我有两个YAML文件,“A”和“B”,我希望将A的内容插入到B中,要么拼接到现有的数据结构中,就像数组一样,要么作为元素的子元素,就像某个散列键的值一样。
这可能吗?怎么做?如果不是,是否有指向规范引用的指针?
我有两个YAML文件,“A”和“B”,我希望将A的内容插入到B中,要么拼接到现有的数据结构中,就像数组一样,要么作为元素的子元素,就像某个散列键的值一样。
这可能吗?怎么做?如果不是,是否有指向规范引用的指针?
当前回答
我举了一些例子供你参考。
import yaml
main_yaml = """
Package:
- !include _shape_yaml
- !include _path_yaml
"""
_shape_yaml = """
# Define
Rectangle: &id_Rectangle
name: Rectangle
width: &Rectangle_width 20
height: &Rectangle_height 10
area: !product [*Rectangle_width, *Rectangle_height]
Circle: &id_Circle
name: Circle
radius: &Circle_radius 5
area: !product [*Circle_radius, *Circle_radius, pi]
# Setting
Shape:
property: *id_Rectangle
color: red
"""
_path_yaml = """
# Define
Root: &BASE /path/src/
Paths:
a: &id_path_a !join [*BASE, a]
b: &id_path_b !join [*BASE, b]
# Setting
Path:
input_file: *id_path_a
"""
# define custom tag handler
def yaml_import(loader, node):
other_yaml_file = loader.construct_scalar(node)
return yaml.load(eval(other_yaml_file), Loader=yaml.SafeLoader)
def yaml_product(loader, node):
import math
list_data = loader.construct_sequence(node)
result = 1
pi = math.pi
for val in list_data:
result *= eval(val) if isinstance(val, str) else val
return result
def yaml_join(loader, node):
seq = loader.construct_sequence(node)
return ''.join([str(i) for i in seq])
def yaml_ref(loader, node):
ref = loader.construct_sequence(node)
return ref[0]
def yaml_dict_ref(loader: yaml.loader.SafeLoader, node):
dict_data, key, const_value = loader.construct_sequence(node)
return dict_data[key] + str(const_value)
def main():
# register the tag handler
yaml.SafeLoader.add_constructor(tag='!include', constructor=yaml_import)
yaml.SafeLoader.add_constructor(tag='!product', constructor=yaml_product)
yaml.SafeLoader.add_constructor(tag='!join', constructor=yaml_join)
yaml.SafeLoader.add_constructor(tag='!ref', constructor=yaml_ref)
yaml.SafeLoader.add_constructor(tag='!dict_ref', constructor=yaml_dict_ref)
config = yaml.load(main_yaml, Loader=yaml.SafeLoader)
pk_shape, pk_path = config['Package']
pk_shape, pk_path = pk_shape['Shape'], pk_path['Path']
print(f"shape name: {pk_shape['property']['name']}")
print(f"shape area: {pk_shape['property']['area']}")
print(f"shape color: {pk_shape['color']}")
print(f"input file: {pk_path['input_file']}")
if __name__ == '__main__':
main()
输出
shape name: Rectangle
shape area: 200
shape color: red
input file: /path/src/a
更新2
你可以把它们结合起来,像这样
# xxx.yaml
CREATE_FONT_PICTURE:
PROJECTS:
SUNG: &id_SUNG
name: SUNG
work_dir: SUNG
output_dir: temp
font_pixel: 24
DEFINE: &id_define !ref [*id_SUNG] # you can use config['CREATE_FONT_PICTURE']['DEFINE'][name, work_dir, ... font_pixel]
AUTO_INIT:
basename_suffix: !dict_ref [*id_define, name, !product [5, 3, 2]] # SUNG30
# ↓ This is not correct.
# basename_suffix: !dict_ref [*id_define, name, !product [5, 3, 2]] # It will build by Deep-level. id_define is Deep-level: 2. So you must put it after 2. otherwise, it can't refer to the correct value.
其他回答
加上上面@Joshbode的初始回答,我对代码片段进行了一些修改,以支持UNIX风格的通配符模式。
不过我还没有在windows中进行测试。为了便于维护,我面临着将大型yaml中的数组拆分到多个文件中的问题,并正在寻找一种解决方案,以便在基本yaml的同一个数组中引用多个文件。因此,下面的解决方案。解决方案不支持递归引用。它只支持在基本yaml中引用的给定目录级别中的通配符。
import yaml
import os
import glob
# Base code taken from below link :-
# Ref:https://stackoverflow.com/a/9577670
class Loader(yaml.SafeLoader):
def __init__(self, stream):
self._root = os.path.split(stream.name)[0]
super(Loader, self).__init__(stream)
def include(self, node):
consolidated_result = None
filename = os.path.join(self._root, self.construct_scalar(node))
# Below section is modified for supporting UNIX wildcard patterns
filenames = glob.glob(filename)
# Just to ensure the order of files considered are predictable
# and easy to debug in case of errors.
filenames.sort()
for file in filenames:
with open(file, 'r') as f:
result = yaml.load(f, Loader)
if isinstance(result, list):
if not isinstance(consolidated_result, list):
consolidated_result = []
consolidated_result += result
elif isinstance(result, dict):
if not isinstance(consolidated_result, dict):
consolidated_result = {}
consolidated_result.update(result)
else:
consolidated_result = result
return consolidated_result
Loader.add_constructor('!include', Loader.include)
使用
a:
!include a.yaml
b:
# All yamls included within b folder level will be consolidated
!include b/*.yaml
不,标准YAML不包括任何类型的“import”或“include”语句。
可能在问问题时不支持,但你可以将其他YAML文件导入其中:
imports: [/your_location_to_yaml_file/Util.area.yaml]
虽然我没有任何在线参考资料,但这对我来说很有用。
您的问题没有要求使用Python解决方案,但这里有一个使用PyYAML的解决方案。
PyYAML允许您将自定义构造函数(例如!include)附加到YAML加载器。我已经包含了一个可以设置的根目录,以便这个解决方案支持相对和绝对文件引用。
基于类的解决方案
这是一个基于类的解决方案,避免了我原始响应的全局根变量。
请参阅以下要点,了解一个类似的、更健壮的Python 3解决方案,该解决方案使用元类注册自定义构造函数。
import yaml
import os
class Loader(yaml.SafeLoader):
def __init__(self, stream):
self._root = os.path.split(stream.name)[0]
super(Loader, self).__init__(stream)
def include(self, node):
filename = os.path.join(self._root, self.construct_scalar(node))
with open(filename, 'r') as f:
return yaml.load(f, Loader)
Loader.add_constructor('!include', Loader.include)
一个例子:
foo.yaml
a: 1
b:
- 1.43
- 543.55
c: !include bar.yaml
bar.yaml
- 3.6
- [1, 2, 3]
现在可以使用以下方法加载文件:
>>> with open('foo.yaml', 'r') as f:
>>> data = yaml.load(f, Loader)
>>> data
{'a': 1, 'b': [1.43, 543.55], 'c': [3.6, [1, 2, 3]]}
结合其他答案,这里是一个简短的解决方案,没有重载Loader类,它可以与任何加载器操作文件:
import json
from pathlib import Path
from typing import Any
import yaml
def yaml_include_constructor(loader: yaml.BaseLoader, node: yaml.Node) -> Any:
"""Include file referenced with !include node"""
# noinspection PyTypeChecker
fp = Path(loader.name).parent.joinpath(loader.construct_scalar(node)).resolve()
fe = fp.suffix.lstrip(".")
with open(fp, 'r') as f:
if fe in ("yaml", "yml"):
return yaml.load(f, type(loader))
elif fe in ("json", "jsn"):
return json.load(f)
else:
return f.read()
def main():
loader = yaml.SafeLoader # Works with any loader
loader.add_constructor("!include", yaml_include_constructor)
with open(...) as f:
yml = yaml.load(f, loader)
PyTypeChecker的存在是为了防止pep检查警告预期类型'ScalarNode',得到'节点'而不是通过节点:yaml。节点到loader.construct_scalar()。
如果yaml。加载输入流不是文件流,因为loader.name在这种情况下不包含路径:
class Reader(object):
...
def __init__(self, stream):
...
if isinstance(stream, str):
self.name = "<unicode string>"
...
elif isinstance(stream, bytes):
self.name = "<byte string>"
...
else:
self.name = getattr(stream, 'name', "<file>")
...
在我的用例中,我知道只包含YAML文件,所以解决方案可以进一步简化:
def yaml_include_constructor(loader: yaml.Loader, node: yaml.Node) -> Any:
"""Include YAML file referenced with !include node"""
with open(Path(loader.name).parent.joinpath(loader.construct_yaml_str(node)).resolve(), 'r') as f:
return yaml.load(f, type(loader))
Loader = yaml.SafeLoader # Works with any loader
Loader.add_constructor("!include", yaml_include_constructor)
def main():
with open(...) as f:
yml = yaml.load(f, Loader=Loader)
甚至是使用lambda的一行代码:
Loader = yaml.SafeLoader # Works with any loader
Loader.add_constructor("!include",
lambda l, n: yaml.load(Path(l.name).parent.joinpath(l.construct_scalar(n)).read_text(), type(l)))