如果前三个句子包含关键字，如何过滤字符串

Question

我有一个名为 df 的 pandas 数据框。它有一个名为article 的列。 article 列包含 600 个字符串，每个字符串代表一篇新闻文章。 我只想保留前四个句子包含关键字“COVID-19”AND（“China”或“Chinese”）的文章。但我无法自己找到一种方法来执行此操作。

（在字符串中，句子以\n分隔。示例文章如下所示：）

\nChina may be past the worst of the COVID-19 pandemic, but they aren’t taking any chances.\nWorkers in Wuhan in service-related jobs would have to take a coronavirus test this week, the government announced, proving they had a clean bill of health before they could leave the city, Reuters reported.\nThe order will affect workers in security, nursing, education and other fields that come with high exposure to the general public, according to the edict, which came down from the country’s National Health Commission.\ .......

Answer 1

首先我们定义一个函数来根据你的关键字是否出现在给定的句子中返回一个布尔值：

def contains_covid_kwds(sentence):
    kw1 = 'COVID19'
    kw2 = 'China'
    kw3 = 'Chinese'
    return kw1 in sentence and (kw2 in sentence or kw3 in sentence)

然后我们通过将此函数（使用Series.apply）应用于df.article 列的句子来创建一个布尔序列。

请注意，我们使用 lambda 函数来截断传递给contains_covid_kwds 的句子，直到第五次出现'\n'，即您的前四个句子（有关其工作原理的更多信息here）：

series = df.article.apply(lambda s: contains_covid_kwds(s[:s.replace('\n', '#', 4).find('\n')]))

然后我们将布尔序列传递给df.loc，以便本地化序列被评估为True的行：

filtered_df = df.loc[series]

Answer 2

您可以使用 pandas 的 apply 方法并按照我的方式进行操作。

string = "\nChina may be past the worst of the COVID-19 pandemic, but they aren’t taking any chances.\nWorkers in Wuhan in service-related jobs would have to take a coronavirus test this week, the government announced, proving they had a clean bill of health before they could leave the city, Reuters reported.\nThe order will affect workers in security, nursing, education and other fields that come with high exposure to the general public, according to the edict, which came down from the country’s National Health Commission."
df = pd.DataFrame({'article':[string]})

def findKeys(string):
    string_list = string.strip().lower().split('\n')
    flag=0
    keywords=['china','covid-19','wuhan']

    # Checking if the article has more than 4 sentences
    if len(string_list)>4:
        # iterating over string_list variable, which contains sentences.
        for i in range(4):
            # iterating over keywords list
            for key in keywords:
                # checking if the sentence contains any keyword
                if key in string_list[i]:
                    flag=1
                    break
    # Else block is executed when article has less than or equal to 4 sentences
    else:
        # Iterating over string_list variable, which contains sentences
        for i in range(len(string_list)):
            # iterating over keywords list
            for key in keywords:
                # Checking if sentence contains any keyword
                if key in string_list[i]:
                    flag=1
                    break
    if flag==0:
        return False
    else:
        return True

然后在df上调用pandas的apply方法：-

df['Contains Keywords?'] = df['article'].apply(findKeys)

Answer 3

首先，我创建一个系列，其中仅包含原始 `df['articles'] 列中的前四个句子，并将其转换为小写，假设搜索应该与大小写无关。

articles = df['articles'].apply(lambda x: "\n".join(x.split("\n", maxsplit=4)[:4])).str.lower()

然后使用简单的布尔掩码仅过滤在前四个句子中找到关键字的那些行。

df[(articles.str.contains("covid")) & (articles.str.contains("chinese") | articles.str.contains("china"))]

Answer 4

这里：

found = []
s1 = "hello"
s2 = "good"
s3 = "great"
for string in article:
    if s1 in string and (s2 in string or s3 in string):
        found.append(string)