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

我正在一些网页中搜索关键字。再次感谢@Abdou 帮助我处理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))

如您所见，我请求存储在df.data 中的URL，并从mykeywords 中搜索关键字并将二进制结果存储在extra_column 中。该脚本基本上产生以下内容：

#                            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。但我想知道我找到了哪些关键字——无需为mykeywords 中的每个关键字单独运行contains_keywords()。有没有办法为每个关键字创建一个新列并将结果（1 = 找到关键字）存储在DataFrame 中？即：我需要在df 中为每个关键字添加额外的列。

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))