python - 递归组合字典

Question

好吧，这让我费尽心思。我有两个包含对象组的字典，如下所示:

groups = {
    'servers': ['unix_servers', 'windows_servers'],
    'unix_servers': ['server_a', 'server_b', 'server_group'],
    'windows_servers': ['server_c', 'server_d'],
    'server_group': ['server_e', 'server_f']
}

hosts = {
    'server_a': '10.0.0.1',
    'server_b': '10.0.0.2',
    'server_c': '10.0.0.3',
    'server_d': '10.0.0.4',
    'server_e': '10.0.0.5',
    'server_f': '10.0.0.6'
}

我正在寻找的输出是:

d3 = {
    'servers': {
        'unix_servers': {
            'server_a': '10.0.0.1',
            'server_b': '10.0.0.2',
            'server_group': {
                'server_e': '10.0.0.5',
                'server_f': '10.0.0.6'
            }
        },
        'windows_servers': {
            'server_c': '10.0.0.3',
            'server_d': '10.0.0.4'
        }
    }
}

我遇到的问题是我事先不知道有多少递归级别，因为嵌套组在理论上可以无限继续下去。此外，我无法确定哪些键应该是组合字典中的顶级键。

我目前有以下内容:

def resolve(d1, d2):
for k, v in d1.items():
    for i in v:
        if i in d2.keys():
            d3[k] = {i: d2[i]}

返回:

{
  "servers": {
    "unix_servers": {
      "server_a": "10.0.0.1",
      "server_b": "10.0.0.2",
      "server_group": {
        "server_e": "10.0.0.5",
        "server_f": "10.0.0.6"
      }
    },
    "windows_servers": {
      "server_c": "10.0.0.3",
      "server_d": "10.0.0.4"
    }
  },
  "unix_servers": {
    "server_b": "10.0.0.2"
  },
  "windows_servers": {
    "server_d": "10.0.0.4"
  },
  "server_group": {
    "server_f": "10.0.0.6"
  }
}

这很接近，但它显然缺少递归并且不处理键的嵌套。主要是在这里寻找指针，递归逻辑还没有为我点击...

Answer 1

我认为这可以满足您的需求:

def resolve(groups, hosts):
    # Groups that have already been resolved
    resolved_groups = {}
    # Group names that are not root
    non_root = set()
    # Make dict with resolution of each group
    result = {}
    for name in groups:
        result[name] = _resolve_rec(name, groups, hosts, resolved_groups, non_root)
    # Remove groups that are not root
    for name in groups:
        if name in non_root:
            del result[name]
    return result

def _resolve_rec(name, groups, hosts, resolved_groups, non_root):
    # If group has already been resolved finish
    if name in resolved_groups:
        return resolved_groups[name]
    # If it is a host finish
    if name in hosts:
        return hosts[name]
    # New group resolution
    resolved = {}
    for child in groups[name]:
        # Resolve each child
        resolved[child] = _resolve_rec(child, groups, hosts, resolved_groups, non_root)
        # Mark child as non-root
        non_root.add(child)
    # Save to resolved groups
    resolved_groups[name] = resolved
    return resolved

以你的例子:

groups = {
    'servers': ['unix_servers', 'windows_servers'],
    'unix_servers': ['server_a', 'server_b', 'server_group'],
    'windows_servers': ['server_c', 'server_d'],
    'server_group': ['server_e', 'server_f']
}

hosts = {
    'server_a': '10.0.0.1',
    'server_b': '10.0.0.2',
    'server_c': '10.0.0.3',
    'server_d': '10.0.0.4',
    'server_e': '10.0.0.5',
    'server_f': '10.0.0.6'
}

d3 = {
    'servers': {
        'unix_servers': {
            'server_a': '10.0.0.1',
            'server_b': '10.0.0.2',
            'server_group': {
                'server_e': '10.0.0.5',
                'server_f': '10.0.0.6'
            }
        },
        'windows_servers': {
            'server_c': '10.0.0.3',
            'server_d': '10.0.0.4'
        }
    }
}


print(resolve(groups, hosts) == d3)
# True

请注意，对于格式错误的输入，这可能会陷入无限递归，例如，如果您的 A 组包含 B 组，但 ​​B 组包含组 A。