

    return myChoice


import random

def linearSearch(myList):
    target = int(input("--->"))
    for i in range(len(myList)):
        if myList[i] == target:
            return i
        return -1


#This will generate the five random numbers from 0 to 9
def createList():
    newList = []
    while True:
        try:
            num = input("Give me five numbers: ")
            num = [int(num) for num in input().split(' ')]
            print(num)
            if any([num < 0 for num in a]):
                Exception

            print("Thank you")
            break
        except:
            print("Invalid. Try again...")

    for i in range(5):
        newList.append(random.randint(0,9))
    return newList


#Option two to display the list
def displayList():
    myList = newList
    print("Your list is: ", newList)



#Option 3 to search the list
def searchList():
    target = int(input("--->"))
    result = linearSearch(myList,target)
    if result == -1:
        print("Not found...")
    else:
        print("Found at", result)




main()

然而，当它提示用户输入五个数字时，无论你输入什么内容，它都会显示无效。有人可以举个例子说明我如何解决这个问题吗？

Answer 1

小心输入解析。将输入保留为字符串并处理字符串，或者最初将其转换为int并继续处理整数。

myChoice = str(input())
# input 1
if myChoice in ['1', '2', '3', '4']:
  print(myChoice)

或

myChoice = int(input())
# input 1
if myChoice in [1, 2, 3, 4]:
  print(myChoice)

Answer 2

在createList中，您将input两次，然后引用未定义的变量a

num = input("Give me five numbers: ")
num = [int(num) for num in input().split(' ')]
...
if any([num < 0 for num in a]):
     Exception

应该是

text_input = input("Give me five numbers: ")
numbers = [int(num) for num in text_input.split(' ')]
if any([num < 0 for num in numbers]):
    raise ValueError

你的程序有许多其他的奇怪之处，但也许你还在研究这些：

您不会对createList中输入的数字执行任何操作，而是返回随机数列表。您的描述还建议您想要一个随机数列表，那么为什么要输入数字？
您创建变量myList和newList，但除非您将其声明为global或明确将其作为参数传递，否则您的函数将无法访问这些变量
linearSearch可以使用list.index更简单地编写：

def linearSearch(myList, target):
    if target in myList:
        return myList.index(target)
    else:
        return -1

Answer 3

好的 - 我在多个地方发现了逻辑和语法错误。我会尽量列出它们。

在displayMenu()函数中注释掉了redundent选项块。
修复了main（）函数。您没有捕获返回的值/列表功能
将myList定义为需要传递的全局列表
linearSearch(myList,target)有额外的用户输入和调整逻辑上班。它只适用于在第一个索引处查找数字。理想情况下，您可以使用list.index(value)方法获取索引＃
在createList()函数
调整函数以接收全局列表以使其正常工作。

我尽可能少地调整了你的代码。注释掉不需要的行。我相信它按预期工作。

工作代码：

def main():
    myList = []
    choice = displayMenu()
    while choice != '4':
        if choice == '1':
            myList =createList()
        elif choice == '2':
            #print(myList)
            displayList(myList)
        elif choice == '3':
            searchList(myList)
        choice = displayMenu()

    print("Thanks for playing!")

def displayMenu():
    myChoice = '0'
    if myChoice != '1' and myChoice != '2' and \
            myChoice != '3' and myChoice != '4':
         print ("""Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit
                        """)
         myChoice = str(input("Enter option-->"))
         """

         if myChoice != 1 and myChoice != 2 and \
            myChoice != 3 and myChoice != 4:
             print("Invalid option. Please select again.")
            """

    return myChoice


import random

def linearSearch(myList,target):
    #target = int(input("--->"))
    found = False
    for i in range(len(myList)):
        if myList[i] == target:
            found = True
            return i

    if found == False:
        return -1


#This will generate the five random numbers from 0 to 9
def createList():
    print "Creating list..............."
    newList = []
    """
    while True:
        try:
            num = input("Give me five numbers: ")
            num = [int(num) for num in input().split(' ')]
            print(num)

            if any([num < 0 for num in a]):
                Exception

            print("Thank you")
            break

        except:
            print("Invalid. Try again...")

    """
    for i in range(5):        
        newList.append(random.randint(0,9))
        #print newList
    return newList


#Option two to display the list
def displayList(myList):
    #myList = newList
    print("Your list is: ", myList)



#Option 3 to search the list
def searchList(myList):
    target = int(input("--->"))
    #print("Searching for" , target)
    result = linearSearch(myList,target)
    if result == -1:
        print("Not found...")
    else:
        print("Found at", result)

main()

输出

>>> 
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->1
Creating list...............
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->2
('Your list is: ', [1, 4, 6, 9, 3])
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->3
--->4
('Found at', 1)
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->9
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->10
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->3
--->10
Not found...
Please choose
                        1. Create a new list of 5 integers
                        2. Display the list
                        3. Search the list
                        4. Quit

Enter option-->4
Thanks for playing!
>>>

最后的笔记，您可能希望从简单的事情开始，例如创建一个菜单，为不同的选择打印不同的句子，看它是否有效，然后根据您的意愿复杂化。

Answer 4

更改

myChoice = input("Enter option-->")

到

myChoice = str(input("Enter option-->"))

函数输入返回一个整数，与str的比较总是返回False。

验证用户输入时出错

4 个答案: