How do you read in a dataframe with lists using pd.read_clipboard?

Question

Here's some data from another question:

                          positive                 negative          neutral
1   [marvel, moral, bold, destiny]                       []   [view, should]
2                      [beautiful]      [complicated, need]               []
3                      [celebrate]   [crippling, addiction]            [big]

What I would do first is to add quotes across all words, and then:

import ast

df = pd.read_clipboard(sep='\s{2,}')
df = df.applymap(ast.literal_eval)

Is there a smarter way to do this?

Answer 1

Lists of strings

For basic structures you can use yaml without having to add quotes:

import yaml
df = pd.read_clipboard(sep='\s{2,}').applymap(yaml.load)

type(df.iloc[0, 0])
Out: list

---

Lists of numeric data

Under certain conditions, you can read your lists as strings and the convert them using literal_eval (or pd.eval, if they are simple lists).

For example,

           A   B
0  [1, 2, 3]  11
1  [4, 5, 6]  12

First, ensure there are at least two spaces between the columns, then copy your data and run the following:

import ast 

df = pd.read_clipboard(sep=r'\s{2,}', engine='python')
df['A'] = df['A'].map(ast.literal_eval)    
df
    
           A   B
0  [1, 2, 3]  11
1  [4, 5, 6]  12

df.dtypes

A    object
B     int64
dtype: object

Notes

• for multiple columns, use applymap in the conversion step:

  df[['A', 'B', ...]] = df[['A', 'B', ...]].applymap(ast.literal_eval)
  

• if your columns can contain NaNs, define a function that can handle them appropriately:

  parser = lambda x: x if pd.isna(x) else ast.literal_eval(x)
  df[['A', 'B', ...]] = df[['A', 'B', ...]].applymap(parser)
  

• if your columns contain lists of strings, you will need something like yaml.load (requires installation) to parse them instead if you don't want to manually add quotes to the data. See above.

Answer 2

I did it this way:

df = pd.read_clipboard(sep='\s{2,}', engine='python')
df = df.apply(lambda x: x.str.replace(r'[\[\]]*', '').str.split(',\s*', expand=False))

PS i'm sure - there must be a better way to do that...

Answer 3

Another alternative is

In [43]:  df.applymap(lambda x: x[1:-1].split(', '))
Out[43]: 
                         positive                negative         neutral
1  [marvel, moral, bold, destiny]                      []  [view, should]
2                     [beautiful]     [complicated, need]              []
3                     [celebrate]  [crippling, addiction]           [big]

Note that this assumes the first and last character in each cell is [ and ]. It also assumes there is exactly one space after the commas.

Answer 4

Another version:

df.applymap(lambda x:
            ast.literal_eval("[" + re.sub(r"[[\]]", "'", 
                                          re.sub("[,\s]+", "','", x)) + "]"))

Answer 5

Per help from @MaxU

df = pd.read_clipboard(sep='\s{2,}', engine='python')

Then:

>>> df.apply(lambda col: col.str[1:-1].str.split(', '))
                         positive                negative         neutral
1  [marvel, moral, bold, destiny]                      []  [view, should]
2                     [beautiful]     [complicated, need]              []
3                     [celebrate]  [crippling, addiction]           [big]

>>> df.apply(lambda col: col.str[1:-1].str.split()).loc[3, 'negative']
['crippling', 'addiction']

And per the notes from @unutbu who came up with a similar solution:

assumes the first and last character in each cell is [ and ]. It also assumes there is exactly one space after the commas.