Question

我有一个YAML文件，我想存储不同的＆＃34;玩家＆＃34;在我的游戏中。

YAML文件如下所示：

Person:
  Name:
  Age:
  Nationality:
  Footed:
  Position:

创建播放器后，YAML文件应如下所示：

Person:
   Name: Rich
   Age: 23 
   Nationality: British
   Footed: Right
   Position: Forward

到目前为止，我的代码是这样的：

import yaml

name = input('What is your name? ')
age = int(input('What is your age? '))
nationality = input('What is your nationality? ')
footed = input('What foot? ')
position = input('What is your position? ')

with open('test.yml', 'a') as outfile:
    outfile.write(yaml.dump({'Name' : name, 'Age' : age, 'Nationality' : nationality,
    'Footed' : footed, 'Position' : position}))

但是当我运行它并给用户输入时，yaml文件最终看起来像：

{Age: 2, Footed: r, Name: r, Nationality: r, Position: r}

如何将其添加到YAML文件而不是追加，以及如何将其垂直构建而不是水平构建？最后，如果我想添加10/20 / n的玩家数量，我希望YAML文件将它们全部存储在彼此之下，这样我就可以单独调用每个玩家

Answer 1

当您按原样写出YAML时，首先将数据结构写入内存中的文件，然后将内存文件内容作为字符串检索，然后将其写入文件。这是低效和缓慢的。

您还应该阅读您拥有的YAML文件，更新数据结构并将其转储：

import yaml

file_name = 'test.yml'

with open(file_name) as infile:
    data = yaml.load(infile)

person = data['Person']
person['name'] = input('What is your name? ')
person['age'] = int(input('What is your age? '))
person['nationality'] = input('What is your nationality? ')
person['footed'] = input('What foot? ')
person['position'] = input('What is your position? ')

with open(file_name, 'w') as outfile:
    yaml.dump(data, stream=outfile, default_flow_style=False, indent=3)

default_flow_style参数可确保您的键值对列在彼此之下。

使用PyYAML，infile中的任何注释都将丢失，映射中键的顺序可能会被扰乱。如果这是一个问题，我建议您使用ruamel.yaml包（免责声明：我是该包的作者），并将代码更改为：

import ruamel.yaml

file_name = 'test.yml'

with open(file_name) as infile:
    data = ruamel.yaml.round_trip_load(infile)

person = data['Person']
person['name'] = input('What is your name? ')
person['age'] = int(input('What is your age? '))
person['nationality'] = input('What is your nationality? ')
person['footed'] = input('What foot? ')
person['position'] = input('What is your position? ')

with open(file_name, 'w') as outfile:
    yaml.round_trip_dump(data, stream=outfile, indent=3)

如果您想存储多个玩家。确保您的顶层数据结构是一个序列，或者是从某个唯一值（例如Person的名称）映射。在这种情况下，使用不同的输入文件作为模板，并通过读取输入文件来更新输出文件，附加到列表cq。更新dict并写出文件。只要名称是唯一的，使用顶级映射/字典就可以更容易地完成此操作。

Answer 2

使用"w"打开文件，而不是"a"。
您还必须撰写Person。
使用选项default_flow_style=False。

代码：

with open('test.yml', 'w') as outfile:
    outfile.write(yaml.dump(
      {"Person": {
        'Name' : name,
        'Age' : age,
        'Nationality' : nationality,
        'Footed' : footed,
        'Position' : position}
      },
      default_flow_style=False))

输出

Person:
   Name: Rich
   Age: 23 
   Nationality: British
   Footed: Right
   Position: Forward

Answer 3

你可以这样做：

import pyaml as yaml

name = input('What is your name? ')
age = int(input('What is your age? '))
nationality = input('What is your nationality? ')
footed = input('What foot? ')
position = input('What is your position? ')

person = {'Person':{
              'Name': name, 
              'Age': age, 
              'Nationality': nationality, 
              'Footed': footed, 
              'Position': position}
         }

with open('test.yml', 'a') as outfile:
    yaml.dump(person, outfile, indent=4)

使用Python和YAML

3 个答案: