如何从嵌套列表和 return 那些列表中找到包含较高值的列表？

Question

我有这个包含重复条目的嵌套列表：

[['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']]

我想通过 i[3] 过滤嵌套列表，所以最终输出将是这样的

[['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'],
 ['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']]

我尝试了 for 循环，但我不知道如何获得重复列表的最大值

Answer 1

这个问题可能有更优雅/pythonic 的解决方案，但这是一个可能的途径：

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))

Answer 2

像这样：

_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']

Answer 3

这是我能想到的最 pythonic 的方法。我的做法是先对列表的列表进行排序，按sublist[3]排序，这意味着当我们遍历列表时，我们最终会在遇到重复项之前遇到具有最大评论数的子列表。这个技巧将用于构建最终列表。

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']]

基本上它会将所有不存在的元数据添加到 final_list。为什么这行得通？因为您在循环时遇到的第一个元 是具有最高评论数 的元。所以一旦添加了那个，它的复制品就无法添加，我们就完成了。

注意：这不会保留评论本身的顺序。它只会确保只保留评论数最高的评论，以防出现同名的重复评论。

如何从嵌套列表和 return 那些列表中找到包含较高值的列表？

How to find the list that contains the higher value from nested list and return those lists?

python

nested-lists