Adding variables "by name" to local namespace

Question

I know that I can add a variable to the global namespace "by name" with something like:

def getNewVar(name,val):
    globals()[name]=val

The things is, I would like to do this in a local namespace. I have already tried and failed in using locals() instead of globals(). Someone probably wants to know why I would do something like this. OK, in my use case, the argument to the function is actually a dictionary, and I would like to do something like:

def processDictEntries(dict):
    for varname in dict.keys():
        locals()[varname]=dict[varname]  # won't work, of course

This way, further down in the function, I won't have to keep typing

result1=dict['var1']+5.
result2=dict['var2']*dict['var7']

over and over again. I can just type

result1=var1+5
result2=var2*var7

And, if there is a way to do this in a small loop like I have written, then I don't have to do:

var1=dict['var1']
var2=dict['var2'] etc.

either. I'm really just looking for economy of code; it's a big dictionary.

BTW, the entries of the dictionary are never altered; it is strictly input. And, yes, I know that if the input dictionary lacks one of the variables the function needs, I will be in touble, but I think that can be dealt with. Thanks!

mgilson · Accepted Answer · 2013-04-18 20:00:31Z

2

If you can change how you call the function:

def processDictEntries(var1=None,var2=None,**kwargs):
    #do stuff with var1,var2 ...

And then call the function as:

processDictEntries(**dict)

and of course, if you can't do that, you can always use processDictEntries as a wrapper:

def _processDictEntries(var1=None,var2=None,**kwargs):
    ...

def processDictEntries(d):
    return _processDictEntries(**d)

^{As a side note, it's not a good idea to name a variable dict as then you shadow the builtin function}

answered Apr 18, 2013 at 20:00

mgilson

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