将字典列表与每个列表中可变数量的字典进行比较答案

【问题标题】：Compare list of dictionaries with variable number of dictionaries in each list将字典列表与每个列表中可变数量的字典进行比较
【发布时间】：2020-04-21 15:17:24
【问题描述】：

我有两个字典列表：

list1 = [
    {'vehicle': 1, 'mileage': 25, 'speed': 80}, 
    {'vehicle': 2, 'mileage': 35, 'speed': 70},
    {'vehicle': 3, 'mileage': 40, 'speed': 90},
    {'vehicle': 5, 'mileage': 40, 'speed': 90}
]

list2 = [
    {'vehicle': 1, 'mileage': 35, 'speed': 80}, 
    {'vehicle': 2, 'mileage': 35, 'speed': 70},
    {'vehicle': 3, 'mileage': 40, 'speed': 80},
    {'vehicle': 4, 'mileage': 40, 'speed': 80}
]

如果 list2 上有同名车辆并且相应车辆的里程和速度大于或等于 list2 中的，我必须从 list1 打印字典。在这个例子中，输出应该是：

[{'vehicle': 1, 'mileage': 25, 'speed': 80}, 
 {'vehicle': 2, 'mileage': 35, 'speed': 70}]

非常感谢任何帮助。

【问题讨论】：

你试过什么？您可以嵌套两个循环并检查当前值是否符合您的要求。如果这是真的，请将值添加到一个空列表中，否则继续。

标签： python python-3.x dictionary

【解决方案1】：

您可以创建一个查找字典来搜索 O(1) 中的相应匹配项：

list1 = [{'vehicle': 1, 'mileage': 25, 'speed': 80},
         {'vehicle': 2, 'mileage': 35, 'speed': 70},
         {'vehicle': 3, 'mileage': 40, 'speed': 90},
         {'vehicle': 5, 'mileage': 40, 'speed': 90}]

list2 = [{'vehicle': 1, 'mileage': 35, 'speed': 80},
         {'vehicle': 2, 'mileage': 35, 'speed': 70},
         {'vehicle': 3, 'mileage': 40, 'speed': 80},
         {'vehicle': 4, 'mileage': 40, 'speed': 80}]


lookup = {e['vehicle'] : e for e in list1 }

result = []
for e in list2 :
    if e['vehicle'] in lookup:
        d = lookup[e['vehicle']]
        if e['mileage'] >= d['mileage'] and e['speed'] >= d['speed']:
            result.append(d)

print(result)

输出

[{'vehicle': 1, 'mileage': 25, 'speed': 80}, {'vehicle': 2, 'mileage': 35, 'speed': 70}]

使用list comprehension 的替代方法：

def better(e, d, keys=('mileage', 'speed')):
    return all(e[k] >= d[k] for k in keys)


result = [lookup[e['vehicle']] for e in list2 if e['vehicle'] in lookup and better(e, lookup[e['vehicle']])]

print(result)

这两种方法的总体复杂度为O(n)。

【讨论】：

【解决方案2】：

我会使用熊猫数据框。不是最有效但非常易读

import pandas as pd
df1 = pd.DataFrame(list1)
df2 = pd.DataFrame(list2)
result_df = df1[(df1.vehicle.isin(df2.vehicle)) & (df1.speed <= df2.speed) & (df1.mileage <= df2.mileage)]
result_list = result_df.to_dict('records')

print(result_list)

输出：

[{'mileage': 25, 'speed': 80, 'vehicle': 1},
 {'mileage': 35, 'speed': 70, 'vehicle': 2}]

【讨论】：

【解决方案3】：

您可以将字典列表转换为嵌套字典，其中vehicle 是关键：

list1 = [
    {"vehicle": 1, "mileage": 25, "speed": 80},
    {"vehicle": 2, "mileage": 35, "speed": 70},
    {"vehicle": 3, "mileage": 40, "speed": 90},
    {"vehicle": 5, "mileage": 40, "speed": 90},
]

list2 = [
    {"vehicle": 1, "mileage": 35, "speed": 80},
    {"vehicle": 2, "mileage": 35, "speed": 70},
    {"vehicle": 3, "mileage": 40, "speed": 80},
    {"vehicle": 4, "mileage": 40, "speed": 80},
]

dict1 = {x["vehicle"]: x for x in list1}
# {1: {'vehicle': 1, 'mileage': 25, 'speed': 80}, 2: {'vehicle': 2, 'mileage': 35, 'speed': 70}, 3: {'vehicle': 3, 'mileage': 40, 'speed': 90}, 5: {'vehicle': 5, 'mileage': 40, 'speed': 90}}

dict2 = {x["vehicle"]: x for x in list2}
# {1: {'vehicle': 1, 'mileage': 35, 'speed': 80}, 2: {'vehicle': 2, 'mileage': 35, 'speed': 70}, 3: {'vehicle': 3, 'mileage': 40, 'speed': 80}, 4: {'vehicle': 4, 'mileage': 40, 'speed': 80}}

然后取键的交集并使用list comprehension将结果作为字典列表返回：

result = [
    dict1[k]
    for k in dict1.keys() & dict2.keys()
    if dict2[k]["mileage"] >= dict1[k]["mileage"]
    and dict2[k]["speed"] >= dict1[k]["speed"]
]

print(result)

输出：

[{'vehicle': 1, 'mileage': 25, 'speed': 80}, {'vehicle': 2, 'mileage': 35, 'speed': 70}]

【讨论】：