Optional argument in lambda function

Question

I have a function:

cost(X, model, reg = 1e-3, sparse) 

And I need to pass this function to another one under the form:

f(X, model) f(X, model, reg = reg) 

I am using lambda to do this:

f = lambda X, model: cost(X, model, sparse = np.random.rand(10,10)) 

And python complains that lambda got an unexpected argument reg. How do I do this correctly?

If I do the other way:

f = lambda X, model, reg: cost(X, model, reg = reg, sparse = np.random.rand(10,10)) 

Then it's not working in the first case.

Answer 1

Lambda's take the same signature as regular functions, and you can give reg a default:

f = lambda X, model, reg=1e3: cost(X, model, reg=reg, sparse=np.random.rand(10,10)) 

What default you give it depends on what default the cost function has assigned to that same parameter. These defaults are stored on that function in the cost.__defaults__ structure, matching the argument names. It is perhaps easiest to use the inspect.getargspec() function to introspect that info:

from inspect import getargspec  spec = getargspec(cost) cost_defaults = dict(zip(spec.args[-len(defaults:], spec.defaults)) f = lambda X, model, reg=cost_defaults['reg']: cost(X, model, reg=reg, sparse=np.random.rand(10,10)) 

Alternatively, you could just pass on any extra keyword argument:

f = lambda X, model, **kw: cost(X, model, sparse=np.random.rand(10,10), **kw) 

Answer 2

have you tried something like

f = lambda X, model, **kw: cost(X, model, sparse = np.random.rand(10,10), **kw)

then reg (and any other named argument you want to pass through (other than sparse)) should work fine.