Yaml 将 python 字典转储为不带单引号的映射

Question

我想将 dict 转储到没有引号的 yaml 文件中。基本上就像一个映射

import yaml

windows_list = []
server_list = ['abc-def-01', 'pqr-str-02']
site_list = ['dev', 'prod']

server_dict = dict(zip(server_list, site_list))

for k,v in server_dict.items():
    a = '{ SERVER: '+k+', SITE: '+v+' }'
    windows_list.append(a)

final_dict = {'service':'something','servers': windows_list}

with open('config.yml', 'w') as yaml_file:
    yaml.dump(final_dict, yaml_file, default_flow_style=False)

config.yml 的输出如下：

service: some name
servers:
- '{ SERVER: abc-def-01, SITE: dev }'
- '{ SERVER: pqr-str-02, SITE: prod }'

我不想要它周围的引号。我希望它是一个没有引号的映射

期望的输出

service: some name
servers:
- { SERVER: abc-def-01, SITE: dev }
- { SERVER: pqr-str-02, SITE: prod }

我阅读了this post，它说如果您有特殊字符，则无法删除引号，但我想知道是否有某种解决方法来实现我想要的。

Answer 1

需要引号是因为您创建和转储的字符串看起来像流式 映射。换句话说，当您加载您想要的内容时，您会得到不同的 数据结构比您“手动”创建的（或从您获得的输出中加载）。

如果您遇到这种情况，最好的办法是加载您想要获取的内容，然后 转储 Python 数据结构，以便您知道需要创建什么。这也验证了你想要的是否真的是 YAML：

import sys
import ruamel.yaml

yaml_str = """\
service: some name
servers:
- { SERVER: abc-def-01, SITE: dev }
- { SERVER: pqr-str-02, SITE: prod }
"""

yaml = ruamel.yaml.YAML(typ='safe')
data = yaml.load(yaml_str)
print(data)

给出：

{'service': 'some name', 'servers': [{'SERVER': 'abc-def-01', 'SITE': 'dev'}, {'SERVER': 'pqr-str-02', 'SITE': 'prod'}]}

如您所见，该序列由字典组成，因此您需要追加：

import sys
import ruamel.yaml


windows_list = []
server_list = ['abc-def-01', 'pqr-str-02']
site_list = ['dev', 'prod']

server_dict = dict(zip(server_list, site_list))

for k,v in server_dict.items():
    a = { 'SERVER': k, 'SITE': v }
    windows_list.append(a)

final_dict = {'service':'something', 'servers': windows_list}

yaml = ruamel.yaml.YAML()
yaml.default_flow_style = None
yaml.dump(final_dict, sys.stdout)

结果：

service: something
servers:
- {SERVER: abc-def-01, SITE: dev}
- {SERVER: pqr-str-02, SITE: prod}

顺便说一句。自 2006 年 9 月起，YAML 文件的推荐扩展名为 .yaml。

Answer 2

如果要输出字典，不要构造字符串。改为构造一个字典：

import yaml

windows_list = []
server_list = ['abc-def-01', 'pqr-str-02']
site_list = ['dev', 'prod']

server_dict = dict(zip(server_list, site_list))

for k,v in server_dict.items():
    windows_list.append({'SERVER': k, 'SITE': v})

final_dict = {'service':'something','servers': windows_list}

with open('config.yml', 'w') as yaml_file:
    yaml.dump(final_dict, yaml_file, default_flow_style=False)

这将输出

servers:
- SERVER: abc-def-01
  SITE: dev
- SERVER: pqr-str-02
  SITE: prod
service: something

如果您希望项目以流样式表示，您可以使用自定义表示器：

import yaml

windows_list = []
server_list = ['abc-def-01', 'pqr-str-02']
site_list = ['dev', 'prod']

class ServerMap:
    def __init__(self, server, site):
        self.server = server
        self.site = site

def servermap_representer(dumper, data):
    return dumper.represent_mapping('tag:yaml.org,2002:map',
            {'SERVER': data.server, 'SITE': data.site}, [True, True])

yaml.add_representer(ServerMap, servermap_representer)

server_dict = dict(zip(server_list, site_list))

for k,v in server_dict.items():
    windows_list.append(ServerMap(k, v))

final_dict = {'service':'something','servers': windows_list}

with open('config.yml', 'w') as yaml_file:
    yaml.dump(final_dict, yaml_file, default_flow_style=False)

输出：

servers:
- {SERVER: abc-def-01, SITE: dev}
- {SERVER: pqr-str-02, SITE: prod}
service: something

Answer 3

试试这个：

for k,v in server_dict.items():
    a = { 'SERVER': f'{k}', 'SITE': f'{v}' }
    windows_list.append(a)

final_dict = {'service':'something','servers': windows_list}

with open('config.yml', 'w') as yaml_file:
    yaml.dump(final_dict, yaml_file, default_flow_style=False)

它将以 yaml 格式/结构准确生成您需要的内容：

servers:
- SERVER: abc-def-01
  SITE: dev
- SERVER: pqr-str-02
  SITE: prod
service: something

HTH