使用str.split和 * 列表组件 *：

df["List Stuff"] = [x for l in df.pop("Aleatory Stuff").str.split(", ")
                    for x in l if x in list_0]

输出：

print(df)

   ID List Stuff
0   1       rice
1   2  breakfast
2   3       test
3   4       Ball
4   5  cellphone

你可以尝试这样做。你可以使用str.split(", ")来拆分字符串列Aleatory stuff，其中", "是你的大小写中单词之间的分隔符。一旦项目被拆分，你可以使用set.intersection(set)来查找每一行和列表之间的公共项目。
你可以用一个类似的例子来检查我的代码：

import pandas as pd 

test_data = [
    {"id": 1, "aleatory_stuff": "banana, rice, potato"},
    {"id": 2, "aleatory_stuff": "cucumber, breakfast"}
]

test_list = ["rice", "cucumber"]

test_df = pd.DataFrame(test_data)
test_df["list_stuff"] = test_df["aleatory_stuff"].str.split(", ").apply(lambda x: list(set(x).intersection(set(test_list)))[0])

print(test_df)

id        aleatory_stuff list_stuff
0   1  banana, rice, potato       rice
1   2   cucumber, breakfast   cucumber

将extractall用于聚合：

import re

pattern = '|'.join(map(re.escape, list_0))
# 'Ball|cellphone|breakfast|rice|test'

df['List Stuff'] = (df['Aleatory Stuff'].str.extractall(f'({pattern})')
                    [0].groupby(level=0).agg(', '.join)
                   )

• 注意：如果您想要不区分大小写的匹配，请将flags=re.I作为参数添加到extractall。*

输出：

ID           Aleatory Stuff List Stuff
0   1     banana, rice, potato       rice
1   2      cucumber, breakfast  breakfast
2   3     Heaven, Bucket, test       test
3   4  Ball, Pan, Candy, shirt       Ball
4   5  draw, cellphone, bubble  cellphone

让我们使用str.findall来查找匹配regex模式的所有示例

pat = r'(?i)\b(%s)\b' % '|'.join(list_0)
df['List Stuff'] = df['Aleatory Stuff'].str.findall(pat).str.join(',')

ID           Aleatory Stuff List Stuff
0   1     banana, rice, potato       rice
1   2      cucumber, breakfast  breakfast
2   3     Heaven, Bucket, test       test
3   4  Ball, Pan, Candy, shirt       Ball
4   5  draw, cellphone, bubble  cellphone

PS：正则表达式标记(?i)用于不区分大小写的匹配