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6 年前
回复了 user633183 创建的主题 » 用python解构dict和objects

不要一开始就把争论压扁。当你写一个8元函数的时候 User ,你一定会犯错误,比如按错误的顺序传递论点。

以下哪项将生成您想要的用户?

  1. User(24, "Jon Doe", "http://mywebsite.com", "I am an actor", "example@example.com", "M", "+12345678", "johndoe")
  2. User(24, "Jon Doe", "http://mywebsite.com", "I am an actor", "example@example.com", "+12345678", "M", "johndoe")

不可能知道!如果函数采用描述符,则不存在此问题-

class User:
  def __init__ (self, desc = {}):
    self.desc = desc # whitelist items, if necessary

  def __str__ (self):
    # invent our own "destructuring" syntax
    [ name, age, gender ] = \
      destructure(self.desc, 'name', 'age', 'gender')

    return f"{name} ({gender}) is {age} years old"

# create users with a "descriptor"
u = User({ 'age': 2, 'gender': 'M' })
v = User({ 'gender': 'F', 'age': 3 })
x = User({ 'gender': 'F', 'name': 'Alice', 'age': 4 })

print(u) # None (M) is 2 years old
print(v) # None (F) is 3 years old
print(x) # Alice (F) is 4 years old

我们可以定义自己的 destructure AS -

def destructure (d, *keys):
  return [ d[k] if k in d else None for k in keys ]

这仍然可能导致长链,但顺序取决于调用方,因此它不像原始问题中的8元函数那样脆弱-

[ name, age, gender ] = \
  destructure(self.desc, 'name', 'age', 'gender')

# works the same as

[ gender, name, age ] = \
  destructure(self.desc, 'gender', 'name', 'age')

另一种选择是使用关键字参数-

class User:
  def __init__ (self, **desc):
    self.desc = desc # whitelist items, if necessary

  def __str__ (self):
    [ name, age, gender ] = \
      destructure(self.desc, 'name', 'age', 'gender')

    return f"{name} ({gender}) is {age} years old"

# create users with keyword arguments
u = User(age = 2, gender = 'M')
v = User(gender = 'F', age = 3)
x = User(gender = 'F', name = 'Alice', age = 4)

print(u) # None (M) is 2 years old
print(v) # None (F) is 3 years old
print(x) # Alice (F) is 4 years old