Python中的字符串关键字搜索

时间:2018-10-08 16:51:16

标签: python string python-3.x list search

我正在尝试在列表的任何索引中找到关键字并获取该索引。我使用BeautifulSoup4创建了一个小型网页抓取工具,以抓取同人小说的数据。

由于并非所有幻想小说都列出了流派或人物,甚至更新日期(如果它们是新出版的),所有信息都将位于不同的索引中。

因此,我需要搜索,例如'Words:',并获取整个字符串的索引,即'Words:1,854'== list [3]或类似的东西,并将其保存为变量words = list [3]稍后再调用,以便稍后将其放入excel文件中的正确列中。这是我当前的抓取工具,目前仅设置为抓取一页,只需减小“ u”的原始值即可添加更多要抓取的页面。

import requests
from bs4 import BeautifulSoup
# import time
# from random import randint
# import xlsxwriter
# import urllib3
# from tinydb import TinyDB, Query

total = 0
u = int(1127)

while u < 2000:
    u = u+1
    url = 'https://www.fanfiction.net/Naruto-Crossovers/1402/0/?&srt=1&lan=1&r=10&p=' + str(u)
    page = requests.get(url)
    soup = BeautifulSoup(page.content, 'html.parser')

    raw = soup.find_all('div', class_='z-indent z-padtop')
    for n in range(len(raw)):
        stats = raw[n]
        info = stats.div
        text = info.text
        formatted = text.split(' - ')
        print(formatted[1:(len(formatted))])

2 个答案:

答案 0 :(得分:0)

然后解决方案可能是这样的(检查功能find_keyword

import requests
from bs4 import BeautifulSoup
# import time
# from random import randint
# import xlsxwriter
# import urllib3
# from tinydb import TinyDB, Query

total = 0
u = int(1127)

results = []
while u < 1130: #decreased u due to testing time
    u = u+1
    url = 'https://www.fanfiction.net/Naruto-Crossovers/1402/0/?&srt=1&lan=1&r=10&p=' + str(u)
    page = requests.get(url)
    soup = BeautifulSoup(page.content, 'html.parser')

    raw = soup.find_all('div', class_='z-indent z-padtop')
    for n in range(len(raw)):
        stats = raw[n]
        info = stats.div
        text = info.text
        formatted = text.split(' - ')
        if formatted:
            results.append(formatted)
print(results)

# function to search for a keyword
def find_keyword(list, keyword):
    results = []
    for element in list:
        value = ''
        for tag in element:
            if tag.find(keyword) >= 0:
                value = tag
        results.append(value)

    return(results)

words_list = find_keyword(results, 'Words') #example of how to search and build list for keyword
print(words_list)

答案 1 :(得分:0)

This is the code I came up with, it wordks wonderfully. The find function was essential.

# For later use, searches for keywords and adds them to the specified list
    def assign_stats(keyword, stat_list):
        k = 13
        b = 0
        t = 0
        while k >= 1:
            if t == len(formatted):
                t = 0
            check = formatted[t]
            value = check.find(keyword)
            if value != -1:
                # values = formatted[t]
                stat_list.append(check)
                b = 1
            elif k < 2 and b == 0:
                stat_list.append('')

            t = t + 1
            k = k - 1


    # For later use, searches for keywords and adds them to the specified list
    def assign_stats_status(keyword, stat_list):
        k = 13
        b = 0
        t = 0
        while k >= 1:
            if t == len(formatted):
                t = 0
            check = formatted[t]
            value = check.find(keyword)
            if value != -1:
                # values = formatted[t]
                stat_list.append(check)
                b = 1
            elif k < 2 and b == 0:
                stat_list.append('In-Progress')
            t = t + 1
            k = k - 1


    # For later use, searches for specified indexes of story data lists and adds them to specified list
    def assign_stats_concrete(index, stat_list):
        stat_list.append(formatted[index])

    # Searches for keywords/indexes for the specified story stat lists
    assign_stats('Words', words)
    assign_stats_concrete(2, rating)
    assign_stats('English', language)
    assign_stats('Chapters', chapters)
    assign_stats('Reviews', reviews)
    assign_stats('Favs', favorites)
    assign_stats('Follows', follows)
    assign_stats('Updated', updated)
    assign_stats_status('Complete', status)
    assign_stats('Published', published)
    assign_stats_concrete(1, crossover)