电梯仿真的python线程

Question

我试图做一个电梯模拟，因为我在CareerCup上看到了一个有趣的问题。我的问题是我想让电梯“花点时间”从一层到另一层。现在，它只是立即移动到“访问”列表中的下一层。我不知道如何对它进行编程，以便在电梯移动时“取件请求”可以进来。我想这可能需要穿线时间。睡觉（）功能。如何让一个线程向电梯发出随机请求，另一个线程让电梯尝试满足所有请求？到目前为止，我得到的是：

import time
from random import *
import math

class Elevator:
    def __init__(self, num_floors):
        self.going_up = False
        self.going_down = False
        self.docked = True
        self.curr_floor = 0
        self.num_floors = num_floors
        self.floors_to_visit = []
        self.people_waiting = []

    def print_curr_status(self):
        for i in range(self.num_floors):
            if i == self.curr_floor:
                print('.  []')
            else:
                print('.')
        print ("to_visit: ", self.floors_to_visit)

    def handle_call_request(self, person):
        if not self.going_up and not self.going_down:
            self.floors_to_visit = [person.curr_floor] + self.floors_to_visit
            self.going_up = True
            self.docked = False
            self.people_waiting.append(person)
        else:
            self.floors_to_visit.append(person.curr_floor)
            self.people_waiting.append(person)

    def handle_input_request(self, floor_num):
        self.floors_to_visit.append(floor_num)

    def go_to_next(self):
        if not self.floors_to_visit:
            self.print_curr_status()
            return
        self.curr_floor = self.floors_to_visit.pop(0)
        for i,person in enumerate(self.people_waiting):
            if person.curr_floor == self.curr_floor:
                person.riding = True
                person.press_floor_num()
                self.people_waiting.pop(i)
        return


class Person:
    def __init__(self, assigned_elevator, curr_floor):
        self.curr_floor = curr_floor
        self.desired_floor = math.floor(random() * 10)
        self.assigned_elevator = assigned_elevator
        self.riding = False

    def print_floor(self):
        print(self.desired_floor)

    def call_elevator(self):
        self.assigned_elevator.handle_call_request(self)

    def press_floor_num(self):
        self.assigned_elevator.handle_input_request(self.desired_floor)


my_elevator = Elevator(20)

while True:
    for i in range(3):
        some_person = Person(my_elevator, math.floor(random() * 10))
        some_person.call_elevator()
    my_elevator.go_to_next()
    my_elevator.print_curr_status()
    time.sleep(1)

Answer 1

不必担心。你可以引入两个新的变量：一个跟踪电梯启动的时间，另一个跟踪电梯的运行时间。那么只要检查一下电梯运行的时间是否足够长。您可以调用函数time.time()；它将以秒为单位返回1970年1月1日以来的时间（因为您只关心差异，这无关紧要；您只需要一个时间增量的函数）。虽然，这个函数通常不能给出比1秒更精确的时间段。如果您觉得在您的机器上不准确，那么您可以使用datetime。在

class Elevator:
    def __init__(self, num_floors):
        self.start_time = 0
        self.ride_duration = 1
        ...

    def call_elevator(self):
         self.start_time = time.time()
         self.assigned_elevator.handle_call_request(self)

    def go_to_next(self):
        if time.time() - self.start_time < self.ride_duration:
            return  # Do nothing.
        else:
            ...

您可能需要重构代码以满足您的需要，并添加一些逻辑，说明在使用电梯时应该做什么，等等