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在Python中尝试pop异常

python

Chevady Ju · 技术社区 · 7 年前

t1.py主程序

t3.py子程序

不一定,但有时在主程序中可能会遇到下面的错误有什么解决办法吗?

------错误消息--------

res=list_links.pop(random.randint(1,len(list_links)))#根据列表大小随机弹出每个项目

索引器错误:弹出索引超出范围

# -*- coding: utf-8 -*-
"""
Created on Fri Oct  5 21:51:34 2018

@author: chevady
"""
from linebot.models import *
from t3 import apple_newsss
apple_newss_content0, apple_newss_content1, apple_newss_content2 = apple_newsss()

print(apple_newss_content0)

print(apple_newss_content1)

print(apple_newss_content2)

t3.py型

# -*- coding: utf-8 -*-
import requests
import re
import os
import random
from bs4 import BeautifulSoup
from flask import Flask, request, abort
#from imgurpython import ImgurClient
from argparse import ArgumentParser

def apple_newsss():

    target_url = 'http://www.appledaily.com.tw/realtimenews/section/new/'
    print('Start parsing appleNews....')
    rs = requests.session()
    res = rs.get(target_url, verify=False)
    soup = BeautifulSoup(res.text, 'html.parser')

    list_links = [] # Create empty list

    for a in soup.select("div[class='abdominis rlby clearmen']")[0].findAll(href=True): # find links based on div
            if a['href']!= None and a['href'].startswith('https://'):
                    list_links.append(a['href']) #append to the list
                    print(a['href']) #Check links

#for l in list_links: # print list to screen (2nd check)
#    print(l)


    print("\n")

    random_list = [] #create random list if needed..
    random.shuffle(list_links) #random shuffle the list
    apple_newss_content0 = ''
    apple_newss_content1 = ''
    apple_newss_content2 = ''
    for i in range(3): # specify range (5 items in this instance)
        res = list_links.pop(random.randint(1, len(list_links))) # pop of each item randomly based on the size of the list
        random_list.append(res)
    #print(res)
    #print(random_list)
    print("\n")
    apple_newss_content0=random_list[0]
    apple_newss_content1=random_list[1]
    apple_newss_content2=random_list[2]

    print(apple_newss_content0)
    print(apple_newss_content1)
    print(apple_newss_content2)

    return apple_newss_content0, apple_newss_content1, apple_newss_content2

谢谢!!!

2 回复 | 直到 7 年前

Rakesh 7 年前

使用 len(list_links)-1

前任:

for i in range(3): # specify range (5 items in this instance)
    res = list_links.pop(random.randint(1, len(list_links)-1)) # pop of each item randomly based on the size of the list
    random_list.append(res)

bruno desthuilliers 7 年前

res = list_links.pop(random.randint(1, len(list_links)))

表达式 random.randint(1, len(list_links)) len(list_links) ,但是(像大多数其他语言FWIW一样)python列表索引是基于零的,所以您需要 random.randint(0, len(list_links) - 1)

list_links 其中至少有3个项,因此如果它是空的或在for循环期间变为空,则仍然会得到一个错误。

编辑

而且,您的代码比需要的复杂得多。如果你想要的是3个随机不同的网址 列出\u链接

# make sure we only have distinct urls
list_links = list(set(list_links))
# make sure the urls are in arbitrary order
random.shuffle(list_links) 
# returns the x (x <= 3) first urls 
return list_links[:3]

min(3, len(linked_lists) 实际上是URL,所以调用者不应该总是依赖于3个URL的返回,但这并不是必须的:每次你发现自己在写东西的时候 somevar1, somevar2, somevarN

urls = apple_newsss()
for url in urls:
    print(url)