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May 2, 2013 · 0. If you want to create a nested dictionary given a list (arbitrary length) for a path and perform a function on an item that may exist at the end of the path, this handy little recursive function is quite helpful: def ensure_path(data, path, default=None, default_func=lambda x: x): """. Function:
Mar 12, 2009 · Basically, sometimes I want to think of a nested dictionary as a flat dictionary, and sometimes I want to think of it indeed as a complex hierarchy. I could wrap this all in a class, but it seems like someone might have done this already. Alternatively, it seems like there might be some really elegant syntactical constructions to do this.
No, those are nested dictionaries, so that is the only real way (you could use get() but it's the same thing in essence). However, there is an alternative. However, there is an alternative. Instead of having nested dictionaries, you can use a tuple as a key instead:
May 3, 2017 · Iterating through a dictionary only gives you the keys. You told python to expect a bunch of tuples, and it tried to unpack something that wasn't a tuple (your code is set up to expect each iterated item to be of the form (key,value), which was not the case (you were simply getting key on each iteration).
I'm building some Python code to read and manipulate deeply nested dicts (ultimately for interacting with JSON services, however it would be great to have for other purposes) I'm looking for a way...
I just wanted to note (as this is one of the top results for converting from a nested dictionary to a pandas dataframe) that there are other ways of nesting dictionaries that can be also be converted to a dataframe (e.g. nesting via columns). e.g. the following nested dictionary
The set method in ( Setting a value in a nested python dictionary given a list of indices and value ) seems more robust to missing parental keys. To copy it over: def nested_set(dic, keys, value): for key in keys[:-1]: dic = dic.setdefault(key, {}) dic[keys[-1]] = value.
Jan 9, 2020 · In python, append is only for lists, not dictionaries. This should do what you want: d['A']['b'] = 3. Explanation: When you write d['A'] you are getting another dictionary (the one whose key is A), and you can then use another set of brackets to add or access entries in the second dictionary. answered Jan 9, 2020 at 20:51.
Feb 20, 2017 · 1. You can not access "team" as key because dict["standing"] holds the list value. In order to print all the teams, do: for standing in my_json['standing']: print standing['team'] where my_json holds your dict object. In case you want to get the list of all teams, you may use list comprehension expression as:
While you can use the various tools that Python has to count the elements in this trivial dictionary, the more interesting and productive thing is to think about the structure of the data in the first place. The basic data structures of Python are lists, sets, tuples, and dictionaries. Any of these data structures can 'hold', by reference, any ...
