"I have seen some programmers use it like a struct." I think namedtuple is entirely too tempting to use for that purpose, given that I think it's the wrong sol

"I have seen some programmers use it like a struct." I think namedtuple is entirely too tempting to use for that purpose, given that I think it's the wrong solution. I don't want my struct to be iterable and indexable. I think len(my_struct) should fail. The namedtuple is a great solution for when you have a tuple API (like the old output from os.stat() or time.gmtime()) and want to transition to an attribute API. I would prefer a "namedstruct" for a tuple-less version of the same API, to help avoid the attractive nuisance of namedtuple repurposed as a struct creator. "The other day I was looking for a way to convert a Python dictionary into an object and I came across some code that did something like this:" >>> from collections import namedtuple >>> Parts = {'id_num':'1234', 'desc':'Ford Engine', 'cost':1200.00, 'amount':10} >>> parts = namedtuple('Parts', Parts.keys())(**Parts) (As a minor tweak, that last line can use 'namedtuple('Parts', Parts)(Parts)'.) I'll point out that the namedtuple-less alternative is a class like this: class PartsClass(object): def __init__(self, d): self.__dict__.update(d) parts = PartsClass(Parts) There are some important differences: 0) my version can't be done inline, 1) my PartsClass doesn't have the useful repr() of a namedtuple(), 2) my PartsClass attributes can be modified while a namedtuple's attributes are read-only, 3) the inline nametuple is dreadfully slow - 600x slower than making a class! >>> timeit.timeit(number=1000, stmt=lambda: namedtuple('Parts', Parts)(**Parts)) 1.4882779121398926 >>> timeit.timeit(number=1000, stmt=lambda: PartsClass(Parts)) 0.0022149085998535156 4) hoisting the Parts namedtuple creation results in about 10% faster code than creating a class instance: >>> timeit.timeit(stmt=lambda: PartsClass(Parts)) 1.1018929481506348 >>> PartsTuple = namedtuple('Parts', Parts) >>> timeit.timeit(stmt=lambda: PartsTuple(**Parts)) 1.0174322128295898 5) the tuple API requires the * * notation, while the class notation can take the dictionary directly, and 6) object instance attribute lookup is faster than a namedtuple property lookup: >>> parts_instance = PartsClass(Parts) >>> timeit.timeit(stmt=lambda: parts_instance.desc) 0.19052886962890625 >>> parts_tuple = PartsTuple(**Parts) >>> timeit.timeit(stmt=lambda: parts_tuple.desc) 0.3053739070892334 If there are even two attribute lookups per object then it's faster to use a class instance (1.10+2×0.19 = 1.48) than a namedtuple (1.01+2×0.305=1.62)