Question

我正在尝试使用Python进行简单的YAML到JSON转换器，但它看起来似乎不正确。我是一个本地的Perl / Ruby程序员，所以我有三个脚本和一个输入文件：

testinput.yaml

---
default:
  default:
    line_one: '[I]<description>[/I]'
    line_three: '<creator>'
    line_two: '<title> [<type>]'
    link_to: '<citation>'

和一个Python / Perl / Ruby脚本，每个在我的脑海中完全相同：

y2j.rb

require 'rubygems'
require 'json'
require 'yaml'
yml = YAML.load_file('testinput.yaml')
json = JSON.dump(yml)

puts json

y2j.pl

use JSON;
use YAML;
my $filename = "testinput.yaml";
my $yaml  = YAML::LoadFile($filename);
print encode_json($yaml);

y2j.py

import yaml
import json


stream = open("testinput.yaml", 'r')
data = yaml.load_all(stream)
json = json.dumps(data)

print(json)
enter code here

然后输出：

ruby y2j.rb

{"default":{"default":{"link_to":"<citation>","line_two":"<title> [<type>]","line_three":"<creator>","line_one":"[I]<description>[/I]"}}}

perl y2j.pl

{"default":{"default":{"line_three":"<creator>","line_two":"<title> [<type>]","link_to":"<citation>","line_one":"[I]<description>[/I]"}}}

（到目前为止，很好）

python y2j.py

Traceback (most recent call last):
  File "y2j.py", line 7, in <module>
    json = json.dumps(data)
  File "/usr/lib64/python2.6/json/__init__.py", line 230, in dumps
    return _default_encoder.encode(obj)
  File "/usr/lib64/python2.6/json/encoder.py", line 367, in encode
    chunks = list(self.iterencode(o))
  File "/usr/lib64/python2.6/json/encoder.py", line 317, in _iterencode
    for chunk in self._iterencode_default(o, markers):
  File "/usr/lib64/python2.6/json/encoder.py", line 323, in _    iterencode_default
    newobj = self.default(o)
  File "/usr/lib64/python2.6/json/encoder.py", line 344, in default
    raise TypeError(repr(o) + " is not JSON serializable")
TypeError: <generator object load_all at 0x15a81e0> is not JSON serializable

我在这里缺少一些完全明显的东西吗？

Answer 1

根据错误消息，

yaml.load_all为generator。请参阅the documentation中的以下示例：

>>> for data in yaml.load_all(documents):
...     print data

{'description': 'A set of handgear with sparks that crackle across its knuckleguards.\n',
'name': "The Set of Gauntlets 'Pauraegen'"}
{'description': 'A set of gauntlets that gives off a foul, acrid odour yet remains untarnished.\n',
'name': "The Set of Gauntlets 'Paurnen'"}
{'description': 'A set of handgear, freezing with unnatural cold.\n',
'name': "The Set of Gauntlets 'Paurnimmen'"}

请注意，此代码迭代生成器以访问其内容。

相反，您应该load数据（或者，为了降低安全风险，safe_load）。我认为你的Python版本应该如下：

import json

import yaml

with open("testinput.yaml") as stream:
    yaml_data = yaml.safe_load(stream)
json_data = json.dumps(yaml_data)

print(json_data)

请注意以下事项：

import按the style guide;
使用with上下文管理器来处理文件;和
使用名称json_data以避免遮蔽json库。

如果将在文件中包含多个文档，您可以尝试例如yaml_data = list(yaml.load_all(stream))。

YAML-JSON转换器;有缺陷的东西？

1 个答案: