CODEWITHSUNDEEP
pythonpandasdataframe
Mob_Abominator
Mob_Abominator

Reputation: 354

how do I split a column into seperate columns in a csv file?

So I'm working on a movie genre data set and the dataset has all the genres in a single column but I want to split them.

here's how the data set looks like:

   genres
----------------------------------------------
   [{'id': 16, 'name': 'Animation'}, {'id': 35, 'name': 'Comedy'}, {'id': 10751, 'name': 'Family'}]
   [{'id': 35, 'name': 'Comedy'}, {'id': 10749, 'name': 'Romance'}]
   [{'id': 35, 'name': 'Comedy'}, {'id': 18, 'name': 'Drama'}, {'id': 10749, 'name': 'Romance'}]
   [{'id': 35, 'name': 'Comedy'}]
   [{'id': 28, 'name': 'Action'}, {'id': 80, 'name': 'Crime'}, {'id': 18, 'name': 'Drama'}, {'id': 53, 'name': 'Thriller'}]

So what I want to do is get only the first genre so the new column should look like:

  genres
_____________
  Animation
  Comedy
  Comedy
  Comedy
  Action

enter image description here

I hope this is clear enough to understand my problem.

Upvotes: 0

Views: 337

Answers (3)

ansev
ansev

Reputation: 30940

Use DataFrame.apply. The first dictionary in the list is selected in each cell. From that dictionary the name field is selected:

df['genres']=df['genres'].apply(lambda x: x[0]['name'])
print(df)

   ID     genres
0   0  Animation
1   1     Comedy
2   2     Comedy
3   3     Comedy
4   4     Action

or 

df['genres']=df['genres'].apply(lambda x: eval(x)[0]['name'])

TRY THIS

def decode_str_dict(x):
    try:
        out=eval(x)[0]['name']
    except Exception:
        try:
            out=eval(x)['name']
        except Exception:
            try:
                out=eval(x)
            except Exception:
                out=x
    return out



df['genres'].apply(decode_str_dict)

Upvotes: 5

Michael Gardner
Michael Gardner

Reputation: 1813

This works if the values are considered a string.

from ast import literal_eval

df['genres'] = df.genres.map(lambda x: literal_eval(x)[0]['name'])

Result:

Out[294]: 
  ID     genres
1  0  Animation
2  1     Comedy
3  2     Comedy
4  3     Comedy
5  4     Action

Upvotes: 3

chlebek
chlebek

Reputation: 2451

df['genres'] = df['genres'].map(lambda x:[i['name'] for i in x])
df['first_genre'] = df['genres'][0]
df = df[['name','first_genre']]

Upvotes: 3

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