Convert a columns of string to list in pandas

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Solution 1

You can use ast.literal_eval, which will give you a tuple:

import ast
df.LABELS = df.LABELS.apply(ast.literal_eval)

If you do want a list, use:

df.LABELS.apply(lambda s: list(ast.literal_eval(s)))

Solution 2

Use str.strip and str.split:

df['LABELS'] = df['LABELS'].str.strip('()').str.split(',')

But if no NaNs here, list comprehension working nice too:

df['LABELS'] = [x.strip('()').split(',') for x in df['LABELS']]

Solution 3

Sorry I was late to the party. So for other latecomers I got this to work based on the above replies:

df['hashtags'] = df.apply(lambda row:  row['hashtags'].strip('[]').replace('"', '').replace(' ', '').split(',')   , axis=1)

I loaded a csv with some columns looking like this ...,['hashtag1','hashtag2'],... and the Panda DataFrame loaded it as a string object. I used the above code and it converted to list. I then used "explode" to flatten the data.

Solution 4

Alternatively, you might consider regular expressions:

pattern = re.compile("[0-9]\.[0-9]")
df.LABELS.apply(pattern.findall)

Solution 5

You can try this (assuming your csv is called filename.csv):

df = pd.read_csv('filename.csv')

df['LABELS'] = df.LABELS.apply(lambda x: x.strip('()').split(','))

>>> df
   ID                               LABELS
0   1  [1.0, 2.0, 2.0, 3.0, 3.0, 1.0, 4.0]
1   2  [1.0, 2.0, 2.0, 3.0, 3.0, 1.0, 4.0]
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Guido Muscioni
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Guido Muscioni

Doing R&D in healthcare.

Updated on January 08, 2021

Comments

  • Guido Muscioni
    Guido Muscioni over 3 years

    I have a problem with the type of one of my column in a pandas dataframe. Basically the column is saved in a csv file as a string, and I wanna use it as a tuple to be able to convert it in a list of numbers. Following there is a very simple csv:

    ID,LABELS
    1,"(1.0,2.0,2.0,3.0,3.0,1.0,4.0)"
    2,"(1.0,2.0,2.0,3.0,3.0,1.0,4.0)"
    

    If a load it with the function "read_csv" I get a list of strings. I have tried to convert to a list, but I get the list version of a string:

    df.LABELS.apply(lambda x: list(x))
    

    returns:

    ['(','1','.','0',.,.,.,.,.,'4','.','0',')']
    

    Any idea on how to be able to do it?

    Thank you.

  • BENY
    BENY almost 6 years
    I will say this is the faster solution among 3 :-)