It's simple: you're calling. https://docs.python.org/3/library/stdtypes.html#dict.setdefault and indicate the key first, and the second value, returned and installed in the dictionary, if there is no such key:>>> d = dict() >>> d.setdefault('Вася', "Пупкин") 'Пупкин' >>> d["Вася"] 'Пупкин' >>> d["12"] Traceback (most recent call last): File "<stdin>", line 1, in <module> KeyError: '12' Another option would be https://docs.python.org/3/library/collections.html#collections.defaultdict ♪ It enables all those who do not exist in the vocabulary to give meaning to default:>>> from collections import defaultdict >>> d = defaultdict(int) >>> d['12'] 0 >>> d = defaultdict(str) >>> d['12'] '' >>> d = defaultdict(list) >>> d['12'] [] >>> d = defaultdict(lambda x=None: "Вася") >>> d['12'] 'Вася' If you take a simple dictionary, it's an analog for him. defaultdict will:>>> d = dict() >>> if '12' not in d: >>> d['12'] = int() >>> d['12'] 0

In the cycle, you're changing the line. 'словарь'on each iteration receives a symbol in the meaning K and the key to the dictionary

We need to do this:print(', '.join(a.keys())) Helped?

Let's turn around.{l:dict(zip(range(1,5), tariff[i:i+5])) for i, l in zip(range(0, len(tariff)+1, 4), Branch)}

We can use it. zip_longest, but you need to specify the parameter fillvalue (thanks for the SergFSM reminder) so that you do not have a breakdown of the parameters:from itertools import zip_longest different_a = {'a': 1, 'b': 2, 'c': 3} different_b = {'a': 4, 'b': 5} for (key1, value1), (key2, value2) in zip_longest(different_a.items(), different_b.items(), fillvalue=(None, None)): print("{0}-{1}\t\t\t\t{2}-{3}" .format(key1, value1, key2, value2)) Conclusion:a-1 a-4 b-2 b-5 c-3 None-None

There's no need for this task, just a list of n+1. Operation + is wrong, we need to move all the yellows. Modified i is used negligently.n, k = map(int, input().split(" ")) L = [1] * (n+1) for i in range(k): operation, a, b = map(str, input().split(' ')) a = int(a) b = int(b) if operation == '+': L[a] += L[b] L[b] = 0 elif operation == '*': L[a], L[b] = L[b], L[a] elif operation == '/': lob = L[b]//2 L[a] += L[b] - lob L[b] = lob mx = max(L) for i, x in enumerate(L): if x == mx: print(i, end = " ")

I'd rather not remove it from the list, but reassemble the list:a = [{"word": "c", "count": 4}, {"word": "dog", "count": 2}, {"word": "_restore", "count": 1}, {"word": "-f", "count": 5}] a = [obj for obj in a if len(obj['word']) != 1 and obj['word'][0] not in ('_', '-')] print(a) Besides, you've got it written on condition. And (sighs)в которых значение длины ключа "word" равно 1 и первый символ равен одному из этих знаков ['-', '']in code ORI gave an example of an option. ORBut since I don't remove, I collect a list with the right elements, that's the condition:size not one and first symbol not '-' or ''

You can decide in the forehead:dict_1 = [{'id': 1, 'value': 11}, {'id': 2, 'value': 12}] dict_2 = [{'id': 1, 'anotherValue': 13}, {'id': 2, 'anotherValue': 14}] tmp = {} for obj in [dict_1, dict_2]: for elem in obj: if elem['id'] not in tmp: tmp[elem['id']] = elem else: tmp[elem['id']].update(elem) res = [value for value in tmp.values()] print(res)

animals = ['snail', 'fish', 'cow'] result = [{"id": n, "animal": animal} for n, animal in enumerate(animals)] # [{'id': 0, 'animal': 'snail'}, {'id': 1, 'animal': 'fish'}, {'id': 2, 'animal': 'cow'}]

Let's go on the list. a♪ For each part of this list (Speaker) d♪ Take the key d["color"]Break it up. split()We'll take the second element. [1] And we'll check that he's not on the list. b♪ If that's the case, the vocabulary is placed in the results list. c♪c = [d for d in a if d["color"].split()[1] not in b]

Let's figure out what's going on.Here you compare the dictionaries completely.if text == text2: print(True) i.e. and keys and values, and if two vocabularies are presented at the inlet (sing off, no values)text = {'test': 1, 'test2': 2} text2 = {'test': 1, 'test2': 3} body ifIt's not gonna work, it's gonna go further. else♪In this part of the code:else: c = set(text.keys()) ^ set(text2.keys()) print(c) You only work with the keys and the same input data. text text2), variable c will contain empty set()which, in principle, you do.If you're looking for the difference only with the keys, you'll only work with the keys, providedif text.keys() == text2.keys(): print(True) else: c = set(text.keys()) ^ set(text2.keys()) print(c) Subject to this conclusion True♪If you're looking for a difference values under the same keys, c = set(text.keys()) ^ set(text2.keys()) Not good.

Change method get_coverthat he didn't return the dictionary, but only the key to that dictionary. def get_cover(self, obj): return GalleryImagesSerializer(obj.photos.first()).data.get('photo')

Just change the exterior (or internal) skirts, you don't have a code now.from ast import literal_eval dic = '{"pos": 7, 4: 5, "go": 2, "tpjc": None}' dic = literal_eval(dic) print(type(dic), dic) Conclusion:<class 'dict'> {'pos': 7, 4: 5, 'go': 2, 'tpjc': None} Through json It's usually better, but in this case the format is not appropriate for him.

Well, the algorithm is simple.You break the text into words split()And then you check whether this word is in the dictionary as a key and if not add a word with a value of 0.You add one key to your word.It's the best thing to do:words = dict() ... words[word] = word.get(word, 0) + 1 word.get(word, 0) tries to extract the key. word And if there's no such key in the dictionary, it's back 0.

I put it in the id line just to keep the view with zero at the beginning.animals = ['lion', 'dog', 'dog'] id = ['0042','0043','0003'] собрать словарь res = dict() for obj in zip(animals, id): res[obj[0]] = res.get(obj[0], []) # если в словаре отсутствует ключ - создать запись с пустым списком res[obj[0]].append(obj[1]) # добавить в список новый элемент отсортировать значения словаря for key in res: res[key].sort() print(res)

In English SO, there's interesting https://stackoverflow.com/a/55355928/8324991 In your case:import pandas as pd df = pd.DataFrame({'all_rel':[ {'Иван': 0.358, 'Михаил':0.25, 'Илья': 0.456}, {'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426}, {'Иван': 0.568, 'Михаил':0.145, 'Илья': 0.464}, {'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426}, ]}) pd.json_normalize(df['all_rel']) Conclusion: Иван Михаил Илья 0 0.358 0.250 0.456 1 0.698 0.125 0.426 2 0.568 0.145 0.464 3 0.698 0.125 0.426 There are other interesting options on the reference, but it's kind of fast and simple.The conversion data with the main date may be combined by index:df = df.drop(columns='all_rel') .join(pd.json_normalize(df['all_rel'])) Or so, just as the respected MaxU says:df = df.join(pd.json_normalize(df.pop('all_rel')))

in cycle instead values() Use it. items()which contains information on the key and the meaning, not just the meaningglobal_dict = {'B': {'old': 60691.96, 'now': 61928.03, 'change': 2.04}, 'E': {'old': 4418.98, 'now': 4605.43, 'change': 4.22}, 'D': {'old': 188.5, 'now': 192.49, 'change': 2.12}} max_percent = 4.22 for xer in global_dict.items(): if xer[1]['change'] == max_percent: print(xer) break

The dictionary itself is not sorted because it depends on how the key is processed.But you can put the dictionary on the list and quarry the list.my_list = sorted(list(my_dict.items()))

It's not very clear that you've got your input data. And why don't you make one request to the OBD immediately of both fields and create a dictionary at once?But in general, if you need to put two lists in a dictionary, it's like,user_xp = [1,2,3] users = ["Вася", "Петя", "Маша"] top_users = dict(zip(users, user_xp)) {'Вася': 1, 'Маша': 3, 'Петя': 2} And what you're writing now is you're all top_users[user] Appropriation of the last part of the list user_xp♪

It is possible to use a recurring function:def set_value(d): if not isinstance(d, dict): return d if d.get("value"): return d r = [x.get("value") for x in b_dict["values"] if x.get("id") == d.get("id")] if r and r[0]: d["value"] = r[0] return d def apply_recursive(func, obj): if isinstance(obj, list): return [apply_recursive(func, elem) for elem in obj] elif isinstance(obj, dict): obj = func(obj) return {k: apply_recursive(func, v) for k, v in obj.items()} else: return func(obj) challenge:_ = apply_recursive(set_value, a_dict) result:In [106]: a_dict Out[106]: {'tests': [{'id': 1, 'title': 'S', 'value': 'passed'}, {'id': 2, 'title': 'D', 'value': 'failed'}, {'id': 3, 'title': 'P', 'value': 'failed', 'values': [{'id': 31, 'title': 'M', 'value': 'passed', 'values': [{'id': 311, 'title': 'P', 'values': [{'id': 3111, 'title': 'f', 'value': 'passed'}, {'id': 3112, 'title': 'f', 'value': ''}], 'value': 'passed'}]}, {'id': 32, 'title': 'S', 'value': '', 'values': [{'id': 321, 'title': 'P', 'values': [{'id': 3211, 'title': 's', 'value': 'passed'}, {'id': 3212, 'title': 'h', 'value': 'passed'}], 'value': 'failed'}]}]}, {'id': 4, 'title': 'S', 'value': 'passed', 'values': [{'id': 41, 'title': 'C', 'value': 'failed'}, {'id': 42, 'title': 'I', 'value': 'passed'}]}]}