"unpacking" a passed dictionary into the function's name space in Python?

Question

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} 

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] ) 

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)     

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

Here's the obvious manual way:

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} 

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 

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

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

Answer 1

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)