将 pandas 列中的列表转换为字符串

Posted 2023-03-12

【中文标题】将 pandas 列中的列表转换为字符串

【英文标题】：Convert List of lists in a pandas column to string

【发布时间】：2019-01-14 20:52:48

【问题描述】：

如何将包含列表列表的 pandas df 列转换为字符串。 df 中“类别”列的 sn-p

[['Electronics', 'Computers & Accessories', 'Cables & Accessories', 'Cables & Interconnects', 'USB Cables'], ['Video Games', 'Sony PSP']]
[['Video Games', 'PlayStation 3', 'Accessories', 'Controllers', 'Gamepads']]
[['Cell Phones & Accessories', 'Accessories', 'Chargers', 'Travel Chargers'], ['Video Games', 'Nintendo DS']]

我尝试了以下代码：

df.loc[:,"categories"]=[item for sublist in df.loc[:,"categories"] for item in sublist]

但它给了我一个错误。有没有其他方法可以做到这一点？

ValueError：值的长度与索引的长度不匹配

预期列：

'Electronics', 'Computers & Accessories', 'Cables & Accessories', 'Cables & Interconnects', 'USB Cables','Video Games', 'Sony PSP'
'Video Games', 'PlayStation 3', 'Accessories', 'Controllers', 'Gamepads'
'Cell Phones & Accessories', 'Accessories', 'Chargers', 'Travel Chargers','Video Games', 'Nintendo DS'

【问题讨论】：

你的预期输出是什么？

用预期的输出修改了原来的 qn

【参考方案1】：

使用带有join的嵌套生成器：

df["categories"]=[', '.join(item for sublist in x for item in sublist) for x in df["categories"]]

如果性能在较大的DataFrame 中很重要：

from  itertools import chain

df["categories"] = [', '.join(chain.from_iterable(x)) for x in df["categories"]]

print (df)
                                          categories
0  Electronics, Computers & Accessories, Cables &...
1  Video Games, PlayStation 3, Accessories, Contr...
2  Cell Phones & Accessories, Accessories, Charge...

时序：（实际数据应该不同，最好先测试一下）：

df = pd.concat([df] * 10000, ignore_index=True)


In [45]: %timeit df["c1"]=[', '.join(item for sublist in x for item in sublist) for x in df["categories"]]
39 ms ± 706 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [46]: %timeit df["c2"]=[', '.join(chain.from_iterable(x)) for x in df["categories"]]
22.1 ms ± 258 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [47]: %timeit df['c3'] = df["categories"].apply(lambda x: ', '.join(str(r) for v in x for r in v))
66.7 ms ± 695 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

【参考方案2】：

你需要列表理解

df['col'] = df.col.apply(lambda x: ', '.join(str(r) for v in x for r in v))

输出：

    col
0   Electronics, Computers & Accessories, Cables &...
1   Video Games, PlayStation 3, Accessories, Contr...
2   Cell Phones & Accessories, Accessories, Charge...