使用预定义列表获取 pandas 列中匹配单词的计数

Question

我有一个 DataFrame 包含 index 和 text 列。

例如：

index | text
1     | "I have a pen, but I lost it today."
2     | "I have pineapple and pen, but I lost it today."

现在我有一个很长的列表，我想将text 中的每个单词与列表进行匹配。

假设：

long_list = ['pen', 'pineapple']

我想创建一个FunctionTransformer 以将long_list 中的单词与列值的每个单词匹配，如果匹配，则返回计数。

index | text                                             | count
1     | "I have a pen, but I lost it today."             | 1
2     | "I have pineapple and pen, but I lost it today." | 2

我是这样做的：

def count_words(df):
    long_list = ['pen', 'pineapple']
    count = 0
    for c in df['tweet_text']:
        if c in long_list:
            count = count + 1
            
    df['count'] = count   
    return df

count_word = FunctionTransformer(count_words, validate=False)

我如何开发我的另一个 FunctionTransformer 的一个例子是：

def convert_twitter_datetime(df):
    df['hour'] = pd.to_datetime(df['created_at'], format='%a %b %d %H:%M:%S +0000 %Y').dt.strftime('%H').astype(int)
    return df

convert_datetime = FunctionTransformer(convert_twitter_datetime, validate=False)

Answer 1

灵感来自@Quang Hoang 的回答

import pandas as pd
import sklearn as sk

y=['pen', 'pineapple']

def count_strings(X, y):
    pattern = r'\b{}\b'.format('|'.join(y))
    return X['text'].str.count(pattern)

string_transformer = sk.preprocessing.FunctionTransformer(count_strings, kw_args={'y': y})
df['count'] = string_transformer.fit_transform(X=df)

结果

    text                                              count
1   "I have a pen, but I lost it today."                1
2   "I have pineapple and pen, but I lost it today.     2

对于以下df2：

#df2
      text
1     "I have a pen, but I lost it today. pen pen"
2     "I have pineapple and pen, but I lost it today."

我们得到

string_transformer.transform(X=df2)
#result
1    3
2    2
Name: text, dtype: int64

这表明，我们将函数转换为sklearn 样式的对象。为了进一步抽象这一点，我们可以将列名作为关键字参数传递给count_strings。

Answer 2

用| 连接列表中的元素。查找与.str.findall() 匹配的元素并应用.str.len() 进行计数

 p='|'.join(long_list)
df=df.assign(count=(df.text.str.findall(p)).str.len())
                                             text   count
0              "I have a pen, but I lost it today."      1
1  "I have pineapple and pen, but I lost it today."      2

Answer 3

熊猫有str.count:

# matching any of the words
pattern = r'\b{}\b'.format('|'.join(long_list))

df['count'] = df.text.str.count(pattern)

输出：

   index                                              text  count
0      1              "I have a pen, but I lost it today."      1
1      2  "I have pineapple and pen, but I lost it today."      2