问题描述
我正在尝试编写一种算法,该算法遍历整个节点集合并返回其报酬之和。每个奖励只能计算一次。该算法的输入将是节点开头的URL,例如http://fake.url/a。
{
"children":[
"http://fake.url/b","http://fake.url/c"
],"reward":1
}
这是我尝试过的:
import multiprocessing
import requests
import json
my_q = multiprocessing.Queue()
my_list =['http://fake.url/']
reward_sum = 0
def enqueue(q):
for data in my_list:
q.put(data)
def get_it(q):
while not q.empty():
item = q.get()
print(item)
response = requests.get(item)
kids = json.loads(response.content)
print(f'URL: {item} --> {kids["reward"]}')
for kid in kids['children']:
print(kid)
q.put(kid)
p1 = multiprocessing.Process(target=enqueue,args=(my_q,))
p2 = multiprocessing.Process(target=get_it,))
p1.start()
p2.start()
p1.join()
p2.join()
以上工作原理:
- 我正在使用多处理程序。
- 我正在正确接触孩子并获得奖励。
- 我正在得到这样的输出:
http://fake.url/a
URL: http://fake.url/a --> 1
{'children': ['http://fake.url/b','http://fake.url/c'],'reward': 1}
http://fake.url/b
http://fake.url/c
http://fake.url/b
URL: http://fake.url/b --> 2
{'children': ['http://fake.url/d','http://fake.url/e'],'reward': 2}
http://fake.url/d
http://fake.url/e
http://fake.url/c
URL: http://fake.url/c --> 3
{'children': ['http://fake.url/f','http://fake.url/g'],'reward': 3}
http://fake.url/f
http://fake.url/g
http://fake.url/d
URL: http://fake.url/d --> 4
{'reward': 4}
http://fake.url/e
URL: http://fake.url/e --> 5
{'reward': 5}
http://fake.url/f
URL: http://fake.url/f --> 6
{'children': ['http://fake.url/h'],'reward': 6}
http://fake.url/h
http://fake.url/g
我需要帮助的问题是什么?
解决方法
def get_it(q):
rewards_total = 0
seen = set()
while not q.empty():
item = q.get()
print(item)
if item in seen:
continue
seen.add(item)
response = requests.get(item)
kids = json.loads(response.content)
rewards_total += kids["reward"]
print(f'URL: {item} --> {kids["reward"]}')
for kid in kids['children']:
print(kid)
q.put(kid)
return rewards_total