在Python 3.6中添加迭代时重复元素
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我正在尝试编写一部分代码,该代码从两个不同的列表中获取元素并进行匹配,如下所示,但是由于某种原因,我一直在输出列表中重复获取元素.
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,第一个列表是这样的：
[[11:05, Christina Holtzer, not-assigned],[11:05, Christina Holtzer, Beate Adenauer],[11:05, Andrej Hlavac, not-assigned]]
解决方法:
我真的不知道您要遵循的逻辑,这甚至不是我的意思,我的意思是,您可能无法专注于逻辑,因为您太忙于处理那些索引问题了.因此,我随意修改了一些代码以显示重要内容,如果您使用的是python,则应利用此功能,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,只需输入您要考虑的内容,思考逻辑的任务就会变得更加简单.
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)
输出为：
[[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,但是所有请求者都拥有或没有运算符.
同样,我不知道这种逻辑是否是您所追求的逻辑,但是我希望您看到通过这种方法可以更轻松地思考问题(真正重要的问题),并且让其他人更容易阅读.
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