在有特定鍵值對的列表中獲取字典

Question

假設我在列表中有這本字典：

 [
    {
        "Name": "Person",
        "Confidence": 97.56156921386719,
        "Instances": [
            {
                "BoundingBox": {
                    "Width": 0.6137702465057373,
                    "Height": 0.9175498485565186,
                    "Left": 0.22297996282577515,
                    "Top": 0.0739903450012207
                },
                "Confidence": 94.51961517333984
            },
            {
                "BoundingBox": {
                    "Width": 0.46570318937301636,
                    "Height": 0.6405649781227112,
                    "Left": 0.11447866261005402,
                    "Top": 0.34079012274742126
                },
                "Confidence": 82.2153549194336
            }
        ],
        "Parents": []
    },
    {
        "Name": "Human",
        "Confidence": 97.56156921386719,
        "Instances": [],
        "Parents": []
    },
    {
        "Name": "Clothing",
        "Confidence": 91.08417510986328,
        "Instances": [],
        "Parents": []
    }
    }]

如何僅檢索鍵值為"Name":"Clothing"的字典？ 是否可以從字典中檢索它？ 關鍵“服裝”可能出現在列表的任何部分。

    {
        "Name": "Clothing",
        "Confidence": 91.08417510986328,
        "Instances": [],
        "Parents": []
    }
  

Answer 1

您可以使用以下代碼獲取每個具有Name of Clothing ：

names = [
    {
        "Name": "Person",
        "Confidence": 97.56156921386719,
        "Instances": [
            {
                "BoundingBox": {
                    "Width": 0.6137702465057373,
                    "Height": 0.9175498485565186,
                    "Left": 0.22297996282577515,
                    "Top": 0.0739903450012207
                },
                "Confidence": 94.51961517333984
            },
            {
                "BoundingBox": {
                    "Width": 0.46570318937301636,
                    "Height": 0.6405649781227112,
                    "Left": 0.11447866261005402,
                    "Top": 0.34079012274742126
                },
                "Confidence": 82.2153549194336
            }
        ],
        "Parents": []
    },
    {
        "Name": "Human",
        "Confidence": 97.56156921386719,
        "Instances": [],
        "Parents": []
    },
    {
        "Name": "Clothing",
        "Confidence": 91.08417510986328,
        "Instances": [],
        "Parents": []
    }
]

for name in names:
    if name['Name'] == 'Clothing':
        print(name)

如果您希望它簡短，可以使用以下代碼：

x = [y for y in names if name['Name'] == 'Clothing']

Answer 2

您可以使用這個簡單的list comprehension ：

[print(dictionary) for dictionary in lst if 'Name' in dictionary.keys() and dictionary['Name'] == 'Clothing']

輸出：

{'Name': 'Clothing', 'Confidence': 91.08417510986328, 'Instances': [], 'Parents': []}

這是完整的代碼：

lst =  [
    {
        "Name": "Person",
        "Confidence": 97.56156921386719,
        "Instances": [
            {
                "BoundingBox": {
                    "Width": 0.6137702465057373,
                    "Height": 0.9175498485565186,
                    "Left": 0.22297996282577515,
                    "Top": 0.0739903450012207
                },
                "Confidence": 94.51961517333984
            },
            {
                "BoundingBox": {
                    "Width": 0.46570318937301636,
                    "Height": 0.6405649781227112,
                    "Left": 0.11447866261005402,
                    "Top": 0.34079012274742126
                },
                "Confidence": 82.2153549194336
            }
        ],
        "Parents": []
    },
    {
        "Name": "Human",
        "Confidence": 97.56156921386719,
        "Instances": [],
        "Parents": []
    },
    {
        "Name": "Clothing",
        "Confidence": 91.08417510986328,
        "Instances": [],
        "Parents": []
    }
]

[print(dictionary) for dictionary in lst if 'Name' in dictionary.keys() and dictionary['Name'] == 'Clothing']

Answer 3

.get(Key, defaultValue)可以在這里使用。

.get("Name", "") --> 如果鍵Name在字典中不存在，則返回一個空字符串。 doc .get("Name") --> 如果Name不存在，則返回None 。

所以.get("Name") == "Clothing"是選擇字典的標准。

並使用filter()僅選擇滿足條件的那些。

把這一切放在一起

filter(lambda x: x.get("Name") == "Clothing", mylist)

filter 返回一個迭代器，因此您可以將其傳遞給 list 以創建完整列表。 如果您只想迭代它，則無需調用 list。

所以，

list(filter(lambda x: x.get("Name") == "Clothing", mylist))

Answer 4

假設你的列表被稱為my_list ，那么這應該給你你想要的：

my_list_new = [x for x in my_list if ("Name", "Clothing") in x.items()]