Python ruamel.yaml - 在将 yaml 转换为 dict 并将结果保存到 yaml 文件时保留值中的引号

Question

我正在尝试将 YAML 文件加载到字典，然后将字典和安全 YAML 流处理到文件，但遇到以下问题。

源 y.yaml 文件包含一些带双引号的字符串：

---
lev1:
  lev2:
    - name: schema
      templates:
        - name:  temp1
          deploy_onpr: "yes"
          deploy_cl: "no"

我在 dest.yaml 中得到了什么：

---
lev1:
  lev2:
    - name: schema
      new_item: test
      templates:
        - deploy_cl: no
          deploy_onpr: yes
          name: temp1

我们可以看到引号已从 yes 和 no 字符串中删除。

我在 dest.yaml 中期望的是不应从字符串中删除双引号：

---
lev1:
  lev2:
    - name: schema
      new_item: test
      templates:
        - deploy_cl: "no"
          deploy_onpr: "yes"
          name: temp1

代码：

from fnmatch import fnmatch
import json
import os
from json import loads, dumps
import ruamel.yaml

yaml = ruamel.yaml.YAML(typ='safe', pure=True)
yaml.preserve_quotes = True

yaml.explicit_start = True
yaml.default_flow_style = False
yaml.indent(mapping=2, sequence=4, offset=2)

def to_dict(input_ordered_dict):
    """Convert the inventory data stream to dict"""
    print(dumps(input_ordered_dict))
    return loads(dumps(input_ordered_dict))

def _list_files():
    '''
    list all yaml files in the yaml directory recursive
    '''
    pattern = "*.yaml"
    in_folder="./"
    all_files = []
    for path, subdirs, files in os.walk(in_folder):
        for name in files:
            if fnmatch(name, pattern):
                f = os.path.join(path, name)
                all_files.append(f)
        return all_files

def load_files():
    '''
    load directory recursive and generate a dict tree
    data allocated in the structure under the key "file name"
    '''
    dict_tree = {}
    for sfile in _list_files():
        with open(sfile, 'r') as stream:
            print(type(yaml))
            data_loaded = yaml.load(stream)
            print(data_loaded)
            dict_tree[sfile] = data_loaded
    return dict_tree


load_yaml_files = load_files()
inventoty_data = to_dict(load_yaml_files)
inventoty_data['./y.yaml']["lev1"]["lev2"][0]["new_item"]="test"
new_dict=inventoty_data["./y.yaml"]

dst_file="dest.yaml"
with open(dst_file, 'w') as yaml_file:
    yaml.dump(new_dict, yaml_file)

Answer 1

这看起来像 YAML 1.1，其中 yes 和 no 应该是布尔值（因此如果它们是标量字符串，则需要引用）。

preserve_quotes 属性仅在使用默认 round-trip 加载程序和转储程序时有效（使用 SafeLoader 的子类，也可以安全地用于未知的 YAML 源）：

import sys
import ruamel.yaml
from pathlib import Path

    
yaml = ruamel.yaml.YAML()
yaml.explicit_start = True
yaml.indent(sequence=4, offset=2)
yaml.preserve_quotes = True
data = yaml.load(Path('y.yaml'))
yaml.dump(data, sys.stdout)

给出：

---
lev1:
  lev2:
    - name: schema
      templates:
        - name: temp1
          deploy_onpr: "yes"
          deploy_cl: "no"

即使您将 data 作为值分配给普通字典中的键，加载期间创建的字符串的特殊版本将被保留并用引号转储。

如果您需要 add/update Python 中的某些值需要字符串 "abc" 的引号，那么请执行：

DQ = ruamel.yaml.scalarstring.DoubleQuotesScalarString

some_place_in_your_data_structure = DQ("abc")`

Answer 2

以下代码按预期工作：

from fnmatch import fnmatch
import json
import os
from json import loads, dumps
import ruamel.yaml

import sys
import ruamel.yaml
from pathlib import Path

ruamel.yaml.representer.RoundTripRepresenter.ignore_aliases = lambda x, y: True
yaml = ruamel.yaml.YAML()
yaml.explicit_start = True
yaml.indent(sequence=4, offset=2)
yaml.preserve_quotes = True


def _list_files():
    '''
    list all yaml files in the yaml directory recursive
    '''
    pattern = "*.yaml"
    in_folder="./"
    all_files = []
    for path, subdirs, files in os.walk(in_folder):
        for name in files:
            if fnmatch(name, pattern):
                f = os.path.join(path, name)
                all_files.append(f)
        return all_files

def load_files():
    '''
    load directory recursive and generate a dict tree
    data allocated in the structure under the key "file name"
    '''
    dict_tree = {}
    for sfile in _list_files():
        with open(sfile, 'r') as stream:
            data_loaded = yaml.load(stream)
            print(data_loaded)
            dict_tree[sfile] = data_loaded
    return dict_tree


load_yaml_files = load_files()
inventoty_data = load_yaml_files
inventoty_data['./y.yaml']["lev1"]["lev2"][0]["new_item"]="test"
new_dict=inventoty_data["./y.yaml"]

dst_file="dest.yaml"
with open(dst_file, 'w') as yaml_file:
    yaml.dump(new_dict, yaml_file)

输出：

---
lev1:
  lev2:
    - name: schema
      templates:
        - name: temp1
          deploy_onpr: "yes"
          deploy_cl: "no"
      new_item: test

Python ruamel.yaml - 在将 yaml 转换为 dict 并将结果保存到 yaml 文件时保留值中的引号

Python ruamel.yaml - preserving quote in value while convert yaml to dict and save result to yaml file

python

dictionary

yaml

ruamel.yaml