not a Python function

sorry if i'm being dumb about this, but functions with multiple input arguments don't seem to be recognized as functions:

```
julia> using PyCall

julia> @pyimport scipy.optimize as so

julia> so.curve_fit((x,m,b)->m*x+b, 1:10, 2:2:20)
ERROR: PyError (:PyObject_Call)
TypeError(' is not a Python function',)
File "/home/arthurb/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/scipy/optimize/minpack.py", line 604, in curve_fit
args, varargs, varkw, defaults = _getargspec(f)
File "/home/arthurb/.julia/v0.4/Conda/deps/usr/lib/python2.7/site-packages/scipy/_lib/_util.py", line 289, in getargspec_no_self
argspec = inspect.getargspec(func)
File "/home/arthurb/.julia/v0.4/Conda/deps/usr/lib/python2.7/inspect.py", line 816, in getargspec
raise TypeError('{!r} is not a Python function'.format(func))

[inlined code] from /home/arthurb/.julia/v0.4/PyCall/src/exception.jl:81
in pycall at /home/arthurb/.julia/v0.4/PyCall/src/PyCall.jl:356
in call at /home/arthurb/.julia/v0.4/PyCall/src/PyCall.jl:372

julia> so.newton(x -> cos(x) - x, 1)
0.7390851332151607
```

Reproduce the scipy.optimize.curve_fit example and compare it with the working newton call. Start in PyCall/src/PyCall.jl around pycall and call, using the traceback locations as context. Done means Julia anonymous functions with multiple arguments work with Python code that inspects its callable signature, with regression coverage for this example.

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