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Notes about dict.update(): (1) It changes the existing dict, (2) It returns None, and (3) It overwrites existing items (as does the dictionary merge operator dict1 | dict2). – ChaimG Commented Dec 15, 2023 at 15:23
4. If you want to do the job neatly, better use update method of the dictionary like below: my_dict = {1:"a value", 2:"another value"} my_dict.update({1:"your value"}) also, from Python 3.10 on you can do the following: my_dict |= {1:"your value"} still, there is more: my_dict = {**my_dict, 1:"your value"} and you will get the following result:
my_dict.update(other_dict) In this case you don't have to know how many keys are in the other_dict. You'll just be sure that all of them will be updated on my_dict. 2. You can use any iterable of key/value pairs with dict.update
@MartijnPieters I'd say you're just wrong here. For “an expression that expects a value to exist”, but you don't want to update the dict, you would simply use get with a default value.
Python 3.9 and PEP 584 introduces the dict union, for updating one dict from another dict. Dict union will return a new dict consisting of the left operand merged with the right operand, each of which must be a dict (or an instance of a dict subclass). If a key appears in both operands, the last-seen value (i.e. that from the right-hand operand ...
Functions compute new values out of their parameters (including any target object); procedures modify objects and don't return anything (i.e. they return None). So procedures have side effects, functions don't. update is a procedure, hence it doesn't return a value.
I made a simple function, in which you give the key, the new value and the dictionary as input, and it recursively updates it with the value: def update(key,value,dictionary): if key in dictionary.keys(): dictionary[key] = value. return. dic_aux = [] for val_aux in dictionary.values(): if isinstance(val_aux,dict):
What you could do, instead of calling .update() on the existing dict, is add a sub-dict. Its key could be the name of the source from which you gathered the information. The value could be the dict you receive from the source, and if you need to store more than 1 dict of the same source you can store them in a list. Example.
So this does work: Dict.update({ Var1: Var2 }) But you don't need update or its keyword arguments in your case; you can just use assignment instead: Dict[Var1] = Var2. A couple style things: don't name variables with capital letters; that is usually an indication that a value is a constant. Also, don't use "Dict" as a variable name.
Using the built-in update () function is even faster. I tweaked Steven Rumbalski's example above a bit and it shows how update () is the fastest. There are at least two ways to use it (with a list of tuples or with another dictionary). The former (shown below as update_method1) is the fastest.