有时候，在处理数据库时，我们需要执行某些列表操作，这些操作更像是查询语言，例如，将嵌套的列表元素相对于其他索引元素进行分组。本文讨论二元嵌套列表，并将每个嵌套列表元素相对于其其他索引元素进行分组。

1. 列表解析

# initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

# using list comprehension

# to perform binary list grouping

temp = set(map(lambda i: i[1], test_list))

res = [[j[0] for j in test_list if j[1] == i] for i in temp]

# printing result

print("The grouped list is : " + str(res))

输出

The grouped list is : [['G', 'F', 'G'], ['I', 'S'详情页外包], ['B', 'E', 'S', 'T']]

2. 使用itertools.groupby + itemgetter

我们也可以使用groupby函数来执行这个特定的任务。该方法遵循2-3个步骤。首先，序列相对于第二个元素进行排序，现在可以将其馈送以进行分组。最后，我们根据结果的要求打印第一个元素。

from itertools import groupby

from operator import itemgetter

# Initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

# Performing binary list grouping

# using itertools.groupby() + itemgetter()

test_list.sort(key=itemgetter(1))

groups = groupby(test_list, itemgetter(1))

res = [[i[0] for i in val] for (key, val) in groups]

# Printing the resultant list

print("The grouped list is : " + str(res))

输出

The grouped list is : [['G', 'F', 'G'], ['I', 'S'], ['B', 'E', 'S', 'T']]

3. 使用collections.defaultdict()

import collections

# initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

# using collections.defaultdict()

# to perform binary list grouping

res = collections.defaultdict(list)

for val in test_list:

res[val[1]].append(val[0])详情页外包

# printing result

print("The grouped list is : " + str(res.values()))

输出

The grouped list is : dict_values([['G', 'F', 'G'], ['I', 'S'], ['B', 'E', 'S', 'T']])

4. 使用for循环 + sort

# initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

# using list comprehension

# to perform binary list grouping

x = []

for i in test_list:

if i[1] not in x:

x.append(i[1])

x.sort()

res = []

for i in x:

a = []

for j in test_list:

if j[1] == i:

a.append(j[0])

res.append(a)

# printing result

print("The grouped list is : " + str(res))

输出

The grouped list is : [['G', 'F', 'G'], ['I', 'S'], ['B', 'E', 'S', 'T']]

5. 使用numpy

import numpy as np

# initializing list

test_list = [["G", 0], ["F", 0], ["B", 2], ["E", 2],

['I', 1], ['S', 1], ['S', 2], ['T', 2], ['G', 0]]

arr = np.array(test_list)

# Storing all unique values

unique_values = np.unique(arr[:, 1])

result_list = [list(arr[arr[:, 1] == i, 0]) for i in unique_values]

# Printing the result

print("The grouped list is:", result_list)

输出

The grouped list is: [['G', 'F', 'G'], ['I', 'S'], ['B', 'E', 'S', 'T']]

6. 使用dictionary和setdefault

此方法使用字典按条目的第二个元素对条目进行分组，在线美工该元素充当字典中的键。setdefault方法用于创建一个空列表作为新键的值，或者将当前项的第一个元素附加到与该键关联的列表中。然后将生成的字典值转换为列表以获得最终结果。

# initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

# Using dictionary and setdefault() to perform binary list grouping

group_dict = {}

for item in test_list:

group_dict.setdefault(item[1], []).append(item[0])

res = list(group_dict.values())

# printing result

print ("The grouped list is : " + str(res))

输出

The grouped list is : [['G', 'F', 'G'], ['B', 'E', 'S', 'T'], ['I', 'S']]

7. 使用pandas

它初始化一个名为test_list的列表，其中包含两个元素的子列表。然后，它将列表转换为pandas DataFrame，按每个子列表的第二个元素对DataFrame进行分组，应用lambda函数选择每个子列表的第一个元素，将结果pandas Series转换为列表，并打印分组后的列表。

import pandas as pd

# initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

# converting list to DataFrame

df = pd.DataFrame(test_list, columns=['value', 'group'])

# grouping by group

grouped = df.groupby('group')['value'].apply(lambda x: x.tolist()).tolist()

# printing result

print("The grouped list is: " + str(grouped))

输出

The grouped list is: [['G', 'F', 'G'], ['I', 'S'], ['B', 'E', 'S', 'T']]

8. 使用列表和索引

# Initializing list

test_list = [["G", 0], ["F", 0], ["B", 2],

["E", 2], ['I', 1], ['S', 1],

['S', 2], ['T', 2], ['G', 0]]

num_groups = max([elem[1] for elem in test_list]) + 1

grouped = [[] for _ in range(num_groups)]

# Iterating over our input list

for elem in test_list:

grouped[elem[1]].append(elem[0])

grouped = [lst for lst in grouped if lst]

# Printing the result

print("The grouped list is:", grouped)

输出

The grouped list is: [['G', 'F', 'G'], ['I', 'S'], ['B', 'E', 'S', 'T']]