Passing a dictionary to a function as keyword parameters

Question

I’d like to call a function in python using a dictionary with matching key-value pairs for the parameters.

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?

Asked By: Dave Hillier

||

Source

Answer 1

Here ya go – works just any other iterable:

d = {'param' : 'test'}

def f(dictionary):
    for key in dictionary:
        print key

f(d)

Answered By: Patrick Harrington

Answer 2

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)

Answered By: Dave Hillier

Answer 3

In python, this is called "unpacking", and you can find a bit about it in the tutorial. The documentation of it sucks, I agree, especially because of how fantasically useful it is.

Answered By: llimllib

Answer 4

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):

The function can have parameters that are not included in the dictionary
You can not override a function parameter that is already in the dictionary
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'

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

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}

Answered By: David Parks

Passing a dictionary to a function as keyword parameters

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