I think you have mixed up index numbers with keys. Dictionaries are formed like such: {key: value} data.keys() will return a list of keys. In your case: data.keys() [0,1,2] From there, you can call the first item, which is 0 (First item in a list is 0, and then progresses by one). data.keys()[0] 0 If … Read more
Something like this: dictionary = {‘results’: [{‘name’: ‘sarah’, ‘age’: ’18’}]} name = dictionary[‘results’][0][‘names’] + ‘.txt’ open(name, “w”) Opening a file in the write mode, if it doesn’t exist already, will create a new file with that name. 2 solved How to name a file using a value in a list in a dictionary?
You could do this: a = {‘vladimirputin’: {‘milk’: 2.87, ‘parsley’: 1.33, ‘bread’: 0.66}, ‘barakobama’:{‘parsley’: 0.76, ‘sugar’: 1.98, ‘crisps’: 1.09, ‘potatoes’: 2.67, ‘cereal’: 9.21}} b = {} for prez in a: for food in a[prez]: if food not in b: b[food] = {prez: a[prez][food]} else: b[food][prez] = a[prez][food] This gives: {‘bread’: {‘vladimirputin’: 0.66}, ‘cereal’: {‘barakobama’: 9.21}, … Read more
The tricky part is aligning the several table entries and the table headers. For this, we first have to find out how long the longest entry in each column is. pad then can be used to add a number of padding spaces to the given string. fields = [“ident”, “make”, “model”, “disp”, “power”, “luxury”] max_len … Read more
Let’s say that you pass the string XII to your function. That makes len(s) 3, which means that the body of your loop would (if the panic didn’t occur) execute 4 times (when i is 0, 1, 2 and 3 since your condition is i <= len(s). On the iteration where i is 2 (i.e. … Read more
You can use Generics with where TValue : ICloneable constraint: public static Dictionary<TKey, TValue> deepCopyDic<TKey, TValue>(Dictionary<TKey, TValue> src) where TValue : ICloneable { //Copies a dictionary with all of its elements //RETURN: // = Dictionary copy Dictionary<TKey, TValue> dic = new Dictionary<TKey, TValue>(); foreach (var item in src) { dic.Add(item.Key, (TValue)item.Value.Clone()); } return dic; } … Read more
def get_smallest_length(x): return [k for k in x.keys() if len(x.get(k))==min([len(n) for n in x.values()])] def get_largest_sum(x): return [k for k in x.keys() if sum(x.get(k))==max([sum(n) for n in x.values()])] x = {‘a’: [4, 2], ‘c’: [4, 3], ‘b’: [3, 4], ‘e’: [4], ‘d’: [4, 3], ‘g’: [4], ‘f’: [4]} print get_smallest_length(x) print get_largest_sum(x) Returns: [‘e’, ‘g’, … Read more
This will let you loop over your dictionary values: foreach (string key in dictionary.Keys) { string value = dictionary[key]; //Now, do something with value, such as add to database } And this will let you find a specific dictionary value: string key = “a_key”; if (dictionary.ContainsKey(key)) { string value = dictionary[key]; //Now do something with … Read more
Look into -(void)locationManager:(CLLocationManager *)manager didUpdateLocations:(NSArray *)locations; and use this to get your current location: [locations lastObject]; and to get distance use this: distanceFromLocation: 9 solved How to show locations of friends on map near by 20 km from user current location in ios [closed]
Make a dictionary with List<string> as value, and then just add values : foreach(var d in c) { if (!dict.ContainsKey(d.Key)) dict.Add(d.Key, new List<string>()); dict[d.Key].Add(d.Value); } and later you can get comma delimited string from list with string.Join string commaDelimitedList = string.Join(“,”, valueList.ToArray()); solved Dictionary one key many values in code c#
Here’s something you can try. It uses a collections.defaultdict or collections.Counter to count High, Med and Low for each product, then merges the results at the end. from collections import defaultdict, Counter product_issues = [ {“product”: “battery”, “High”: 0, “Med”: 1, “Low”: 0}, {“product”: “battery”, “High”: 1, “Med”: 0, “Low”: 0}, {“product”: “battery”, “High”: 1, … Read more
First of all, just printing the dictionary itself will print everything on one line: >>> dicMyDictionary = {“michael”:”jordan”, “kobe”:”bryant”, “lebron”:”james”} >>> print(dicMyDictionary) {‘kobe’: ‘bryant’, ‘michael’: ‘jordan’, ‘lebron’: ‘james’} If you want to format it in some way: >>> print(‘ ‘.join(str(key)+”:”+str(value) for key,value in dicMyDictionary.iteritems())) kobe:bryant michael:jordan lebron:james ref: str.join, dict.iteritems Or with a for loop … Read more
Try This: def orangecap(match_details): players_data = {} for k, v in match_details.iteritems(): for player_name, score in v.iteritems(): prev_player = player_name if prev_player == player_name: score = players_data.get(player_name, 0) + score players_data[player_name] = score high_score_player = max(players_data, key=lambda i: players_data[i]) print (str(high_score_player), players_data[high_score_player]) 6 solved python program using dictionary
raw = [{‘Name’: ‘Erica’,’Value’:12},{‘Name’:’Sam’,’Value’:8},{‘Name’:’Joe’,’Value’:60}] raw.sort(key=lambda d: d[‘Value’], reverse=True) result = [] for i in range(2): result.append(raw[i][‘Name’]) print(result) # => [‘Joe’, ‘Erica’] print(result) Try this. 2 solved Dictionary Sorting based on lower dictionary value
You can try if let sprites = pokemonDictionary[“sprites”] as? [String:Any] { print(sprites) if let backImgUrl = sprites[“back_default”] as? String { print(backImgUrl) } } Also you should run function named CallUrl in a thread other than the main , as Data(contentsOf:) runs synchronously and blocks the main thread 1 solved Nested Dictionary in my swift application