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Accepted
honeybeeham
Asked: 2023-08-18 05:49:54 +0800 CST

通过 Ansible map('regex_replace') 后比较两个列表

  • 772

我正在尝试比较两个列表以找出差异和重复项。例子：

- name: Debug
  hosts: "localhost"
  gather_facts: no
  connection: local

  vars: 
    list1: 
      - {name: aBC_SOMEITEM-A01_dEF, type: random, policy: one}
      - {name: aBC_SOMEITEM-A02_dEF, type: random, policy: two}
      - {name: aBC_SOMEITEM-A03_dEF, type: random, policy: three}

    list2: 
      - {name: "SOMEITEM-A01", type: "random", policy: "one"}
      - {name: "SOMEITEM-A02", type: "random", policy: "two"}
      - {name: "SOMEITEM-A03", type: "random", policy: "three"}
      - {name: "SOMEITEM-A03", type: "random", policy: "three"}
      - {name: "SOMEITEM-A04", type: "random", policy: "four"}

    list3: "{{ list1 | map('ansible.builtin.regex_replace', 'aBC_|_dEF', '') | list }}"

  tasks: 
  - debug: 
      msg: |
        {{ list1 | to_yaml(default_flow_style=none) }}
        {{ list2 | to_yaml(default_flow_style=none) }}
        {{ list3 | to_yaml(default_flow_style=none) }}

        {{ list2 | difference(list3) | to_yaml(default_flow_style=none) }}

list3过滤器后的输出map('regex_replace')如下所示：

['{''name'': ''SOMEITEM-A01'', ''type'': ''random'', ''policy'': ''one''}', '{''name'':
    ''SOMEITEM-A02'', ''type'': ''random'', ''policy'': ''two''}', '{''name'': ''SOMEITEM-A03'',
    ''type'': ''random'', ''policy'': ''three''}']
  1. 如何规范化 list3 以使差异的结果输出为
- {name: SOMEITEM-A04, policy: three, type: random}
  1. list2有重复的字典，我怎样才能找到重复的？例如
- {name: "SOMEITEM-A03", type: "random", policy: "three"}

我从这里查看了两个答案，但无法弄清楚 - Ansible: How to getlicate items from list?

（我不介意编写自定义过滤器插件来完成此任务，因为数据集可能非常大）

ansible

1 个回答

  1. Best Answer

    创建姓名列表

      list1_names: "{{ list1|map(attribute='name')|
                         map('regex_replace', '^.*_(.*)_.*$', '{name: \\1}')|
                         map('from_yaml') }}"

    给出

      list1_names:
    - name: SOMEITEM-A01
    - name: SOMEITEM-A02
    - name: SOMEITEM-A03

    更新列表1

      list1_update: "{{ list1|zip(list1_names)|map('combine') }}"

    给出

      list1_update:
    - {name: SOMEITEM-A01, policy: one, type: random}
    - {name: SOMEITEM-A02, policy: two, type: random}
    - {name: SOMEITEM-A03, policy: three, type: random}

    使用unique过滤器从list2中删除重复项。然后，使用过滤器差异来删除共同的项目

      list3: "{{ list2|unique|difference(list1_update) }}"

    给你想要的

      list3:
    - {name: SOMEITEM-A04, policy: four, type: random}

    用于测试的完整剧本示例

    - hosts: localhost

  vars:

    list1: 
      - {name: aBC_SOMEITEM-A01_dEF, type: random, policy: one}
      - {name: aBC_SOMEITEM-A02_dEF, type: random, policy: two}
      - {name: aBC_SOMEITEM-A03_dEF, type: random, policy: three}

    list2: 
      - {name: SOMEITEM-A01, type: random, policy: one}
      - {name: SOMEITEM-A02, type: random, policy: two}
      - {name: SOMEITEM-A03, type: random, policy: three}
      - {name: SOMEITEM-A03, type: random, policy: three}
      - {name: SOMEITEM-A04, type: random, policy: four}

    list1_names: "{{ list1|map(attribute='name')|
                           map('regex_replace', '^.*_(.*)_.*$', '{name: \\1}')|
                           map('from_yaml') }}"
    list1_update: "{{ list1|zip(list1_names)|map('combine') }}"

    list3: "{{ list2|unique|difference(list1_update) }}"

  tasks:

    - debug:
        var: list1_names
    - debug:
        var: list1_update|to_yaml
    - debug:
        var: list3|to_yaml

    问：“list2 有一个重复的字典，我怎样才能找到重复的？”

    答：计算频率。例如，

        l2_keys: "{{ list2.0.keys() }}"
    freq: "{{ list2|
              map('dict2items')|
              map('map', attribute='value')|
              map('join', ',')|
              community.general.counter }}"
    duplicates: |
      {% filter from_yaml %}
      {% for k,v in freq.items() %}
      {% if v > 1 %}
      {% set arr=k.split(',') %}
      - {{ dict(l2_keys|zip(arr)) }}
      {% endif %}
      {% endfor %}
      {% endfilter %}

    给出

      l2_keys: [name, type, policy]

  freq:
    SOMEITEM-A01,random,one: 1
    SOMEITEM-A02,random,two: 1
    SOMEITEM-A03,random,three: 2
    SOMEITEM-A04,random,four: 1

  duplicates:
    - {name: SOMEITEM-A03, policy: three, type: random}
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