我有以下名为
input.yaml
的 YAML 文件:
cities:
1: [0,0]
2: [4,0]
3: [0,4]
4: [4,4]
5: [2,2]
6: [6,2]
highways:
- [1,2]
- [1,3]
- [1,5]
- [2,4]
- [3,4]
- [5,4]
start: 1
end: 4
我使用 PyYAML 加载它并打印结果,如下所示:
import yaml
f = open("input.yaml", "r")
data = yaml.load(f)
f.close()
print(data)
结果是以下数据结构:
{ 'cities': { 1: [0, 0]
, 2: [4, 0]
, 3: [0, 4]
, 4: [4, 4]
, 5: [2, 2]
, 6: [6, 2]
}
, 'highways': [ [1, 2]
, [1, 3]
, [1, 5]
, [2, 4]
, [3, 4]
, [5, 4]
]
, 'start': 1
, 'end': 4
}
如您所见,每个城市和高速公路都以列表的形式表示。但是,我希望它们被表示为一个元组。因此,我使用推导式手动将它们转换为元组:
import yaml
f = open("input.yaml", "r")
data = yaml.load(f)
f.close()
data["cities"] = {k: tuple(v) for k, v in data["cities"].items()}
data["highways"] = [tuple(v) for v in data["highways"]]
print(data)
但是,这似乎是一种黑客行为。有没有办法指示 PyYAML 直接将它们读取为元组而不是列表?
对于你正在尝试做的事情,我不会称你所做的事情为黑客行为。根据我的理解,您的替代方法是在 YAML 文件中使用特定于 python 的标签,以便在加载 yaml 文件时正确表示它。但是,这需要您修改 yaml 文件,如果该文件很大,可能会非常烦人并且不理想。
查看 PyYaml 文档 进一步说明了这一点。最终,您希望在您想要表示的结构前面放置一个
!!python/tuple
。要获取样本数据,它需要:
YAML 文件:
cities:
1: !!python/tuple [0,0]
2: !!python/tuple [4,0]
3: !!python/tuple [0,4]
4: !!python/tuple [4,4]
5: !!python/tuple [2,2]
6: !!python/tuple [6,2]
highways:
- !!python/tuple [1,2]
- !!python/tuple [1,3]
- !!python/tuple [1,5]
- !!python/tuple [2,4]
- !!python/tuple [3,4]
- !!python/tuple [5,4]
start: 1
end: 4
示例代码:
import yaml
with open('y.yaml') as f:
d = yaml.load(f.read())
print(d)
将输出:
{'cities': {1: (0, 0), 2: (4, 0), 3: (0, 4), 4: (4, 4), 5: (2, 2), 6: (6, 2)}, 'start': 1, 'end': 4, 'highways': [(1, 2), (1, 3), (1, 5), (2, 4), (3, 4), (5, 4)]}
取决于您的 YAML 输入来自“hack”的位置,这是一个很好的解决方案,特别是如果您使用
yaml.safe_load()
而不是不安全的 yaml.load()
。如果您的 YAML 文件中的“叶子”序列需要是元组,您可以这样做 ¹:
import pprint
import ruamel.yaml
from ruamel.yaml.constructor import SafeConstructor
def construct_yaml_tuple(self, node):
seq = self.construct_sequence(node)
# only make "leaf sequences" into tuples, you can add dict
# and other types as necessary
if seq and isinstance(seq[0], (list, tuple)):
return seq
return tuple(seq)
SafeConstructor.add_constructor(
u'tag:yaml.org,2002:seq',
construct_yaml_tuple)
with open('input.yaml') as fp:
data = ruamel.yaml.safe_load(fp)
pprint.pprint(data, width=24)
打印:
{'cities': {1: (0, 0),
2: (4, 0),
3: (0, 4),
4: (4, 4),
5: (2, 2),
6: (6, 2)},
'end': 4,
'highways': [(1, 2),
(1, 3),
(1, 5),
(2, 4),
(3, 4),
(5, 4)],
'start': 1}
如果您需要处理更多材料,其中序列需要再次成为“正常”列表,请使用:
SafeConstructor.add_constructor(
u'tag:yaml.org,2002:seq',
SafeConstructor.construct_yaml_seq)
¹ 这是使用 ruamel.yaml 一个 YAML 1.2 解析器完成的,我是该解析器的作者。如果您只需要支持 YAML 1.1 和/或由于某种原因无法升级,您应该能够对旧版 PyYAML 执行相同的操作
我遇到了与问题相同的问题,我对这两个答案不太满意。在浏览 pyyaml 文档时我发现 确实是两个有趣的方法
yaml.add_constructor
和yaml.add_implicit_resolver
。
隐式解析器通过将字符串与正则表达式进行匹配,解决了必须用
!!python/tuple
标记所有条目的问题。我还想使用元组语法,所以写 tuple: (10,120)
而不是写一个列表 tuple: [10,120]
然后得到
转换为元组,我个人觉得很烦人。我也不想安装外部库。这是代码:
import yaml
import re
# this is to convert the string written as a tuple into a python tuple
def yml_tuple_constructor(loader, node):
# this little parse is really just for what I needed, feel free to change it!
def parse_tup_el(el):
# try to convert into int or float else keep the string
if el.isdigit():
return int(el)
try:
return float(el)
except ValueError:
return el
value = loader.construct_scalar(node)
# remove the ( ) from the string
tup_elements = value[1:-1].split(',')
# remove the last element if the tuple was written as (x,b,)
if tup_elements[-1] == '':
tup_elements.pop(-1)
tup = tuple(map(parse_tup_el, tup_elements))
return tup
# !tuple is my own tag name, I think you could choose anything you want
yaml.add_constructor(u'!tuple', yml_tuple_constructor)
# this is to spot the strings written as tuple in the yaml
yaml.add_implicit_resolver(u'!tuple', re.compile(r"\(([^,\W]{,},){,}[^,\W]*\)"))
最后执行以下命令:
>>> yml = yaml.load("""
...: cities:
...: 1: (0,0)
...: 2: (4,0)
...: 3: (0,4)
...: 4: (4,4)
...: 5: (2,2)
...: 6: (6,2)
...: highways:
...: - (1,2)
...: - (1,3)
...: - (1,5)
...: - (2,4)
...: - (3,4)
...: - (5,4)
...: start: 1
...: end: 4""")
>>> yml['cities']
{1: (0, 0), 2: (4, 0), 3: (0, 4), 4: (4, 4), 5: (2, 2), 6: (6, 2)}
>>> yml['highways']
[(1, 2), (1, 3), (1, 5), (2, 4), (3, 4), (5, 4)]
与我没有测试过的
save_load
相比,load
可能存在潜在的缺点。
这对我有用 -
config.yaml
cities:
1: !!python/tuple [0,0]
2: !!python/tuple [4,0]
3: !!python/tuple [0,4]
4: !!python/tuple [4,4]
5: !!python/tuple [2,2]
6: !!python/tuple [6,2]
highways:
- !!python/tuple [1,2]
- !!python/tuple [1,3]
- !!python/tuple [1,5]
- !!python/tuple [2,4]
- !!python/tuple [3,4]
- !!python/tuple [5,4]
start: 1
end: 4
main.py
import yaml
def tuple_constructor(loader, node):
# Load the sequence of values from the YAML node
values = loader.construct_sequence(node)
# Return a tuple constructed from the sequence
return tuple(values)
# Register the constructor with PyYAML
yaml.SafeLoader.add_constructor('tag:yaml.org,2002:python/tuple',
tuple_constructor)
# Load the YAML file
with open('config.yaml', 'r') as f:
data = yaml.load(f, Loader=yaml.SafeLoader)
print(data)
输出:
{'cities': {1: (0, 0), 2: (4, 0), 3: (0, 4), 4: (4, 4), 5: (2, 2), 6: (6, 2)},
'highways': [(1, 2), (1, 3), (1, 5), (2, 4), (3, 4), (5, 4)],
'start': 1,
'end': 4}
在某些情况下可能不安全,但对我来说一个简单的解决方法是以字符串表示形式存储元组列表。读回时,使用 eval() 将字符串转换为元组列表。
atoms = [(7, 13, 14, 15)] # list of tuple
# when creating dict for YAML dump
ddict[grp] = str(atoms) # convert list of tuples to string
# then after reading the YAML file
ddict[grp] = list(eval(ddict[grp])) # list() for slight safety