Python参数绑定器

Dus*_*etz 60 python partial-application

如何将参数绑定到Python方法以存储一个用于以后调用的nullary仿函数?与C++类似boost::bind.

例如:

def add(x, y):
    return x + y

add_5 = magic_function(add, 5)
assert add_5(3) == 8
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小智 73

functools.partial 返回一个可调用的包装函数,其中一些或所有参数都被冻结.

import sys
import functools

print_hello = functools.partial(sys.stdout.write, "Hello world\n")

print_hello()
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Hello world
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以上用法等同于以下内容lambda.

print_hello = lambda *a, **kw: sys.stdout.write("Hello world\n", *a, **kw)
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Mat*_*vor 48

我对boost :: bind并不过分熟悉,但是partial函数functools可能是一个好的开始:

>>> from functools import partial

>>> def f(a, b):
...     return a+b

>>> p = partial(f, 1, 2)
>>> p()
3

>>> p2 = partial(f, 1)
>>> p2(7)
8
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  • 这是一个比接受的答案更好的例子,因为它显示了部分绑定. (6认同)

jfs*_*jfs 11

如果functools.partial没有,则可以轻松模拟:

>>> make_printer = lambda s: lambda: sys.stdout.write("%s\n" % s)
>>> import sys
>>> print_hello = make_printer("hello")
>>> print_hello()
hello
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要么

def partial(func, *args, **kwargs):
    def f(*args_rest, **kwargs_rest):
        kw = kwargs.copy()
        kw.update(kwargs_rest)
        return func(*(args + args_rest), **kw) 
    return f

def f(a, b):
    return a + b

p = partial(f, 1, 2)
print p() # -> 3

p2 = partial(f, 1)
print p2(7) # -> 8

d = dict(a=2, b=3)
p3 = partial(f, **d)
print p3(), p3(a=3), p3() # -> 5 6 5
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Ale*_*lex 9

lambdas允许您使用较少的参数创建一个新的未命名函数并调用该函数!

>>> def foobar(x,y,z):
...     print "%d, %d, %d" % (x,y,z)
>>> foobar(1,2,3) # call normal function

>>> bind = lambda x: foobar(x, 10, 20) # bind 10 and 20 to foobar
>>> bind(1) # print 1, 10, 20

>>> bind = lambda: foobar(1,2,3) # bind all elements  
>>> bind()  # print 1, 2, 3
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编辑

https://docs.python.org/2/library/functools.html#functools.partial

如果您计划在函数调用中使用命名参数绑定,这也适用:

>>> from functools import partial
>>> barfoo = partial(foobar, x=10)
>>> barfoo(y=5,z=6)
21
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请注意

>>> barfoo(5,6) 
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: foobar() got multiple values for keyword argument 'x'
>>> f = partial(foobar, z=20)
>>> f(1,1)
22        
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Cla*_*diu 7

这也可行:

def curry(func, *args):
    def curried(*innerargs):
       return func(*(args+innerargs))
    curried.__name__ = "%s(%s, ...)" % (func.__name__, ", ".join(map(str, args)))
    return curried

>>> w=curry(sys.stdout.write, "Hey there")
>>> w()
Hey there
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  • 这不是咖喱。柯里化适用于具有多个参数的函数。Curry 的工作原理如下:给定 `g = curry(f)`,我们有 `f(x1,x2,...) = g(x1)(x2)...` (2认同)