CODEWITHSUNDEEP
stringlistpandasdataframetext
pdavid
pdavid

Reputation: 67

Convert List of lists in a pandas column to string

How do i convert a column of pandas df consisting of a list of lists to a string. A snippet of the column 'categories' in a df

[['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']]

I tried the following code:

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

but its giving me an error. Is there any other way of doing this?

ValueError: Length of values does not match length of index

Expected column: 

'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'

Upvotes: 1

Views: 606

Answers (2)

jezrael
jezrael

Reputation: 863531

Use nested generator with join:

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

If performance is important in larger DataFrame:

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...

Timings: (in real data should be different, best test it first):

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)

Upvotes: 1

harpan
harpan

Reputation: 8641

You need list comprehension 

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

Output:

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

Upvotes: 0

Related Questions