如何使以下python程序（代码）更高效？

Question

假设数据很大，则解决以下问题的任何有效方法。 我解决了问题，但是如何改进代码，这将使其变得高效。 有什么建议么？

数据：

movie_sub_themes = {
'Epic': ['Ben Hur', 'Gone With the Wind', 'Lawrence of Arabia'],
'Spy': ['James Bond', 'Salt', 'Mission: Impossible'],
'Superhero': ['The Dark Knight Trilogy', 'Hancock, Superman'],
'Gangster': ['Gangs of New York', 'City of God', 'Reservoir Dogs'],
'Fairy Tale': ['Maleficent', 'Into the Woods', 'Jack the Giant Killer'],
'Romantic':['Casablanca', 'The English Patient', 'A Walk to Remember'],
'Epic Fantasy': ['Lord of the Rings', 'Chronicles of Narnia', 'Beowulf']}

movie_themes = {
'Action': ['Epic', 'Spy', 'Superhero'],
'Crime' : ['Gangster'],
'Fantasy' : ['Fairy Tale', 'Epic Fantasy'],
'Romance' : ['Romantic']}

themes_keys = movie_themes.keys()
theme_movies_keys = movie_sub_themes.keys()

#Iterate in movie_themes
#Check movie_themes keys in movie_sub_keys
#if yes append the movie_sub_keys into the newdict
newdict = {}
for i in range(len(themes_keys)):
   a = []
   for j in range(len(movie_themes[themes_keys[i]])):
     try:
         if movie_themes[themes_keys[i]][j] in theme_movies_keys:
            a.append(movie_sub_themes[movie_themes[themes_keys[i]][j]])
     except:
         pass
   newdict[themes_keys[i]] = a

# newdict contains nested lists
# Program to unpack the nested list into single list
# Storing the value into theme_movies_data 
theme_movies_data = {}
for k, v in newdict.iteritems():
    mylist_n = [j for i in v for j in i]
    theme_movies_data[k] = dict.fromkeys(mylist_n).keys()

print (theme_movies_data)

输出：

{'Action': ['Gone With the Wind', 'Ben Hur','Hancock, Superman','Mission: Impossible','James Bond','Lawrence of Arabia','Salt','The Dark Knight Trilogy'],
 'Crime': ['City of God', 'Reservoir Dogs', 'Gangs of New York'],
 'Fantasy': ['Jack the Giant Killer','Beowulf','Into the Woods','Maleficent','Lord of the Rings','Chronicles of Narnia'],
 'Romance': ['The English Patient', 'A Walk to Remember', 'Casablanca']}

抱歉，无法正确注释代码。

我更担心跑步时间。

谢谢..

Answer 1

您可以使用关系数据库存储两个表，一个表是电影及其子主题，另一个表将子主题与电影主题相关联。 然后，您可以使用SQL查询数据库，选择所有电影及其相关电影主题的列表。

这种方法将更加有效，因为倾向于为了提高处理速度而编译SQL命令。 关系数据库模型具有很好的可伸缩性，因此可以以最小的开销处理非常大的数据集。

有关在Python中创建和使用简单数据库的示例，请参见此处 。 如果您不熟悉SQL操作，请参见此处以获取有关有用操作的简单教程。

Answer 2

这是我的解决方案（使用defaultdict）：

movie_sub_themes = {
'Epic': ['Ben Hur', 'Gone With the Wind', 'Lawrence of Arabia'],
'Spy': ['James Bond', 'Salt', 'Mission: Impossible'],
'Superhero': ['The Dark Knight Trilogy', 'Hancock, Superman'],
'Gangster': ['Gangs of New York', 'City of God', 'Reservoir Dogs'],
'Fairy Tale': ['Maleficent', 'Into the Woods', 'Jack the Giant Killer'],
'Romantic':['Casablanca', 'The English Patient', 'A Walk to Remember'],
'Epic Fantasy': ['Lord of the Rings', 'Chronicles of Narnia', 'Beowulf']}

movie_themes = {
'Action': ['Epic', 'Spy', 'Superhero'],
'Crime' : ['Gangster'],
'Fantasy' : ['Fairy Tale', 'Epic Fantasy'],
'Romance' : ['Romantic']}

from collections import defaultdict
newdict = defaultdict(list)

for theme, sub_themes_list in movie_themes.items():
    for sub_theme in sub_themes_list:
        newdict[theme] += movie_sub_themes.get(sub_theme, [])       

dict(newdict)

>> {'Action': ['Ben Hur',
  'Gone With the Wind',
  'Lawrence of Arabia',
  'James Bond',
  'Salt',
  'Mission: Impossible',
  'The Dark Knight Trilogy',
  'Hancock, Superman'],
 'Crime': ['Gangs of New York', 'City of God', 'Reservoir Dogs'],
 'Fantasy': ['Maleficent',
  'Into the Woods',
  'Jack the Giant Killer',
  'Lord of the Rings',
  'Chronicles of Narnia',
  'Beowulf'],
 'Romance': ['Casablanca', 'The English Patient', 'A Walk to Remember']}

时间：4.84 µs与14.6 µs