如何从嵌套列表中找到包含较高值的列表并返回这些列表？

3条回答

网友

1楼 · 编辑于 2024-05-18 11:17:17

这是我能想到的最具Python风格的方法。我的方法是首先按sublist[3]对列表列表进行排序，这意味着当我们遍历列表时，我们将在遇到重复列表之前遇到具有最大审阅次数的子列表。此技巧将用于构造最终列表

meta_list = [['Coloring book moana', 'ART_AND_DESIGN', '3.9', 967, '14M', '500,000+', 'Free', '0', 'Everyone', 'Art & Design;Pretend Play', 'January 15, 2018', '2.0.0', '4.0.3 and up'],
 ['Coloring book moana', 'FAMILY', '3.9', 974, '14M', '500,000+', 'Free', '0', 'Everyone', 'Art & Design;Pretend Play', 'January 15, 2018', '2.0.0', '4.0.3 and up'],
 ['Gmail', 'COMMUNICATION', '4.3', 4604324, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Everyone', 'Communication', 'August 2, 2018', 'Varies with device', 'Varies with device'],
 ['Gmail', 'COMMUNICATION', '4.3', 4604483, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Everyone', 'Communication', 'August 2, 2018', 'Varies with device', 'Varies with device'],
 ['Instagram', 'SOCIAL', '4.5', 66577313, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device'],
 ['Instagram', 'SOCIAL', '4.5', 66577446, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device'],
 ['Instagram', 'SOCIAL', '4.5', 66509917, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device']]

# Sort the list by review count and review name - make sure the highest review is first
meta_list.sort(key=lambda x: (int(x[3]), x[0]), reverse=True)

# This is the list we'll use to store the final data in
final_list = []
# Go through all the items in the meta_list
for meta in meta_list:
    
    if not meta[0] in [item[0] for item in final_list]:
        '''
        If another meta with the same name (0th index)
        doesn't already exist in final_list, add it
        '''
        final_list.append(meta)

输出-

[['Instagram',
  'SOCIAL',
  '4.5',
  66577446,
  'Varies with device',
  '1,000,000,000+',
  'Free',
  '0',
  'Teen',
  'Social',
  'July 31, 2018',
  'Varies with device',
  'Varies with device'],
 ['Gmail',
  'COMMUNICATION',
  '4.3',
  4604483,
  'Varies with device',
  '1,000,000,000+',
  'Free',
  '0',
  'Everyone',
  'Communication',
  'August 2, 2018',
  'Varies with device',
  'Varies with device'],
 ['Coloring book moana',
  'FAMILY',
  '3.9',
  974,
  '14M',
  '500,000+',
  'Free',
  '0',
  'Everyone',
  'Art & Design;Pretend Play',
  'January 15, 2018',
  '2.0.0',
  '4.0.3 and up']]

基本上，它将所有不存在的元添加到final_list。为什么这样做有效？因为循环时遇到的第一个meta是审核计数最高的。所以，一旦那一个被添加，它的副本就不能被添加，我们就完成了

注意：这不会保留评论本身的顺序。它只会确保保留审查次数最高的审查，以防有同名的重复

网友
2楼 · 编辑于 2024-05-18 11:17:17

大概是这样的：
_DATA = [ ['Coloring book moana', 'ART_AND_DESIGN', '3.9', 967, '14M', '500,000+', 'Free', '0', 'Everyone', 'Art & Design;Pretend Play', 'January 15, 2018', '2.0.0', '4.0.3 and up'], ['Coloring book moana', 'ART_AND_DESIGN', '3.9', 974, '14M', '500,000+', 'Free', '0', 'Everyone', 'Art & Design;Pretend Play', 'January 15, 2018', '2.0.0', '4.0.3 and up'], ['Gmail', 'COMMUNICATION', '4.3', 4604324, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Everyone', 'Communication', 'August 2, 2018', 'Varies with device', 'Varies with device'], ['Gmail', 'COMMUNICATION', '4.3', 4604483, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Everyone', 'Communication', 'August 2, 2018', 'Varies with device', 'Varies with device'], ['Instagram', 'SOCIAL', '4.5', 66577313, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device'], ['Instagram', 'SOCIAL', '4.5', 66577446, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device'], ['Instagram', 'SOCIAL', '4.5', 66509917, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device'] ] def print_highest(data): list_map = {} for d in data: key = str(d[0:3] + d[4:]) if key not in list_map: list_map[key] = d continue if d[3] > list_map[key][3]: list_map[key] = d for l in list_map.values(): print(l) print_highest(_DATA)
输出：
['Coloring book moana', 'ART_AND_DESIGN', '3.9', 974, '14M', '500,000+', 'Free', '0', 'Everyone', 'Art & Design;Pretend Play', 'January 15, 2018', '2.0.0', '4.0.3 and up'] ['Gmail', 'COMMUNICATION', '4.3', 4604483, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Everyone', 'Communication', 'August 2, 2018', 'Varies with device', 'Varies with device'] ['Instagram', 'SOCIAL', '4.5', 66577446, 'Varies with device', '1,000,000,000+', 'Free', '0', 'Teen', 'Social', 'July 31, 2018', 'Varies with device', 'Varies with device']

网友
3楼 · 编辑于 2024-05-18 11:17:17

对于这个问题，可能有一个更优雅的/类似Python的解决方案，但这里有一个可能的途径：

my_list = [...] # Nested list here

def compare_duplicates(nested_list, name_index=0, compare_index=3):
    max_values = dict() # Used two dictionaries for readability
    final_indexes = dict()

    for i, item in enumerate(nested_list):
        name, value = item[name_index], item[compare_index]

        if value > max_values.get(name, 0):
            max_values[name] = value
            final_indexes[name] = i

    return [nested_list[i] for i in final_indexes.values()]

print(compare_duplicates(my_list))

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