在系列中搜索一个单词。在新列中返回那个词和其他 N 个词？

Question

好的，我需要帮助。我创建了一个函数来搜索特定单词的字符串。如果函数找到search_word，它将返回单词 the 和它之前的 N 个单词。该函数适用于我的测试字符串，但我无法弄清楚如何将该函数应用于整个系列？

我的目标是在 search_word 存在时在包含 n_words_prior 的数据框中创建一个新列。

n_words_prior = []
test = "New School District, Dale County"

def n_before_string(string, search_word, N):
    global n_words_prior 
    n_words_prior = []
    found_word = string.find(search_word)
    if found_word == -1: return ""
    sentence= string[0:found_word]
    n_words_prior = sentence.split()[N:]
    n_words_prior.append(search_word)
    return n_words_prior

当前数据框如下所示：

data = [['Alabama', 'New School District, Dale County'], 
        ['Alaska', 'Matanuska-Susitna Borough'], 
        ['Arizona', 'Pima County - Tuscon Unified School District']]
df = pd.DataFrame(data, columns = ['State', 'Place']) 

改进后的函数将采用输入 'Place','County',-1 并创建以下结果。

improved_function(column, search_word, N)

new_data = [['Alabama', 'New School District, Dale County','Dale County'], 
        ['Alaska', 'Matanuska-Susitna Borough', ''], 
        ['Arizona', 'Pima County - Tuscon Unified School District','Pima County']]
new_df = pd.DataFrame(new_data, columns = ['State', 'Place','Result']) 

我认为嵌入这个函数会有所帮助，但它只会让事情变得更加混乱。

def fast_add(place, search_word):
    df[search_word] = df[Place].str.contains(search_word).apply(lambda search_word: 1 if search_word == True else 0)

Answer 1

def fun(sentence, search_word, n):
    """Return search_word and n preceding words from sentence."""
    words = sentence.split()
    for i,word in enumerate(words):
        if word == search_word:
            return ' '.join(words[i-n:i+1])
    return ''

例子：

df['Result'] = df.Place.apply(lambda x: fun(x, 'County', 1))

结果：

     State                                         Place       Result
0  Alabama              New School District, Dale County  Dale County
1   Alaska                     Matanuska-Susitna Borough             
2  Arizona  Pima County - Tuscon Unified School District  Pima County