您可能会发现以下逻辑适用于您的数据：

df["Column_2"] = df["Column_1"].str.extract(r'\w+ (\S+(?: \S+)*) \b(?:small|medium|large)\b')

上面的模式从第二个术语开始匹配，直到到达small，medium或large关键字。这里是一个工作正则表达式demo。

看看你发布的例子，拆分列值并返回“中间”项就足够了。你可以做一个简单的函数来封装逻辑并将其应用到 Dataframe 。

from math import floor

df = pd.DataFrame(
    {'Columns_1':
     ["Lyreco A-Type small 2i",
      "Lyreco C-Type small 4i",
      "Lyreco N-Part medium", 
      "Lyreco AKG MT 4i small",
      "Lyreco AKG/ N-Type medium 4i",
      "Lyreco C-Type medium 2i",
      "Lyreco C-Type/ SNU medium 2i",
      "Lyreco K-part small 4i",
      "Lyreco K-Part medium", 
      "Lyreco SNU small 2i",
      "Lyreco C-Part large 2i",
      "Lyreco N-Type large 4i"
     ]
    }
)

def f(row):
    blocks = row['Columns_1'].split()
    mid_index = 1 if len(blocks) <= 4 else floor(len(blocks)/2)
    return ' '.join(blocks[1:mid_index+1])

df['Columns_2'] = df.apply(f, axis=1)

print(df)

输出：

Columns_1    Columns_2
0         Lyreco A-Type small 2i       A-Type
1         Lyreco C-Type small 4i       C-Type
2           Lyreco N-Part medium       N-Part
3         Lyreco AKG MT 4i small       AKG MT
4   Lyreco AKG/ N-Type medium 4i  AKG/ N-Type
5        Lyreco C-Type medium 2i       C-Type
6   Lyreco C-Type/ SNU medium 2i  C-Type/ SNU
7         Lyreco K-part small 4i       K-part
8           Lyreco K-Part medium       K-Part
9            Lyreco SNU small 2i          SNU
10        Lyreco C-Part large 2i       C-Part
11        Lyreco N-Type large 4i       N-Type

df.columns = ['column_1']

df["column_2"] = [col.split(" ")[1] for col in df.column_1]