将列中的值追加到包含列表的另一列的开头

Question

我有一个包含以下值的数据框

    Keyword             Synonyms
0   fuzz                [hair, copper, pig, bull, cop, blur]
1   napkins             [table]
2   rationalizations    [systematization, rationalization]
3   grandmasters        [grandmaster]
4   rehearsing          [practice]

为了在以后的步骤中创建下拉菜单，我如何将“关键字”列中的值附加到同义词列中每个列表的开头？

我的代码如下：

df['Synonyms'] = pd.concat([df['Keyword'].append(df['Synonyms'])

Answer 1

列表理解如何？在python3中，您可以使用扩展的可迭代解压缩功能来有效地创建新列表。

# df['Synonyms'] = [[x, *y] for x, y in df[['Keyword'], 'Synonyms']].values]
df['Synonyms'] = [[x, *y] for x, y in zip(df['Keyword'], df['Synonyms'])]
df

            Keyword                                           Synonyms
0              fuzz         [fuzz, hair, copper, pig, bull, cop, blur]
1           napkins                                   [napkins, table]
2  rationalizations  [rationalizations, systematization, rationaliz...
3      grandmasters                        [grandmasters, grandmaster]
4        rehearsing                             [rehearsing, practice]

---

这是map的有趣替代品：

k = iter(df['Keyword'].tolist())
df['Synonyms'] = df['Synonyms'].map(lambda x: [next(k), *x])
df

            Keyword                                           Synonyms
0              fuzz         [fuzz, hair, copper, pig, bull, cop, blur]
1           napkins                                   [napkins, table]
2  rationalizations  [rationalizations, systematization, rationaliz...
3      grandmasters                        [grandmasters, grandmaster]
4        rehearsing                             [rehearsing, practice]

Answer 2

如果是这种情况

df['Keyword'].apply(lambda x : [x])+df['Synonyms']