A dictionary is a mapping from a key to a value. You need to provide a value, at the very least something that signifies ‘no value’: tag_dic[i] = None 1 solved How to add items to dictionary in a for loop in Python?
Because your first dict’s key is number. Not ‘nom’. for k in test: print test[k][‘nom’] solved parser dict in python [closed]
As @Hitobat mentioned in commend – you have list with dictionary inside so you have to use [0] to get this dictionary. Or you have to use for-loop if you have more elements on list data = [{‘_id’: ‘5f563c1bf8eaa9d98eca231f’, ‘allEnabledDs’: None, ‘allEnabledIdSor’: None, ‘correlationFilterEntitySource’: True, ‘created_at’: ‘2020-09-07T13:56:43.469Z’, ‘dsConnectionList’: None, ‘folderToLabelMapping’: None, ‘idConnectionList’: None, ‘identityResolutionScan’: False, … Read more
Is Regex required? Here’s linq. edit: added a regex option as well since that was requested. ([a-z])([A-Z])”, “$1 $2”) matches lowercase letter and then uppercase letter and returns the matches as $1 and $2 respectively w/ a space in between. Note: this won’t work if you have accented characters like É, is that needed in … Read more
You can use: maketrans–which allows creation of a translation table translate–applies translation table from maketrans to a string Code def encrypt(infile, outfile, translation): ”’ Encrypts by applying translation map to characters in file infile. ”’ # Create translation table table=””.maketrans(translation) # Apply table to all characters in input file # and write to new output … Read more
You could sort the list of dictionaries by the price reversed and then use slicing notation to return the top results. from operator import itemgetter top = 3 data = [{“name”: “bread”, “price”: 100}, {“name”: “wine”, “price”: 138}, {“name”: “meat”, “price”: 15}, {“name”: “water”, “price”: 1}, {“name”: “fish”, “price”: 10}] print(sorted(data, key=itemgetter(‘price’), reverse=True)[:top]) Output: [{‘name’: … Read more
i was able to fix this by setting a new object based on the current key def get_player_key(search_name, requestor): players = get_all_players() found_players = [] for player_id in players: player = players[player_id] if player[‘search_full_name’] == search_name: #Blah Blah Blah solved Iterating through json object with python help needed [closed]
I may have been massively over complicating things here, this appears to do what I want (ignoring the insertion of completely new markets / selection_id’s for now): new_order_list = [] for x in updated_orders: for y in x[‘Live_orders’]: orders_to_update = {} orders_to_update.update({‘Market_id’: x[‘Market_id’], ‘Selection_id’: y[‘selection_id’], ‘live_orders’: y[‘live_orders’]}) new_order_list.append(orders_to_update) for z in new_order_list: for a in … Read more
What you want is a scoped namespace like myVariables to be used as a dictionary, e.g. // this prevents keys like `hasOwnProperty` from being pre-defined on namespace var myVariables = Object.create(null); function test(name) { myVariables[‘a’ + name] = ‘test text’; } test(‘test’); console.log(myVariables.atest); solved How to set specific variable name?
Solution while True: prompt = int(raw_input(” Enter ID of person you would like to search for: “)) if prompt > 100: print(“Please try again”) continue elif prompt < 1: displayPerson(prompt,persons) else: break print(“Terminated.”) Explanation You enter the loop and get an ID from the user. If prompt > 0: display the person Otherwise, break out … Read more
def deleteVal (my_dict, key_val, val): if len(my_dict.get(key_val, “1”))>1: try: my_dict[key_val].remove(val) except ValueError: pass else: my_dict.pop(key_val, None) 9 solved Removing a dictionary key value having multiple value [closed]
SKNearbySearchSettings* searchObject = [SKNearbySearchSettings nearbySearchSettings]; searchObject.coordinate = CLLocationCoordinate2DMake(currentLocation.coordinate.latitude, currentLocation.coordinate.longitude); searchObject.searchTerm=@””; //search for all POIs searchObject.radius = 5000; searchObject.searchMode=SKSearchHybrid; searchObject.searchResultSortType=SKProximitySort; searchObject.searchCategories = @[ @(SKPOICategoryHospital)]; [[SKSearchService sharedInstance] setSearchResultsNumber:10000]; [[SKSearchService sharedInstance]startNearbySearchWithSettings:searchObject]; Use this code for finding nearby hospitals. 0 solved iOS: How to show nearby hospitals in skobbler maps?
Your code seams ok but you have to protect your dictionary from external access. I think the root cause of your problem is this. Change your code private readonly Dictionary<string, double> dict = new Dictionary<string, double>(); Also modify the getValue() method like this: public override Object getValue() { return dict.ToDictionary(m => m.Key, m => m.Value); … Read more
in python 3.x you can use: for k,v in std.items(): print(k,v[‘Uid’]) in python 2.x: for k,v in std.iteritems(): print(k,v[‘Uid’]) python 3.x code: std={} n=int(input(“How many Entries you have to fill:”)) for i in range (n): name=input(“Enter name:”) uid=input(“ID:”) age=input(“Age:”) print(“—-Marks—-“) math=input(“Maths:”) phy=input(“Physics:”) chem=input(“Chemistry:”) std[name]={‘Uid’:uid,’Age’:age,’subject’:{‘math’:math,’phy’:phy,’chem’:chem}} print(‘\n’) for k,v in std.items(): print(k,v[‘Uid’]) input and results for 2 … Read more
import json input=””‘ABCD=123=Durham EFGH=456=Nottingham IJKL=789=Peterborough”’ print(json.dumps({‘data’: [dict(zip((‘name’, ‘id’, ‘place’), l.split(‘=’))) for l in input.split(‘\n’)]})) This outputs: {“data”: [{“name”: “ABCD”, “id”: “123”, “place”: “Durham”}, {“name”: “EFGH”, “id”: “456”, “place”: “Nottingham”}, {“name”: “IJKL”, “id”: “789”, “place”: “Peterborough”}]} 2 solved how to format the text-file data to a specific standard JSON [closed]