Ansible如何按device属性合并两个列表？求正确实现方式

我有两个通过API调用获取的列表，需要按device属性进行合并。以下是两个列表的打印输出：

List1 #

TASK [Print List1] **************************************************************************************************************************************************************************************
ok: [localhost] => {
    "msg": [
        [
            {
                "device": "LTM1_Device",
                "link": "LTM1_Link",
                "ltm_pool": "LTM1_Pool"
            },
            {
                "device": "LTM2_Device",
                "link": "LTM2_Link",
                "ltm_pool": "LTM2_Pool"
            },
            {
                "device": "LTM3_Device",
                "link": "LTM3_Link",
                "ltm_pool": "LTM3_Pool"
            }
        ]
    ]
}

List2 #

TASK [Print List2] ****************************************************************************************************************************************************************************************************
ok: [localhost] => {
    "msg": [
        [
            {
                "device": "LTM1_Device",
                "host": "/Common/LTM1_Host1",
                "ip": "0.0.0.1",
                "port": "5555"
            },
            {
                "device": "LTM1_Device",
                "host": "/Common/LTM1_Host2",
                "ip": "0.0.0.2",
                "port": "5555"
            },
            {
                "device": "LTM2_Device",
                "host": "/Common/LTM2_Host1",
                "ip": "0.0.0.3",
                "port": "5555"
            },
            {
                "device": "LTM2_Device",
                "host": "/Common/LTM2_Host2",
                "ip": "0.0.0.4",
                "port": "5555"
            },
            {
                "device": "LTM3_Device",
                "host": "/Common/LTM3_Host1",
                "ip": "0.0.0.5",
                "port": "5555"
            },
            {
                "device": "LTM3_Device",
                "host": "/Common/LTM3_Host2",
                "ip": "0.0.0.6",
                "port": "5555"
            }
        ]
    ]
}

尝试过的调试代码 #

我用了以下两条debug语句，结果都不符合预期：

- name: Debug
    debug:
        msg: 
            - "{{ (list2 + list1) | groupby('device') | map('last') | map('combine') | list }}"
            - "{{ (list2 + list1) | groupby('device') | map('last') | list }}"

当前输出结果 #

TASK [Debug] ****************************************************************************************************************************************************************************************************
ok: [localhost] => {
    "msg": [
        [
            {
                "device": "LTM1_Device",
                "host": "/Common/LTM1_Host2",
                "ip": "0.0.0.1",
                "link": "LTM1_Link",
                "ltm_pool": "LTM1_Pool",
                "port": "5555"
            },
            {
                "device": "LTM2_Device",
                "host": "/Common/LTM2_Host2",
                "ip": "0.0.0.3",
                "link": "LTM2_Link",
                "ltm_pool": "LTM2_Pool",
                "port": "5555"
            },
            {
                "device": "LTM3_Device",
                "host": "/Common/LTM2_Host2",
                "ip": "0.0.0.5",
                "link": "LTM3_Link",
                "ltm_pool": "LTM3_Pool",
                "port": "5555"
            }
        ],
        [
            [
                {
                    "device": "LTM1_Device",
                    "host": "/Common/LTM1_Host1",
                    "ip": "0.0.0.1",
                    "port": "5555"
                },
                {
                    "device": "LTM1_Device",
                    "host": "/Common/LTM1_Host2",
                    "ip": "0.0.0.2",
                    "port": "5555"
                },
                {
                    "device": "LTM1_Device",
                    "link": "LTM1_Link",
                    "ltm_pool": "LTM1_Pool"
                }
            ],
            [
                {
                    "device": "LTM2_Device",
                    "host": "/Common/LTM2_Host1",
                    "ip": "0.0.0.3",
                    "port": "5555"
                },
                {
                    "device": "LTM2_Device",
                    "host": "/Common/LTM2_Host2",
                    "ip": "0.0.0.4",
                    "port": "5555"
                },
                {
                    "device": "LTM2_Device",
                    "link": "LTM2_Link",
                    "ltm_pool": "LTM2_Pool"
                }
            ],
            [
                {
                    "device": "LTM3_Device",
                    "host": "/Common/LTM3_Host1",
                    "ip": "0.0.0.5",
                    "port": "5555"
                },
                {
                    "device": "LTM3_Device",
                    "host": "/Common/LTM3_Host2",
                    "ip": "0.0.0.6",
                    "port": "5555"
                },
                {
                    "device": "LTM3_Device",
                    "link": "LTM3_Link",
                    "ltm_pool": "LTM3_Pool"
                }
            ]
        ]
    ]
}

第一条语句会丢失List2中每个device对应的第一个host元素，第二条则将元素以嵌套列表形式分开。

期望的合并效果 #

我希望实现如下合并结果（不介意嵌套列表，后续可自行扁平化）：

[
            [
                {
                    "device": "LTM1_Device",
                    "host": "/Common/LTM1_Host1",
                    "ip": "0.0.0.1",
                    "link": "LTM1_Link",
                    "ltm_pool": "LTM1_Pool",
                    "port": "5555"
                },
                {
                    "device": "LTM1_Device",
                    "host": "/Common/LTM1_Host2",
                    "ip": "0.0.0.2",
                    "link": "LTM1_Link",
                    "ltm_pool": "LTM1_Pool",
                    "port": "5555"
                }
            ],
            [
                {
                    "device": "LTM2_Device",
                    "host": "/Common/LTM2_Host1",
                    "ip": "0.0.0.3",
                    "link": "LTM2_Link",
                    "ltm_pool": "LTM2_Pool",
                    "port": "5555"
                },
                {
                    "device": "LTM2_Device",
                    "host": "/Common/LTM2_Host2",
                    "ip": "0.0.0.4",
                    "link": "LTM2_Link",
                    "ltm_pool": "LTM2_Pool",
                    "port": "5555"
                }
            ],
            [
                {
                    "device": "LTM3_Device",
                    "host": "/Common/LTM3_Host1",
                    "ip": "0.0.0.5",
                    "link": "LTM3_Link",
                    "ltm_pool": "LTM3_Pool",
                    "port": "5555"
                },
                {
                    "device": "LTM3_Device",
                    "host": "/Common/LTM3_Host2",
                    "ip": "0.0.0.6",
                    "link": "LTM3_Link",
                    "ltm_pool": "LTM3_Pool",
                    "port": "5555"
                }
            ]
        ]

解决方案 #

可以通过以下步骤实现需求：

1. 先将List1转换为以device为键的字典，方便快速匹配对应属性：

- set_fact:
    list1_device_map: "{{ list1 | items2dict(key_name='device') }}"

2. 对List2按device分组，然后将每组内的每个元素与字典中对应device的属性合并：

- name: 合并并分组列表
  debug:
    msg: >-
      {{ 
        list2 | groupby('device') | map('last') | 
        map('map', 'combine', list1_device_map) | 
        list 
      }}

解释 #

• list1 | items2dict(key_name='device')：把List1的元素转成字典，key是device的值，value是对应的整个字典对象。
• list2 | groupby('device') | map('last')：对List2按device分组，提取每组的元素列表。
• map('map', 'combine', list1_device_map)：对每个分组里的元素，调用combine过滤器，把元素和list1_device_map中对应device的字典合并。

这样就能得到你期望的嵌套列表结构，每个device下的所有host元素都带上了List1中的link和ltm_pool属性。

内容的提问来源于stack exchange，提问作者cilles