python 循环中的柯里化函数

发布于 2024-09-06 19:17:57 字数 836 浏览 3 评论 0原文

所以这里有一些代码可以简化我一直在做的事情：

vars = {
    'a':'alice',
    'b':'bob',
}
cnames = ['charlie', 'cindy']

commands = []

for c in cnames:
    kwargs = dict(vars)
    kwargs['c'] = c
    print kwargs
    commands.append(lambda:a_function(**kwargs))

print commands

def a_function(a=None, b=None, c=None):
    print a
    print b
    print c

for c in commands:
    print "run for "+ repr(c)
    c()

这是它的输出：

{'a': 'alice', 'c': 'charlie', 'b': 'bob'}
{'a': 'alice', 'c': 'cindy', 'b': 'bob'}
[<function <lambda> at 0x1001e9a28>, <function <lambda> at 0x1001e9e60>]
run for <function <lambda> at 0x1001e9a28>
alice
bob
cindy
run for <function <lambda> at 0x1001e9e60>
alice
bob
cindy

我希望先得到 charlie，然后是 cindy，为什么 cindy 显示两次？

原文

So here is some code that simplifies what I've been working on:

vars = {
    'a':'alice',
    'b':'bob',
}
cnames = ['charlie', 'cindy']

commands = []

for c in cnames:
    kwargs = dict(vars)
    kwargs['c'] = c
    print kwargs
    commands.append(lambda:a_function(**kwargs))

print commands

def a_function(a=None, b=None, c=None):
    print a
    print b
    print c

for c in commands:
    print "run for "+ repr(c)
    c()

And here is its output:

{'a': 'alice', 'c': 'charlie', 'b': 'bob'}
{'a': 'alice', 'c': 'cindy', 'b': 'bob'}
[<function <lambda> at 0x1001e9a28>, <function <lambda> at 0x1001e9e60>]
run for <function <lambda> at 0x1001e9a28>
alice
bob
cindy
run for <function <lambda> at 0x1001e9e60>
alice
bob
cindy

I would expect to get charlie, then cindy, why is cindy being displayed twice?

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×纯※雪 2024-09-13 19:17:57

您遇到了一个经典的绑定时间问题，@Mike 的解决方案就是经典的解决方案。一个好的替代方法是编写一个高阶函数：

def makecall(kwargs):
  def callit():
    return a_function(**kwargs)
  return callit

并在循环中使用commands.append(makecall(kwargs))。两种解决方案都遵循相同的原理（通过参数的传递强制进行早期绑定——在我的例子中是一个普通参数，@Mike 中的命名参数的默认值）；选择只是风格或优雅的问题（我，虽然我在超级简单的情况下容忍 lambda ，但只要有最微妙的复杂性介入，我就非常喜欢老式的 def ） ;-)。

You're encountering a classic binding-time problem, and @Mike's solution is the classic one. A good alternative is to write a higher order function:

def makecall(kwargs):
  def callit():
    return a_function(**kwargs)
  return callit

and use commands.append(makecall(kwargs)) in your loop. Both solutions work on the same principle (by forcing early binding through passage of an argument -- a plain argument in my case, a default value for a named argument in @Mike's); the choice is just a matter of style or elegance (me, while I tolerate lambda in super-simple cases, as long as the subtlest complication intervenes I vastly prefer good old def;-).

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