Also, itertools._grouper also needs to be exposed as an iterator. Its type is not exposed by name, but can be accessed like so: type(list(itertools.groupby([0]))[0][1])
Also, itertools._grouper also needs to be exposed as an iterator. Its type is not exposed by name, but can be accessed like so: itertools. groupby( [0]))[0] [1])
type(list(