将带参数的函数传递给Python中的另一个函数?

Is it possible to pass functions with arguments to another function in Python?

是否可能将带参数的函数传递给Python中的另一个函数?

Say for something like:

类似的说:

def perform(function):
    return function()

But the functions to be passed will have arguments like:

但是将要通过的函数将具有如下参数:

action1()
action2(p)
action3(p,r)

6 个解决方案

#1

219

Do you mean this?

你的意思是这个吗?

def perform( fun, *args ):
    fun( *args )

def action1( args ):
    something

def action2( args ):
    something

perform( action1 )
perform( action2, p )
perform( action3, p, r )

#2

This is what lambda is for:

这就是lambda表达式的作用:

def Perform(f):
    f()

Perform(lambda: Action1())
Perform(lambda: Action2(p))
Perform(lambda: Action3(p, r))

#3

You can use the partial function from functools like so.

您可以使用像这样的函数工具的部分函数。

from functools import partial

def perform(f):
    f()

perform(Action1)
perform(partial(Action2, p))
perform(partial(Action3, p, r))

Also works with keywords

也适用于关键字

perform(partial(Action4, param1=p))

#4

Use functools.partial, not lambdas! And ofc Perform is a useless function, you can pass around functions directly.

使用functools。部分,而不是λ!而ofc执行是一个无用的函数，你可以直接传递函数。

for func in [Action1, partial(Action2, p), partial(Action3, p, r)]:
  func()

#5

(months later) a tiny real example where lambda is useful, partial not:
say you want various 1-dimensional cross-sections through a 2-dimensional function, like slices through a row of hills.
quadf( x, f ) takes a 1-d f and calls it for various x.
To call it for vertical cuts at y = -1 0 1 and horizontal cuts at x = -1 0 1,

(几个月后)一个很小的真实示例，lambda是有用的，但不是部分有用的:例如，您希望通过一个二维函数来实现各种一维的横截面，就像在一排小山中切片一样。quadf(x, f)取1-d f，取不同的x，记作y = -1处的垂直切点，x = -1处的水平切点，

fx1 = quadf( x, lambda x: f( x, 1 ))
fx0 = quadf( x, lambda x: f( x, 0 ))
fx_1 = quadf( x, lambda x: f( x, -1 ))
fxy = parabola( y, fx_1, fx0, fx1 )

f_1y = quadf( y, lambda y: f( -1, y ))
f0y = quadf( y, lambda y: f( 0, y ))
f1y = quadf( y, lambda y: f( 1, y ))
fyx = parabola( x, f_1y, f0y, f1y )

As far as I know, partial can't do this --

就我所知，部分不能这么做

quadf( y, partial( f, x=1 ))
TypeError: f() got multiple values for keyword argument 'x'

(How to add tags numpy, partial, lambda to this ?)

(如何向它添加标记numpy、partial和lambda ?)

#6

Here is a way to do it with a closure:

这里有一个结束的方法:

    def generate_add_mult_func(func):
        def function_generator(x):
            return reduce(func,range(1,x))
        return function_generator

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

    def mult(x,y):
        return x*y

    adding=generate_add_mult_func(add)
    multiplying=generate_add_mult_func(mult)

    print adding(10)
    print multiplying(10)

#1

219