python - 在 Python 3.6 中追加迭代时重复元素

Question

我正在尝试编写一部分代码，从两个不同的列表中获取元素并进行匹配，如下所示，但出于某种原因，我不断在输出列表中获取重复的元素。

def assign_tasks(operators, requests, current_time):
    """Assign operators to pending requests.

    Requires:
    - operators, a collection of operators, structured as the output of 
      filesReading.read_operators_file;
    - requests, a list of requests, structured as the output of filesReading.read_requests_file;
    - current_time, str with the HH:MM representation of the time for this update step.
    Ensures: a list of assignments of operators to requests, according to the conditions indicated 
    in the general specification (omitted here for the sake of readability).
    """
    operators = sorted(operators, key=itemgetter(3, 4, 0), reverse=False)
    requests = sorted(requests, key=itemgetter(3), reverse=True)
    isAssigned = 0
    tasks = []
    langr = 0 #Variable that gets the language of the request's file (customer's language)
    lango = 0 #Variable that gets the language of the operator's file (operator's language)
    for i in range(len(requests)-1):
        langr = requests[i][1]                                   #What language does the customer speaks?
        for k in range(len(operators)-1):
            lango = operators[k][1]                              #What language does the operator speaks?
            if langr == lango:                                   #Do they speak the same language?
                for j in range(len(operators[k][2])-1):
                    if (operators[k][2][j] == requests[i][2]) and (operators[k][4] <= 240):     # The operator knows how to solve the client's problem? If yes, then group them together.
                        a = operators[k][2][j]
                        b = requests[i][2]
                        tasks.append([current_time, requests[i][0], operators[k][0]])
                        operator_time = operators[k][4]
                        request_time = requests[i][4]
                        new_operator_time = operator_time + request_time
                        operators[k][4] = new_operator_time
                        isAssigned == True
                        #operators.remove(operators[k])
                        requests.remove(requests[i])
                    else:
                        isAssigned = False
                    if isAssigned == False:
                        tasks.append([current_time, requests[i][0], "not-assigned"])

        operators = sorted(operators, key=itemgetter(3, 4, 0), reverse=False)

    return tasks, operators, requests

我当前的输入是这样的:

operators = [['Atilio Moreno', 'portuguese', ('laptops',), '10:58', 104], ['Leticia Ferreira', 'portuguese', ('laptops',), '11:03', 15], ['Ruth Falk', 'german', ('phones', 'hifi'), '11:06', 150], ['Marianne Thibault', 'french', ('phones',), '11:09', 230], ['Mariana Santana', 'portuguese', ('phones',), '11:11', 230], ['Beate Adenauer', 'german', ('hifi', 'phones'), '11:12', 140], ['Zdenka Sedlak', 'czech', ('phones',), '11:13', 56], ['Romana Cerveny', 'czech', ('phones',), '11:13', 213]]
requests = [['Christina Holtzer', 'german', 'hifi', 'fremium', 7], ['Andrej Hlavac', 'czech', 'phones', 'fremium', 9], ['Dulce Chaves', 'portuguese', 'laptops', 'fremium', 15], ['Otavio Santiago', 'portuguese', 'laptops', 'fremium', 15], ['Dina Silveira', 'portuguese', 'phones', 'fremium', 9], ['Rafael Kaluza', 'slovenian', 'laptops', 'fremium', 13], ['Sabina Rosario', 'portuguese', 'laptops', 'fremium', 10], ['Nuno Rodrigues', 'portuguese', 'laptops', 'fremium', 12], ['Feliciano Santos', 'portuguese', 'phones', 'fremium', 12]]

current_time = "14:55 06:11:2017"
print(assign_tasks(operators, requests, current_time))

我当前的输出是三个列表，例如，第一个是这样的:

[[11:05, Christina Holtzer, not-assigned],[11:05, Christina Holtzer, Beate Adenauer],[11:05, Andrej Hlavac, not-assigned]]

Answer 1

我真的不知道你追求的逻辑，这甚至不是我的观点，我的观点是你可能无法专注于逻辑，因为你太忙于那些索引的事情。因此，我冒昧地稍微修改了您的代码以显示重要内容，如果您使用的是 Python，则应该利用此功能，因为可读性很重要。

from operator import attrgetter

class Person:
    def __init__(self, name, lan):
        self.name = name
        self.lan = lan

    def is_compatible(self, other):
        if other.lan == self.lan:
            return True
        return False

class Requester(Person):
    def __init__(self, *args, problem, mode, time, **kwargs):
        super().__init__(*args, **kwargs)
        self.problem = problem
        self.mode = mode
        self.time = time

class Operator(Person):
    def __init__(self, *args, expertise, hour, time, **kwargs):
        super().__init__(*args, **kwargs)
        self.expertise = expertise
        self.hour = hour
        self.time = time
        self.assigned = False

operators = [
    Operator(name='Atilio Moreno', lan='portuguese', expertise=('laptops',), hour='10:58', time=104),
          .
          .
          .
    Operator(name='Romana Cerveny', lan='czech',  expertise=('phones',), hour='11:13', time=213),
]

requests = [
    Requester(name='Christina Holtzer', lan='german', problem='hifi', mode='fremium', time=7),
          .
          .
          .
    Requester(name='Feliciano Santos', lan='portuguese',  problem='phones',  mode='fremium', time=12),
]

完成此操作后，思考逻辑的任务就变得简单多了，只需输入您的想法即可:

def assign_tasks(operators, requests, current_time):
    operators.sort(key=attrgetter('hour', 'time', 'name'))
    requests.sort(key=attrgetter('mode'))
    tasks = []
    for requester in requests:
        for operator in operators:
            if requester.is_compatible(operator) and requester.problem in operator.expertise and operator.time < 240:
                if not operator.assigned:
                    tasks.append([current_time, requester.name, operator.name])
                    operator.assigned = True
                    operator.time += requester.time
                    break # Breaks out of second for-loop so we go to the next requester
        else: #In case no operator is available
            tasks.append([current_time, requester.name, 'not-assigned'])
    return tasks, operators, requests

tasks, operators, requests = assign_tasks(operators=operators, requests=requests, current_time=0)

print(tasks)

这个的输出是:

 [[0, 'Christina Holtzer', 'Ruth Falk'], [0, 'Andrej Hlavac', 'Zdenka Sedlak'], [0, 'Dulce Chaves', 'Atilio Moreno'], [0, 'Otavio Santiago', 'not-assigned'], [0, 'Dina Silveira', 'not-assigned'], [0, 'Rafael Kaluza', 'not-assigned'], [0, 'Sabina Rosario', 'not-assigned'], [0, 'Nuno Rodrigues', 'not-assigned'], [0, 'Feliciano Santos', 'not-assigned']]

有点长，但是所有的请求者要么有运算符(operator)，要么没有。

同样，我不知道这个逻辑是否是您所追求的逻辑，但我希望您能看到，使用这种方法可以更简单地思考问题(真正重要的是什么)，而且它也是其他人更容易阅读。