比较两个字典并在键匹配的位置追加嵌套对象的 Pythonic 方法

Question

Python 有工作代码的新手，但希望了解是否有更好的方法。

问题陈述 - 一个 python 字典包含有关卷的详细信息，第二个字典包含有关卷快照的详细信息。生成一个 json 文档，每个卷作为键，如果该卷有快照，则将快照字典对象作为嵌套字典添加到相应卷的快照键下。如果该卷没有任何快照，只需在相应卷的快照键下附加一个空对象作为嵌套字典。

是否有更pythonic的方法来解决这个挑战（也许是理解）？
我更新原版词典是不是搞错了？

import json
volume_dict = {
    "vol-04f": {
        "Name": "db_server",
        "State": "in-use",
        "DataType": "Public",
        "Attachments": [
            "i-0fc"
        ]
    },
    "vol-0cc": {
        "Name": "app_server",
        "State": "in-use",
        "DataType": "Public",
        "Attachments": [
            "i-051"
        ]
    }
}

# Snapshot Dictionary
snapshot_dict = {
    "vol-04f": [
        {
            "snap-0086": {
                "Date": 20210911,
                "SnapshotState": "completed"
            }
        },
        {
            "snap-06ff": {
                "Date": 20210910,
                "SnapshotState": "completed"
            }
        },
        {
            "snap-0263": {
                "Date": 20210919,
                "SnapshotState": "completed"
            }
        }
    ]
}

for volume_key, volume_value in volume_dict.items():
    if volume_key in snapshot_dict:
        value = {}
        snapshot_value = snapshot_dict.get(volume_key)
        value["Snapshots"] = snapshot_value
    else:
        value = {}
        value["Snapshots"] = {}
    volume_dict[volume_key].update(value)

print(json.dumps(volume_dict))

期望的输出

{
    "vol-04f": {
        "DataType": "Public",
        "State": "in-use",
        "Name": "db_server",
        "Snapshots": [
            {
                "snap-0086": {
                    "Date": 20210911,
                    "SnapshotState": "completed"
                }
            },
            {
                "snap-06ff": {
                    "Date": 20210910,
                    "SnapshotState": "completed"
                }
            },
            {
                "snap-0263": {
                    "Date": 20210919,
                    "SnapshotState": "completed"
                }
            }
        ],
        "Attachments": [
            "i-0fc"
        ]
    },
    "vol-0cc": {
        "DataType": "Public",
        "State": "in-use",
        "Name": "app_server",
        "Snapshots": {},
        "Attachments": [
            "i-051"
        ]
    }
}

Answer 1

dict.get 有默认值，当找不到键时返回，你可以利用它：

for k, v in volume_dict.items():
    v["Snapshots"] = snapshot_dict.get(k, {})

print(volume_dict)

打印：

{
    "vol-04f": {
        "Name": "db_server",
        "State": "in-use",
        "DataType": "Public",
        "Attachments": ["i-0fc"],
        "Snapshots": [
            {"snap-0086": {"Date": 20210911, "SnapshotState": "completed"}},
            {"snap-06ff": {"Date": 20210910, "SnapshotState": "completed"}},
            {"snap-0263": {"Date": 20210919, "SnapshotState": "completed"}},
        ],
    },
    "vol-0cc": {
        "Name": "app_server",
        "State": "in-use",
        "DataType": "Public",
        "Attachments": ["i-051"],
        "Snapshots": {},
    },
}

Answer 2

Q1

如果你想使用列表推导式，这就是方法

[volume_dict[volume_key].update({"Snapshots": snapshot_dict[volume_key] if volume_key in snapshot_dict else {}})  for volume_key, volume_value in volume_dict.items()]

更简单的方法是

for volume_key, volume_value in volume_dict.items():
    volume_dict[volume_key]["Snapshots"] = snapshot_dict[volume_key] if volume_key in snapshot_dict else {}

Q2

那将根据您的要求。如果以后想用原来的volume_dict，最好弄个副本更新一下。

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