如何对齐/合并 2 个列表？ （Python）

Question

我真的不知道如何描述这一点，但在我可怕的解释之后，我将在预格式化文本中包含一个输出。 所以...我创建了三个列表，我需要使用两个，“employeeSold”和“employeeName”，但是我需要合并/对齐这两个列表，这样我就可以建立一个关于谁获得最多的排名系统，我想出了如何订购对于employeeSold，从最大到最重要，但我最不确定如何将正确的名称链接到正确的值，我认为它需要在订购之前出现，但我真的不知道输入什么，我已经花了想了几个小时。

---

这是我想要的代码的示例输出：

Ranking:
John - 5
Laura - 4
BOJO -1

它将每个整数 (employeeSold) 分配给每个字符串 (employeeName)，然后将用户插入的整数 (employeeSold) 从最大到最重要排序，然后在整数旁边打印他们的名字。

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这是我的代码：

def employee():
  employeeName = input("What is Employee's name?: ")
  employeeID = input("What is Employee's ID?: ")
  employeeSold = int(input("How many houses employee sold?: "))
  nameList.append(employeeName)
  idList.append(employeeID)
  soldList.append(employeeSold)
  nextEmployee = input("Add another employee? Type Yes or No: ")
  if nextEmployee == "2":
    employee()
  else:
    print("Employee Names:")
    print(", ".join(nameList))
    print("Employee's ID: ")
    print(", ".join(idList))
    print("Employee Sold:")
    print( " Houses, ".join( repr(e) for e in soldList ), "Houses" )
    print("Commission: ")
    employeeCommission = [i * 500 for i in soldList]
    print(", ".join( repr(e) for e in employeeCommission ), "" )
    print("Commission Evaluation: ")
    totalCommission = sum(employeeCommission)
    print(totalCommission)
    soldList.sort(reverse=True)
    print("Employee Ranking: ")
    ranking = (", ".join( repr(e) for e in soldList))
    print(ranking)
nameList = []
idList = []
soldList = []

employee()```
--------
Any help would be very much so appreciated.

Answer 1

您可以使用 zip 功能来合并列表，然后自定义键排序。 SortByName 函数返回排序键名称和 SortBySalary 函数返回键薪水

def SortByName(FullList):
    return FullList[0]

def SortBySalary(FullList):
    return FullList[2]

NamesList=['John','Laura','Bojo']
IdList=[1,2,5]
SalaryList=[2000,1000,5000]

ZippedList=zip(NamesList,IdList,SalaryList)
print ZippedList

# result
#[('John', 1, 2000), ('Laura', 2, 1000), ('Bojo', 5, 5000)]

ZippedList.sort(key=SortByName,reverse=True)
print ZippedList

# result
# [('Laura', 2, 1000), ('John', 1, 2000), ('Bojo', 5, 5000)]

ZippedList.sort(key=SortBySalary,reverse=True)
print ZippedList

# result
# [('Bojo', 5, 5000), ('John', 1, 2000), ('Laura', 2, 1000)]

如果你以后想按id对列表进行排序，只需实现id排序功能即可。

def SortByID(FullList):
    return FullList[1]

ZippedList.sort(key=SortByID)

Answer 2

您可以考虑将您的逻辑分解为多个函数和 sorting 一个 zip names 和 num_houses_sold 列表以打印排名正如你所指定的：

def get_int_input(prompt: str) -> int:
    num = -1
    while True:
        try:
            num = int(input(prompt))
            break
        except ValueError:
            print('Error: Enter an integer, try again...')
    return num

def get_yes_no_input(prompt: str) -> bool:
    allowed_responses = {'y', 'yes', 'n', 'no'}
    user_input = input(prompt).lower()
    while user_input not in allowed_responses:
        user_input = input(prompt).lower()
    return user_input[0] == 'y'

def get_single_employee_info(names: list[str], ids: list[int],
                             num_sold_houses: list[int]) -> None:
    names.append(input('What is the employee\'s name?: '))
    ids.append(get_int_input('What is the employee\'s id?: '))
    num_sold_houses.append(
        get_int_input('How many houses did the employee sell?: '))

def get_houses_sold_info(names: list[str], ids: list[int],
                         num_sold_houses: list[int]) -> None:
    get_single_employee_info(names, ids, num_sold_houses)
    add_another_employee = get_yes_no_input('Add another employee [yes/no]?: ')
    while add_another_employee:
        get_single_employee_info(names, ids, num_sold_houses)
        add_another_employee = get_yes_no_input(
            'Add another employee [yes/no]?: ')

def print_entered_info(names: list[str], ids: list[int],
                       num_sold_houses: list[int]) -> None:
    print()
    row_width = 12
    comission_per_house = 500
    header = ['Name', 'Id', 'Houses Sold', 'Commission']
    print(' '.join(f'{h:<{row_width}}' for h in header))
    commission = [n * comission_per_house for n in num_sold_houses]
    for values in zip(*[names, ids, num_sold_houses, commission]):
        print(' '.join(f'{v:<{row_width}}' for v in values))
    print()
    total_commission = sum(commission)
    print(f'Total Commission: {total_commission}')
    print()
    rankings = sorted(zip(num_sold_houses, names), reverse=True)
    print('Ranking:')
    for houses_sold, name in rankings:
        print(f'{name} - {houses_sold}')

def main() -> None:
    print('Houses Sold Tracker')
    print('===================')
    names, ids, num_houses_sold = [], [], []
    get_houses_sold_info(names, ids, num_houses_sold)
    print_entered_info(names, ids, num_houses_sold)

if __name__ == '__main__':
    main()

示例用法：

Houses Sold Tracker
===================
What is the employee's name?: Laura
What is the employee's id?: 1
How many houses did the employee sell?: a
Error: Enter an integer, try again...
How many houses did the employee sell?: 4
Add another employee [yes/no]?: yes
What is the employee's name?: John
What is the employee's id?: 2
How many houses did the employee sell?: 5
Add another employee [yes/no]?: y
What is the employee's name?: BOJO
What is the employee's id?: 3
How many houses did the employee sell?: 1
Add another employee [yes/no]?: no

Name         Id           Houses Sold  Commission  
Laura        1            4            2000        
John         2            5            2500        
BOJO         3            1            500         

Total Commission: 5000

Ranking:
John - 5
Laura - 4
BOJO - 1

顺便说一句，如果你做的事情是真实的，不要使用连续的用户 ID，你可以从监控中获得的信息量是疯狂的，例如，用户获取率，你不这样做'不希望外部人员能够弄清楚。