Description
Since kwarg order will be guaranteed in python 3.6, we could allow this...though would only want to try with a 3.6+ version check, otherwise code could more or less randomly work/not work, depending on dict order.
In [44]: df = pd.DataFrame({'a': [1,2,3]})
In [45]: df.assign(b=1, c=lambda x: x['b'] * 2)