Python 3中的用户输入和排序列表

你几乎做了所有正确的事情,但问题是你在哪里打破了循环。您首先要附加名称,然后检查输入是否只是enter。所以,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()

希望有帮助。

这是实现逻辑的一种方法。

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)]))

解释

• 仅使用 list.append ,则, list.sort 并打印姓名 name 不在中 x 。
• 您似乎希望最多有4个名称。那样的话 break 什么时候 len(x) 达到4。
• 您可以将列表理解用于 str.format 用于最终输出。

您不需要对代码进行太多更改。主要的重组包括将所有异常(如名称重复和无输入)移动到循环的顶部。评论中包含了更多的解释。

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()

我个人更喜欢使用 set 为了以更干净的方式避免重复,我还将两个打印部分合并到一个单独的函数中。因此,我的建议如下,其行为与您的略有不同,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,排序(此处仅用于 print_members 如果成员数量增长过多,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。

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

您可以使用以下方法对列表进行就地排序(列表将替换为已排序的列表) x.sort() .要检查列表中是否有内容,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 . . .

这个 getPositionByBinarySearch 未实现,因此您可以自己实现它。

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:]

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()