CODEWITHSUNDEEP
pythonpandasmatplotlibplotimdb
user8864088
user8864088

Reputation:

pandas: How to plot the pie diagram for the movie counts versus genre of IMDB movies in pandas?

I have the following dataset:

import pandas as pd
import numpy as np 
%matplotlib inline

df = pd.DataFrame({'movie' : ['A', 'B','C','D'], 
                   'genres': ['Science Fiction|Romance|Family', 'Action|Romance',
                              'Family|Drama','Mystery|Science Fiction|Drama']},
                  index=range(4))
df

My attempt

# Parse unique genre from all the movies
gen = []
for g in df['genres']:
    gg = g.split('|')
    gen = gen + gg
    gen = list(set(gen))

print(gen)

df['genres'].value_counts().plot(kind='pie')

I got this image: enter image description here

But I would like to pie chart for each separate genres.

How we get the genres for number count of movies for each unique genres?

Upvotes: 5

Views: 8556

Answers (2)

Lev Zakharov
Lev Zakharov

Reputation: 2427

So, the one-liner solution:

df.genres.str.get_dummies().sum().plot.pie(label='Genre', autopct='%1.0f%%')

Result:

enter image description here

TL;DR

Firstly, convert your categories column to dummies:

df = pd.concat([df.drop('genres', axis=1), df.genres.str.get_dummies()], axis=1)

Result:

  movie  a  b  c  d  e  f  g
0     A  1  1  1  0  0  0  0
1     B  0  0  1  0  1  0  0
2     C  0  0  0  0  0  1  1
3     D  1  1  0  1  1  0  0

Then count number of occurrences for each category:

counts = df.drop('movie', axis=1).sum()

Result:

a    2
b    2
c    2
d    1
e    2
f    1
g    1

And finally plot the pie chart:

counts.plot.pie()

enter image description here

Upvotes: 8

ALollz
ALollz

Reputation: 59579

You can do .str.split() with expand=True, which will give you a DataFrame of all the genres. If you then stack that, you will get the value counts for all of the genres.

df.genres.str.split('|', expand=True).stack().value_counts().plot(kind='pie', label='Genre')

enter image description here

That can be a bit on the slower side to calculate the counts, so a faster implementation for the same plot would be (adding the percentages):

from itertools import chain
from collections import Counter
import matplotlib.pyplot as plt

cts = Counter(chain.from_iterable(df.genres.str.split('|').values))
_ = plt.pie(cts.values(), labels=cts.keys(), autopct='%1.0f%%')
_ = plt.ylabel('Genres')

enter image description here

Upvotes: 6

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