Python 错误';int';对象不可下标,如何将其转换为列表?
我是python新手,还在学习, 我有以下代码:Python 错误';int';对象不可下标,如何将其转换为列表?,python,data-analysis,Python,Data Analysis,我是python新手,还在学习, 我有以下代码: # columns are [0]title [1]year [2]rating [3]length(min) [4]genre [5]budget($mil) [6]box_office_gross($mil) oscar_data = [ ["The Shape of Water", 2017, 6.914, 123, ['sci-fi', 'drama'], 19.4, 195.243464], [&qu
# columns are [0]title [1]year [2]rating [3]length(min) [4]genre [5]budget($mil) [6]box_office_gross($mil)
oscar_data = [
["The Shape of Water", 2017, 6.914, 123, ['sci-fi', 'drama'], 19.4, 195.243464],
["Moonlight", 2016, 6.151, 110, ['drama'], 1.5, 65.046687],
["Spotlight", 2015, 7.489, 129, ['drama', 'crime', 'history'], 20.0, 88.346473],
["Birdman", 2014, 7.604, 119, ['drama', 'comedy'], 18.0, 103.215094]
]
def filter_by_genre(data, genre):
result = []
for row in data:
genres = row[4]
if genre in genres:
result.append(row)
return result
all_genres = [
'sci-fi', 'drama', 'crime', 'history', 'comedy', 'biography',
'thriller', 'war', 'melodrama', 'action', 'adventure', 'western',
'mystery', 'horror'
]
genres_counts = []
for genre in all_genres:
count = len(filter_by_genre(oscar_data, genre))
genres_counts.append(genre)
genres_counts.append(count)
print('Genre | Number')
print('------------------------')
for row in genres_counts:
genre = row[0]
count = row[1]
print('{: <11} | {: >10}'.format(genre, count))
我该怎么做才能把它列出来
谢谢大家! 此处介绍了您的错误:
genres_counts.append(genre)
genres_counts.append(count)
可以将流派和计数作为单独的值追加,而不是作为集合追加。将这两行替换为:
genres_counts.append((genre, count))
更好的方法是使用列表理解:
genres_counts = [(genre, len(filter_by_genre(oscar_data, genre))
for genre in all_genres]
是否要将int转换为一个范围?例如,5将变为列表[0、1、2、3、4]。谢谢,它现在可以工作了!我会尽快接受答案。为什么我需要一个双
(())
?内括号定义了一个2元素元组。此外,请查看编辑。
genres_counts = [(genre, len(filter_by_genre(oscar_data, genre))
for genre in all_genres]