Python 3 中的用户输入和排序列表

Question

我需要帮助生成这样的输出：

1. Enter a member: samantha 
Names: ['Samantha']
2. Enter a member: susan 
Names: ['Samantha', 'Susan']
3. Enter a member: billy 
Names: ['Billy', 'Samantha', 'Susan']
4. Enter a member: billy
4. Enter a member: samantha
4. Enter a member: Jason 
Names: ['Billy', 'Jason', 'Samantha', 'Susan']
5. Enter a member: 

Members:
1. Billy
2. Jason
3. Samantha
4. Susan

我已经努力创建一个可以执行此操作的程序，但无济于事。我将在代码本身的问题中发表评论。提前感谢您的帮助。

def function():

    x = []
    #a = "1." # I tried to number the questions by adding these but it doesnt work
    while True:
        #a += 1
        name = input("Enter name: ").title()
        x.append(name)

        print("Names:", x)
        #if name == name: # this condition is made so that an input that is typed in again doesn't get included in the list
            #name = input("Enter name: ").title()

            # where should I add an insert() here to have the list alphabetically ordered?

        if not name: # pressing enter prints out the second half of the output, listing the members
            #print("\nMembers:", x).sort()
            break

function()

Answer 1

你做的几乎所有事情都是正确的，但问题是你在哪里打破了循环。您首先附加名称，然后检查输入是否只是一个输入。因此，首先检查输入，然后将其附加到列表中。 这是我的代码：

def fun1():
    l = []
    while True:
        s = input("Enter a member: ")
        if not s:
            break
        l.append(s.title())
        print("Names:", l)
    l.sort()
    print("Members:")
    for i in range(0, len(l)):
        print(i+1,end = '')
        print(".", l[i])
fun1()

希望对你有帮助。

Answer 2

这是实现逻辑的一种方式。

def function():

    x = []

    while True:

        name = input('{0}. Enter a member: '.format(len(x)+1)).title()

        if not name in x:
            x.append(name)
            x.sort()
            print('Names:', x)

        if len(x) == 4:
            break

    print('Members:')
    print('\n'.join(['{0}. {1}'.format(i, j) for i, j in enumerate(x)]))

说明

• 仅在name 不在x 中时使用list.append、list.sort 和打印名称。
• 看来您最多需要 4 个名称。在这种情况下，当len(x) 达到 4 时，break。
• 您可以使用带有str.format 的列表推导式作为最终输出。

Answer 3

您不需要对代码进行很多更改。主要重组涉及将所有异常（例如名称重复和无输入）移至循环顶部。更多解释集成在 cmets 中。

def function():
    x = []
    #no need to keep track of the number of members, the list length will give us this information
    #infinite loop that will be interrupted, when no input is given 
    while True:
        #determine length of list, +1 because Python index starts with 0
        n = len(x) + 1
        #ask for input, format prints the number into the bracket position
        name = input("{}. Enter a member: ".format(n))
        #check if name already in list
        if name in x:
            #if name in list, ignore the input and start while loop again
            continue
        #no input - print out members and stop
        if not name: 
            print("Members:")
            #get for each member the index number i
            for i, member in enumerate(x):
                #print index in position {0} and member name in position {1}
                print("{0}. {1}".format(i + 1, member))
            #leave function()
            return
        #append name and sort list
        x.append(name)
        x.sort()        
        #print list
        print("Names:", x)


function()

Answer 4

我个人更喜欢使用set 以避免以更简洁的方式重复，并且我还将两个打印部分合并为一个单独的函数。所以我的建议是针对以下内容，它的作用与你的有点不同，但（在我看来）更简单：

def print_members(members):
    numbered_members = enumerate(sorted(members), 1)
    print("Members:", ", ".join(
        "{}. {}".format(*tup) for tup in numbered_members))


def collect_members():
    members = set()
    while True:
        next_member_num = len(members) + 1
        name = input("{}. Enter a member: ".format(next_member_num)).title()
        if name:
            members.add(name)
            print_members(members)
        else:
            print_members(members)
            return

请注意，如果成员数量增长过多，排序（这里只需要 print_members 函数）将变得非常昂贵，在这种情况下，我建议使用二叉搜索树而不是集合。

Answer 5

我保留了你的大部分结构，但修复了很多东西。为了获得编号，我使用了 % 字符串格式化运算符。你也可以使用 str.format，which many seem to prefer。

您可以使用x.sort() 对列表进行就地排序（列表被排序列表替换）。要检查某些内容是否在列表中，请使用 thing is in mylist 或 thing is not in mylist

def function():

    x = []
    a = 1
    while True:
        prompt = '%d. Enter a member: ' % a
        name = input(prompt)
        name = name.title() # convert first letter to uppercase

        if name.strip() == '':  # enter or empty string entered
            print()
            print('Members:')
            for idx, item in enumerate(x):
                print('%d. %s' % (idx+1, item))
            break
        elif name not in x:
            x.append(name)
            x.sort()  # sort x in place
            print("Names: ", x)
            a += 1


function()

和我的输出：

1. Enter a member: samantha
Names:  ['Samantha']
2. Enter a member: susan
Names:  ['Samantha', 'Susan']
3. Enter a member: billy
Names:  ['Billy', 'Samantha', 'Susan']
4. Enter a member: Jason
Names:  ['Billy', 'Jason', 'Samantha', 'Susan']
5. Enter a member:

Members:
1. Billy
2. Jason
3. Samantha
4. Susan


------------------
(program exited with code: 0)

Press any key to continue . . .

Answer 6

getPositionByBinarySearch 没有实现，你可以自己实现。

def getPositionByBinarySearch(arr, name):
    """
    implement binary search to find the index at which insertion should happen)
    return -1 if insertion is not needed (exact match is found)
    """

def func():
    arr = []
    while True:
        name = input(str(len(x)+1) + '. Enter a number: ')
        if not name:
            print "Members:"
            for i, member in enumerate(members):
                print(str(i+1) + '. ' + member)
            break;
        pos = getPositionByBinarySearch(arr, name)
        if (pos != -1):
            arr = arr[:i] + [name] + arr[i:]

Answer 7

def fun1():
    l = []
    while True:
        s = input("Enter a name or '-1' to quit: ")
        if s=='-1':
            break
        l.append(s)
    l.sort()
    print("The sorted names or numbers(whatever) are:")
    print(l)
fun1()