Python中按组划分的最大值

我有一个python列表。列表列表上的每个值由[类别、类型、项目、分数]表示。对于每个类别和类型，我想返回一个得分最高的项目列表

[["Edibles", "Fruit", "Apple", 3],
"Edibles", "Fruit", "Grapes", 8],
"Edible", "Candy", "Hershey", 4],
"Edible", "Candy", "Snickers", 6],
"NonEdible", "Bikes", "Yamaha", 5],
"NonEdible", "Bikes", "Suzuki", 7],
"NonEdible", "Cars", "Kia", 8],
"NonEdible", "Cars", "Toyota", 9]]

期望输出

[["Edibles", "Fruit", "Grapes", 8],
"Edible", "Candy", "Snickers", 6],
"NonEdible", "Bikes", "Suzuki", 7],
"NonEdible", "Cars", "Toyota", 9]]

我可以通过创建临时列表的多个循环来实现这一点，但是随着输入大小的增加，计算变得非常缓慢。我正在寻找一个有效的解决方案。

您可以使用，但您需要在分组前对列表进行排序：

from itertools import groupby

lst = [["Edibles", "Fruit", "Apple", 3],
["Edibles", "Fruit", "Grapes", 8],
["Edible", "Candy", "Hershey", 4],
["Edible", "Candy", "Snickers", 6],
["NonEdible", "Bikes", "Yamaha", 5],
["NonEdible", "Bikes", "Suzuki", 7],
["NonEdible", "Cars", "Kia", 8],
["NonEdible", "Cars", "Toyota", 9]]

#if lst is already sorted, skip this step:
lst = sorted(lst, key=lambda k: (k[0], k[1]))

out = [max(g, key=lambda k: k[-1]) for _, g in groupby(lst, lambda k: (k[0], k[1]))]

from pprint import pprint
pprint(out)

印刷品：

[['Edible', 'Candy', 'Snickers', 6],
 ['Edibles', 'Fruit', 'Grapes', 8],
 ['NonEdible', 'Bikes', 'Suzuki', 7],
 ['NonEdible', 'Cars', 'Toyota', 9]]

您可以使用，但需要在分组前对列表进行排序：

from itertools import groupby

lst = [["Edibles", "Fruit", "Apple", 3],
["Edibles", "Fruit", "Grapes", 8],
["Edible", "Candy", "Hershey", 4],
["Edible", "Candy", "Snickers", 6],
["NonEdible", "Bikes", "Yamaha", 5],
["NonEdible", "Bikes", "Suzuki", 7],
["NonEdible", "Cars", "Kia", 8],
["NonEdible", "Cars", "Toyota", 9]]

#if lst is already sorted, skip this step:
lst = sorted(lst, key=lambda k: (k[0], k[1]))

out = [max(g, key=lambda k: k[-1]) for _, g in groupby(lst, lambda k: (k[0], k[1]))]

from pprint import pprint
pprint(out)

印刷品：

[['Edible', 'Candy', 'Snickers', 6],
 ['Edibles', 'Fruit', 'Grapes', 8],
 ['NonEdible', 'Bikes', 'Suzuki', 7],
 ['NonEdible', 'Cars', 'Toyota', 9]]

使用 使用数据框架可以轻松地操作、分析和可视化数据。 作为pd进口熊猫 设置数据帧 数据=[[食品，水果，苹果，3]， [食品、水果、葡萄，8]， [食用，糖果，好时，4]， [可食用、糖果、零食，6]， [非食用，自行车，雅马哈，5]， [非食用，自行车，铃木，7]， [非食用，汽车，起亚，8]， [非食用，汽车，丰田，9]] df=pd.DataFramedata groupby max 输出=df.groupby[0,1].aggmax.reset\u索引 0 1 2 3 0个可食用糖果窃笑者6个 1食用水果葡萄8 2辆非食用自行车雅马哈7 3辆非食用车丰田9 如果需要，输出到列表 output.to_numpy 数组[['可食用'，'糖果'，'窃笑'，6]， [‘食用’、‘水果’、‘葡萄’，8]， ['NonEdible'，'Bikes'，'Yamaha'，7]， ['NonEdible'，'Cars'，'Toyota'，9]]，dtype=object 使用 使用数据框架可以轻松地操作、分析和可视化数据。 作为pd进口熊猫 设置数据帧 数据=[[食品，水果，苹果，3]， [食品、水果、葡萄，8]， [食用，糖果，好时，4]， [可食用、糖果、零食，6]， [非食用，自行车，雅马哈，5]， [非食用，自行车，铃木，7]， [非食用，汽车，起亚，8]， [非食用，汽车，丰田，9]] df=pd.DataFramedata groupby max 输出=df.groupby[0,1].aggmax.reset\u索引 0 1 2 3 0个可食用糖果窃笑者6个 1食用水果葡萄8 2辆非食用自行车雅马哈7 3辆非食用车丰田9 如果需要，输出到列表 output.to_numpy 数组[['可食用'，'糖果'，'窃笑'，6]， [‘食用’、‘水果’、‘葡萄’，8]， ['NonEdible'，'Bikes'，'Yamaha'，7]， ['NonEdible'，'Cars'，'Toyota'，9]]，dtype=object 一本简单的字典既快捷又高效

您的列表格式不正确-您没有每个子列表的左括号 你可以用字典一次完成：

input = [["Edibles", "Fruit", "Apple", 3],
    ["Edibles", "Fruit", "Grapes", 8],
    ["Edible", "Candy", "Hershey", 4],
    ["Edible", "Candy", "Snickers", 6],
    ["NonEdible", "Bikes", "Yamaha", 5],
    ["NonEdible", "Bikes", "Suzuki", 7],
    ["NonEdible", "Cars", "Kia", 8],
    ["NonEdible", "Cars", "Toyota", 9]
]

highest_val_dict = {}
for curr_list in input:
    curr_key = (curr_list[0], curr_list[1])  # (category,type) is the key
    curr_item = curr_list[2]
    curr_val = curr_list[3]
    highest_pair = highest_val_dict.get(curr_key, (None, -1))
    if curr_val > highest_pair[1]:
        highest_val_dict[curr_key] = (curr_item, curr_val)

>>> for key, val in highest_val_dict.items():
>>>     print(f'{key[0]}, {key[1]}, {val[0]}, {val[1]}')
Edibles, Fruit, Grapes, 8
Edible, Candy, Snickers, 6
NonEdible, Bikes, Suzuki, 7
NonEdible, Cars, Toyota, 9

一本简单的字典既快捷又高效

您的列表格式不正确-您没有每个子列表的左括号 你可以用字典一次完成：

input = [["Edibles", "Fruit", "Apple", 3],
    ["Edibles", "Fruit", "Grapes", 8],
    ["Edible", "Candy", "Hershey", 4],
    ["Edible", "Candy", "Snickers", 6],
    ["NonEdible", "Bikes", "Yamaha", 5],
    ["NonEdible", "Bikes", "Suzuki", 7],
    ["NonEdible", "Cars", "Kia", 8],
    ["NonEdible", "Cars", "Toyota", 9]
]

highest_val_dict = {}
for curr_list in input:
    curr_key = (curr_list[0], curr_list[1])  # (category,type) is the key
    curr_item = curr_list[2]
    curr_val = curr_list[3]
    highest_pair = highest_val_dict.get(curr_key, (None, -1))
    if curr_val > highest_pair[1]:
        highest_val_dict[curr_key] = (curr_item, curr_val)

>>> for key, val in highest_val_dict.items():
>>>     print(f'{key[0]}, {key[1]}, {val[0]}, {val[1]}')
Edibles, Fruit, Grapes, 8
Edible, Candy, Snickers, 6
NonEdible, Bikes, Suzuki, 7
NonEdible, Cars, Toyota, 9

您可以使用pandas库执行以下操作：

安装熊猫，如：

pip install pandas

您的代码是：

In [2271]: import pandas as pd

In [2272]: l = [["Edibles", "Fruit", "Apple", 3], 
      ...: ["Edibles", "Fruit", "Grapes", 8], 
      ...: ["Edible", "Candy", "Hershey", 4], 
      ...: ["Edible", "Candy", "Snickers", 6], 
      ...: ["NonEdible", "Bikes", "Yamaha", 5], 
      ...: ["NonEdible", "Bikes", "Suzuki", 7], 
      ...: ["NonEdible", "Cars", "Kia", 8], 
      ...: ["NonEdible", "Cars", "Toyota", 9]] 

In [2275]: df = pd.DataFrame(l, columns=['category','type','item','score'])

In [2284]: df.groupby(['category','type'], as_index=False).agg(max).values.tolist()
Out[2284]: 
[['Edible', 'Candy', 'Snickers', 6],
 ['Edibles', 'Fruit', 'Grapes', 8],
 ['NonEdible', 'Bikes', 'Yamaha', 7],
 ['NonEdible', 'Cars', 'Toyota', 9]]

您可以使用pandas库执行以下操作：

安装熊猫，如：

pip install pandas

您的代码是：

In [2271]: import pandas as pd

In [2272]: l = [["Edibles", "Fruit", "Apple", 3], 
      ...: ["Edibles", "Fruit", "Grapes", 8], 
      ...: ["Edible", "Candy", "Hershey", 4], 
      ...: ["Edible", "Candy", "Snickers", 6], 
      ...: ["NonEdible", "Bikes", "Yamaha", 5], 
      ...: ["NonEdible", "Bikes", "Suzuki", 7], 
      ...: ["NonEdible", "Cars", "Kia", 8], 
      ...: ["NonEdible", "Cars", "Toyota", 9]] 

In [2275]: df = pd.DataFrame(l, columns=['category','type','item','score'])

In [2284]: df.groupby(['category','type'], as_index=False).agg(max).values.tolist()
Out[2284]: 
[['Edible', 'Candy', 'Snickers', 6],
 ['Edibles', 'Fruit', 'Grapes', 8],
 ['NonEdible', 'Bikes', 'Yamaha', 7],
 ['NonEdible', 'Cars', 'Toyota', 9]]

---

您可以使用常规dict，将每个唯一键的所有值存储在列表中，只需获取最大值：

data = [
    ["Edibles", "Fruit", "Apple", 3],
    ["Edibles", "Fruit", "Grapes", 8],
    ["Edible", "Candy", "Hershey", 4],
    ["Edible", "Candy", "Snickers", 6],
    ["NonEdible", "Bikes", "Yamaha", 5],
    ["NonEdible", "Bikes", "Suzuki", 7],
    ["NonEdible", "Cars", "Kia", 8],
    ["NonEdible", "Cars", "Toyota", 9]]

dct = {}
for item in data:
    dct.setdefault((item[0], item[1]), []).append((item[-2], item[-1]))

for k, v in dct.items():
    print(list(k) + list(max(v, key=lambda x: x[1])))

输出：

['Edibles', 'Fruit', 'Grapes', 8]
['Edible', 'Candy', 'Snickers', 6]
['NonEdible', 'Bikes', 'Suzuki', 7]
['NonEdible', 'Cars', 'Toyota', 9]

---

您可以使用常规dict，将每个唯一键的所有值存储在列表中，只需获取最大值：

data = [
    ["Edibles", "Fruit", "Apple", 3],
    ["Edibles", "Fruit", "Grapes", 8],
    ["Edible", "Candy", "Hershey", 4],
    ["Edible", "Candy", "Snickers", 6],
    ["NonEdible", "Bikes", "Yamaha", 5],
    ["NonEdible", "Bikes", "Suzuki", 7],
    ["NonEdible", "Cars", "Kia", 8],
    ["NonEdible", "Cars", "Toyota", 9]]

dct = {}
for item in data:
    dct.setdefault((item[0], item[1]), []).append((item[-2], item[-1]))

for k, v in dct.items():
    print(list(k) + list(max(v, key=lambda x: x[1])))

输出：

['Edibles', 'Fruit', 'Grapes', 8]
['Edible', 'Candy', 'Snickers', 6]
['NonEdible', 'Bikes', 'Suzuki', 7]
['NonEdible', 'Cars', 'Toyota', 9]

---

您的多重循环解决方案是什么样子的？根据分数对列表排序，循环并从每种类型中选择一种。时间复杂度将是Onlogn。您的多循环解决方案看起来像什么？根据分数、循环对列表进行排序，并从每种类型中选择一种。时间复杂度仅此而已。仅为此安装pandas有点过头了。它允许您编写最少的代码，在本例中仅需2行。因此，可以选择用最少的行编写干净的代码，而不是编写长的复杂循环。仅为此安装pandas有点过分。它允许您编写最少的代码，在这种情况下，只需2行。因此，可以选择用最少的行编写干净的代码，而不是编写长的复杂循环。