“开箱"在 Python 中将字典传递给函数的名称空间? [英] "unpacking" a passed dictionary into the function's name space in Python?

问题描述

在我所做的工作中，为了方便起见，我经常需要将参数分组为子集:

In the work I do, I often have parameters that I need to group into subsets for convenience:

d1 = {'x':1,'y':2}
d2 = {'a':3,'b':4}

我通过传入多个字典来做到这一点.大多数时候我直接使用传递的字典，即:

I do this by passing in multiple dictionaries. Most of the time I use the passed dictionary directly, i.e.:

def f(d1,d2):
    for k in d1:
        blah( d1[k] )

在一些函数中我需要直接访问变量，事情变得很麻烦；我真的很想在本地命名空间中使用这些变量.我希望能够执行以下操作:

In some functions I need to access the variables directly, and things become cumbersome; I really want those variables in the local name space. I want to be able to do something like:

def f(d1,d2)
    locals().update(d1)
    blah(x)
    blah(y)    

但是不能保证 locals() 返回的字典更新确实更新了命名空间.

but the updates to the dictionary that locals() returns aren't guaranteed to actually update the namespace.

这是明显的手动方式:

def f(d1,d2):
    x,y,a,b = d1['x'],d1['y'],d2['a'],d2['b']
    blah(x)
    return {'x':x,'y':y}, {'a':a,'b':b}

这会导致每个函数的参数列表重复三次.这可以通过装饰器自动化:

This results in three repetitions of the parameter list per function. This can be automated with a decorator:

def unpack_and_repack(f):
    def f_new(d1, d2):
        x,y,a,b = f(d1['x'],d1['y'],d2['a'],d3['b'])
        return {'x':x,'y':y}, {'a':a,'b':b}
    return f_new
@unpack
def f(x,y,a,b):
    blah(x)
    blah(y)
    return x,y,a,b

这导致装饰器重复 3 次，每个函数重复两次，所以如果你有很多函数，效果会更好.

This results in three repetitions for the decorator, plus two per function, so it's better if you have a lot of functions.

有没有更好的办法?也许使用 eval 的东西?谢谢！

Is there a better way? Maybe something using eval? Thanks!

推荐答案

您始终可以将字典作为参数传递给函数.例如，

You can always pass a dictionary as an argument to a function. For instance,

dict = {'a':1, 'b':2}
def myFunc(a=0, b=0, c=0):
    print(a,b,c)
myFunc(**dict)

这篇关于“开箱"在 Python 中将字典传递给函数的名称空间?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！