PANDAS 在一列字符串中查找确切的单词和之前的单词，并将该新列附加到 python (pandas) 列中

Question

在 col_a 中查找目标词和前一个词，并在 col_b_PY 和 col_c_LG 列中追加匹配的字符串

    This code i have tried to achive this functionality but not able to 
get the expected output. if any help appreciated
Here is the below code i approach with regular expressions:

df[''col_b_PY']=df.col_a.str.contains(r"(?:[a-zA-Z'-]+[^a-zA-Z'-]+) 
{0,1}PY")

df.col_a.str.extract(r"(?:[a-zA-Z'-]+[^a-zA-Z'-]+){0,1}PY",expand=True)

数据框如下所示

col_a

Python PY is a general-purpose language LG

Programming language LG in Python PY 

Its easier LG to understand  PY

The syntax of the language LG is clean PY 

期望的输出：

col_a                                       col_b_PY      col_c_LG
Python PY is a general-purpose language LG  Python PY     language LG

Programming language LG in Python PY        Python PY     language LG

Its easier LG to understand  PY            understand PY easier LG

The syntax of the language LG is clean PY   clean  PY     language LG

Answer 1

你可以使用

df['col_b_PY'] = df['col_a'].str.extract(r"([a-zA-Z'-]+\s+PY)\b")
df['col_c_LG'] = df['col_a'].str.extract(r"([a-zA-Z'-]+\s+LG)\b")

或者，提取所有匹配项并用空格连接它们：

df['col_b_PY'] = df['col_a'].str.extractall(r"([a-zA-Z'-]+\s+PY)\b").unstack().apply(lambda x:' '.join(x.dropna()), axis=1)
df['col_c_LG'] = df['col_a'].str.extractall(r"([a-zA-Z'-]+\s+LG)\b").unstack().apply(lambda x:' '.join(x.dropna()), axis=1)

请注意，您需要在正则表达式模式中使用捕获组，以便extract 可以实际提取文本：

将正则表达式pat中的捕获组提取为DataFrame中的列。

注意\b 单词边界是匹配PY / LG 作为一个完整单词所必需的。

另外，如果你只想从一个字母开始匹配，你可以将模式修改为

r"([a-zA-Z][a-zA-Z'-]*\s+PY)\b"
r"([a-zA-Z][a-zA-Z'-]*\s+LG)\b"
   ^^^^^^^^          ^

[a-zA-Z] 将匹配一个字母，[a-zA-Z'-]* 将匹配 0 个或多个字母、撇号或连字符。

Python 3.7 和 Pandas 0.24.2：

pd.set_option('display.width', 1000)
pd.set_option('display.max_columns', 500)

df = pd.DataFrame({
    'col_a': ['Python PY is a general-purpose language LG',
             'Programming language LG in Python PY',
             'Its easier LG to understand  PY',
             'The syntax of the language LG is clean PY',
             'Python PY is a general purpose PY language LG']
    })
df['col_b_PY'] = df['col_a'].str.extractall(r"([a-zA-Z'-]+\s+PY)\b").unstack().apply(lambda x:' '.join(x.dropna()), axis=1)
df['col_c_LG'] = df['col_a'].str.extractall(r"([a-zA-Z'-]+\s+LG)\b").unstack().apply(lambda x:' '.join(x.dropna()), axis=1)

输出：

                                           col_a              col_b_PY     col_c_LG
0     Python PY is a general-purpose language LG             Python PY  language LG
1           Programming language LG in Python PY             Python PY  language LG
2                Its easier LG to understand  PY        understand  PY    easier LG
3      The syntax of the language LG is clean PY              clean PY  language LG
4  Python PY is a general purpose PY language LG  Python PY purpose PY  language LG

Answer 2

检查

df['col_c_LG'],df['col_c_PY']=df['col_a'].str.extract(r"(\w+\s+LG)"),df['col_a'].str.extract(r"(\w+\s+PY)")
df
Out[474]: 
                                        col_a       ...              col_c_PY
0  Python PY is a general-purpose language LG       ...             Python PY
1       Programming language LG in Python PY        ...             Python PY
2             Its easier LG to understand  PY       ...        understand  PY
3   The syntax of the language LG is clean PY       ...              clean PY
[4 rows x 3 columns]