在文本中搜索关键字并为每个找到的关键字创建一个数据框列？

Question

我正在搜索关键字的一些网页。再次感谢@Abdou帮助我与silent error handling！我给你举个例子：

# this is basically what I do 
import pandas as pd 
import requests 


data = [{"URLs" : "https://www.mercedes-benz.de", "electric" : 1}, 
     {"URLs" : "https://www.audi.de", "electric" : 0}, 
     {"URLs" : "https://ww.audo.de", "electric" : 0}, 
     {"URLs" : "NaN", "electric" : 0}] 


def contains_keywords(link, keywords): 
    try: 
     output = requests.get(link).text 
     return int(any(x in output for x in keywords)) 
    except: 
     return "Wrong/Missing URL" 


df = pd.DataFrame(data) 
mykeywords = ('car', 'vehicle', 'automobile') 
df['extra_column'] = df.URLs.apply(lambda l: contains_keywords(l, mykeywords)) 

正如你所看到的，我请求存储在df.data的URL和mykeywords搜索关键词和二进制结果存储在extra_column。该脚本基本上产生如下：

#       URLs electric  extra_column 
# 0 https://www.mercedes-benz.de   1     1 
# 1   https://www.audi.de   0     1 
# 2    https://ww.audo.e   0     0 
# 3       NaN   0 Wrong/Missing URL 

到目前为止，我只知道，如果我找到一个关键字。但我想知道，我发现哪些关键字 - 无需分别为mykeywords中的每个关键字运行contains_keywords()。有没有办法为每个关键字创建一个新列，并将结果（1 =找到的关键字）存储在DataFrame？那就是：我需要df中的每个关键字的其他列。

Answer 1

import pandas as pd 
import requests 


data = [{"URLs" : "https://www.mercedes-benz.de", "electric" : 1}, 
     {"URLs" : "https://www.audi.de", "electric" : 0}, 
     {"URLs" : "https://ww.audo.de", "electric" : 0}, 
     {"URLs" : "NaN", "electric" : 0}] 


def contains_keywords(link, keyword): 
    try: 
     output = requests.get(link).text 
     return int(keyword in output) 
    except: 
     return "Wrong/Missing URL" 


df = pd.DataFrame(data) 
mykeywords = ('car', 'vehicle', 'automobile') 
for keyword in mykeywords: 
    df[keyword] = df.URLs.apply(lambda l: contains_keywords(l, keyword))