问题
The error comes from publishDB = defaultdict(defaultdict({})) I want to make a database like {subject1:{student_id:{assignemt1:marks, assignment2:marks,finals:marks}} , {student_id:{assignemt1:marks, assignment2:marks,finals:marks}}, subject2:{student_id:{assignemt1:marks, assignment2:marks,finals:marks}} , {student_id:{assignemt1:marks, assignment2:marks,finals:marks}}}. I was trying to populate it as DB[math][10001] = a dict and later read out as d = DB[math][10001]. Since, I am on my office computer I can not try different module.
Am I on right track to do so?
回答1:
Such a nested dict structure can be achieved using a recursive defaultdict "tree":
def tree():
return defaultdict(tree)
publishDB = tree()
At each level, the defaultdicts are instantiated with tree which is a zero-argument callable, as required.
Then you can simply assign marks:
publishDB[subject][student][assignment] = mark
回答2:
defaultdict() requires that its first argument be callable: it must be a class that you want an instance of, or a function that returns an instance.
defaultdict({}) has an empty dictionary, which is not callable.
You likely want defaultdict(dict), as dict is a class that returns a dictionary when instantiated (called).
But that still doesn't solve the problem... just moves it to a different level. The outer defaultdict(...) in defaultdict(defaultdict(dict)) has the exact same issue because defaultdict(dict) isn't callable.
You can use a lambda expression to solve this, creating a one-line function that, when called, creates a defaultdict(dict):
defaultdict(lambda: defaultdict(dict))
You could also use the lambda at the lower level if you wanted:
defaultdict(lambda: defaultdict(lambda: {}))
来源:https://stackoverflow.com/questions/40712753/typeerror-first-argument-must-be-callable-defaultdict