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copied!<p>This will print names of all passable arguments, keyword and non-keyword ones:</p> <pre><code>def func(one, two="value"): y = one, two return y print func.func_code.co_varnames[:func.func_code.co_argcount] </code></pre> <p>This is because first <code>co_varnames</code> are always parameters (next are local variables, like <code>y</code> in the example above).</p> <p>So now you could have a function:</p> <pre><code>def getValidArgs(func, argsDict): '''Return dictionary without invalid function arguments.''' validArgs = func.func_code.co_varnames[:func.func_code.co_argcount] return dict((key, value) for key, value in argsDict.iteritems() if key in validArgs) </code></pre> <p>Which you then could use like this:</p> <pre><code>>>> func(**getValidArgs(func, args)) </code></pre> <hr> <p><strong>EDIT</strong>: A small addition: if you <strong>really need only keyword arguments</strong> of a function, you can use the <code>func_defaults</code> attribute to extract them:</p> <pre><code>def getValidKwargs(func, argsDict): validArgs = func.func_code.co_varnames[:func.func_code.co_argcount] kwargsLen = len(func.func_defaults) # number of keyword arguments validKwargs = validArgs[-kwargsLen:] # because kwargs are last return dict((key, value) for key, value in argsDict.iteritems() if key in validKwargs) </code></pre> <p>You could now call your function with known args, but extracted kwargs, e.g.:</p> <pre><code>func(param1, param2, **getValidKwargs(func, kwargsDict)) </code></pre> <p>This assumes that <code>func</code> uses no <code>*args</code> or <code>**kwargs</code> magic in its signature.</p>

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