python - 在python中合并两个yaml文件

Question

我有两个 yaml 文件，如下所述

测试1.yaml

resources:
  server_group_1:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 5] }
      policies: [ { get_param: [server_group_types, 5] } ]

  server_group_2:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 8] }
      policies: [ { get_param: [server_group_types, 8] } ]
output:
  check_1:
    description: Name of the instance
    value: { get_attr: [check_1, vname] }

测试2.yaml

resources:
  server_group_4:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 4] }
      policies: [ { get_param: [server_group_types, 4] } ]

  server_group_9:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 7] }
      policies: [ { get_param: [server_group_types, 7] } ]
output:
  check_6:
    description: Name of the instance
    value: { get_attr: [check_6, vname] }

我想合并这两个文件并创建一个新的输出文件，所以我使用 pyyaml 订单正在更改，在此 link 中发布相同

有人可以帮助合并这些文件而不改变顺序吗？
最终的yaml应该是这样的

最终的.yaml

resources:
  server_group_1:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 5] }
      policies: [ { get_param: [server_group_types, 5] } ]

  server_group_2:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 8] }
      policies: [ { get_param: [server_group_types, 8] } ]

  server_group_4:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 4] }
      policies: [ { get_param: [server_group_types, 4] } ]

  server_group_9:
    type: OS::Nova::ServerGroup
    properties:
      name: { get_param: [server_groups, 7] }
      policies: [ { get_param: [server_group_types, 7] } ]
output:
  check_1:
    description: Name of the instance
    value: { get_attr: [check_1, vname] }

  check_6:
    description: Name of the instance
    value: { get_attr: [check_6, vname] } 

更新

我可以使用 ruamel.yaml 合并文件……这里是更新资源的示例代码

代码:

import ruamel.yaml
yaml = ruamel.yaml.YAML()
#Load the yaml files
with open('/test1.yaml') as fp:
    data = yaml.load(fp)
with open('/test2.yaml') as fp:
    data1 = yaml.load(fp)
#Add the resources from test2.yaml to test1.yaml resources
for i in data1['resources']:
    print i,data1['resources'][i]
    data['resources'].update({i:data1['resources'][i]})
#create a new file with merged yaml
yaml.dump(data,file('/tmp/lal.yaml', 'w'))

Answer 1

下面的示例代码很适合我合并两个 yaml 文件

import ruamel.yaml
yaml = ruamel.yaml.YAML()
#Load the yaml files
with open('/test1.yaml') as fp:
    data = yaml.load(fp)
with open('/test2.yaml') as fp:
    data1 = yaml.load(fp)
#Add the resources from test2.yaml to test1.yaml resources
for i in data1['resources']:
    print i,data1['resources'][i]
    data['resources'].update({i:data1['resources'][i]})
#create a new file with merged yaml
yaml.dump(data,file('/tmp/lal.yaml', 'w'))