Passing a dictionary to a function as keyword parameters

Question

I'd like to call a function in python using a dictionary.

Here is some code:

d = dict(param='test')  def f(param):     print(param)  f(d) 

This prints {'param': 'test'} but I'd like it to just print test.

I'd like it to work similarly for more parameters:

d = dict(p1=1, p2=2) def f2(p1, p2):     print(p1, p2) f2(d) 

Is this possible?

Answer 1

Figured it out for myself in the end. It is simple, I was just missing the ** operator to unpack the dictionary

So my example becomes:

d = dict(p1=1, p2=2) def f2(p1,p2):     print p1, p2 f2(**d) 

Answer 2

In[1]: def myfunc(a=1, b=2): In[2]:    print(a, b)  In[3]: mydict = {'a': 100, 'b': 200}  In[4]: myfunc(**mydict) 100 200 

A few extra details that might be helpful to know (questions I had after reading this and went and tested):

1. The function can have parameters that are not included in the dictionary
2. You can not override a function parameter that is already in the dictionary
3. The dictionary can not have values that aren't in the function.

Examples:

Number 1: The function can have parameters that are not included in the dictionary

In[5]: mydict = {'a': 100} In[6]: myfunc(**mydict) 100 2 

Number 2: You can not override a function parameter that is already in the dictionary

In[7]: mydict = {'a': 100, 'b': 200} In[8]: myfunc(a=3, **mydict)  TypeError: myfunc() got multiple values for keyword argument 'a' 

Number 3: The dictionary can not have values that aren't in the function.

In[9]:  mydict = {'a': 100, 'b': 200, 'c': 300} In[10]: myfunc(**mydict)  TypeError: myfunc() got an unexpected keyword argument 'c' 

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How to use a dictionary with more keys than function arguments:

A solution to #3, above, is to accept (and ignore) additional kwargs in your function (note, by convention _ is a variable name used for something being discarded, though technically it's just a valid variable name to Python):

In[11]: def myfunc2(a=None, **_): In[12]:    print(a)  In[13]: mydict = {'a': 100, 'b': 200, 'c': 300}  In[14]: myfunc2(**mydict) 100 

Another option is to filter the dictionary based on the keyword arguments available in the function:

In[15]: import inspect In[16]: mydict = {'a': 100, 'b': 200, 'c': 300} In[17]: filtered_mydict = {k: v for k, v in mydict.items() if k in [p.name for p in inspect.signature(myfunc).parameters.values()]} In[18]: myfunc(**filtered_mydict) 100 200 

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Example with both positional and keyword arguments:

Notice further than you can use positional arguments and lists or tuples in effectively the same way as kwargs, here's a more advanced example incorporating both positional and keyword args:

In[19]: def myfunc3(a, *posargs, b=2, **kwargs): In[20]:    print(a, b) In[21]:    print(posargs) In[22]:    print(kwargs)  In[23]: mylist = [10, 20, 30] In[24]: mydict = {'b': 200, 'c': 300}  In[25]: myfunc3(*mylist, **mydict) 10 200 (20, 30) {'c': 300}