dictionary

[Solved] What does the get method do on dictionaries? [closed]

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If the key argument is in switcher, the .get() method returns the value for the key. If the key is not in the dictionary, the method returns the optional “nothing”. def numbers_to_strings(argument): switcher = {0: “zero”, 1: “one”, 2: “two”} return switcher.get(argument, “nothing”) Calling the above function with a key that is in the dictionary: … Read more

[Solved] Python dictionary operations

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Yes, you can do this recursively, if all you have is lists and dictionaries, then that is pretty trivial. Test for the object type, and for containers, recurse: def unwrap_data(obj): if isinstance(obj, dict): if ‘data’ in obj: obj = obj[‘data’] return {key: unwrap_data(value) for key, value in obj.items()} if isinstance(obj, list): return [unwrap_data(value) for value … Read more

[Solved] Python. Converting string into directory [closed]

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This is a very weird way to handle data. I would use a nested dicts: exampleDictionary = {‘thing on ground’: {‘backpack’: {‘tea’: ‘Earl Grey’}}} print exampleDictionary[‘thing on ground’] print exampleDictionary[‘thing on ground’][‘backpack’] print exampleDictionary[‘thing on ground’][‘backpack’][‘tea’] Outputs: {‘backpack’: {‘tea’: ‘Earl Grey’}} {‘tea’: ‘Earl Grey’} Earl Grey 2 solved Python. Converting string into directory [closed]

[Solved] Swift Array append not appending values but replacing it

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There is not enough information but I can guess you where you have made mistake. Your class Bookmark is not singleton class so,every time Bookmark() create new instance every time. that means it will create new bookmark object for every instance. What I suggest you is inside func func setBookmark(imageURL:String, title:String, description:String, summary:String, date:String, link:String) … Read more

[Solved] Add every 100 string keys from dictionary in array of strings with comma separator

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You just need to group your array elements and use map to join your keys using joined(separator: “, “): extension Array { func group(of n: IndexDistance) -> Array<Array> { return stride(from: 0, to: count, by: n) .map { Array(self[$0..<Swift.min($0+n, count)]) } } } Testing: let dic = [“f”:1,”a”:1,”b”:1,”c”:1,”d”:1,”e”:1, “g”: 1] let arr = Array(dic.keys).group(of: 2).map{ … Read more

[Solved] Replace items in a dict of nested list and dicts

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A recursive function that creates a new nested dictionary, replacing a key value with a different value def replace_key(elem, old_key, new_key): if isinstance(elem, dict): # nested dictionary new_dic = {} for k, v in elem.items(): if isinstance(v,dict) or isinstance(v, list): new_dic[replace_key(k, old_key, new_key)] = replace_key(v, old_key, new_key) else: new_dic[replace_key(k, old_key, new_key)] = v return new_dic … Read more

[Solved] Access the key of a multilevel JSON file in python

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You can use next with a generator comprehension. res = next(i[‘name’] for i in json_dict[‘ancestors’] if i[‘subcategory’][0][‘key’] == ‘province’) # ‘Lam Dong Province’ To construct the condition i[‘subcategory’][0][‘key’], you need only note: Lists are denoted by [] and the only element of a list may be retrieved via [0]. Dictionaries are denoted by {} and … Read more