Considering that your map values are maps themselves, copying those inner maps will be quite expensive. It’s probably slightly cheaper to use pointers to the inner maps than to use iterator, but both are significantly cheaper than copying the inner maps. However, whether you use pointers or iterators, make sure that they’re not invalidated. The … Read more
You can use a class like this one: import java.util.*; class LegacyGlueifier { private LegacyGlueifier() { } public static String generateLegacyDataset(Map<String, String> data) { final ArrayList<ArrayList<String>> lists = new ArrayList<ArrayList<String>>(); final int width = data.size(); int i = 0; for (Map.Entry<String, String> entry : data.entrySet()) { String[] values = entry.getValue().split(“,”); changeDims(lists, width, values.length + 1); … Read more
The first rule of debugging is to assume that the error message is telling you the truth. In this case it is telling you keyerror u’True. This means that somewhere in your code you’re doing the equivalent of some_dictionary[u’True’]. In the code you posted, there is nothing that is using the literal value True as … Read more
You want to traverse the dictionary, building the current key and accumulating the flat dictionary. For example: def flatten(current, key, result): if isinstance(current, dict): for k in current: new_key = “{0}.{1}”.format(key, k) if len(key) > 0 else k flatten(current[k], new_key, result) else: result[key] = current return result result = flatten(my_dict, ”, {}) Using it: print(flatten(_dict1, … Read more
Try this code to avoid duplication I hope “id” value will be unique in your dictionary. var mydictionary = [“id”: “1”, “quantity”: “”,”sellingPrice”:””] as [String : Any] var arrayOfDictionary = [Dictionary<String, Any>]() //declare this globally let arrValue = arrayOfDictionary.filter{ (($0[“id”]!) as! String).range(of: mydictionary[“id”]! as! String, options: [.diacriticInsensitive, .caseInsensitive]) != nil } if arrValue.count == 0 … Read more
my_dict = {key1:{value1:value2}, key2:{value3:value4}, key3{value5:value6}, key4:{value7:value8}…} result = [] for key1 in my_dict: if sum(my_dict[key1].keys() + my_dict[key1].values())==3: result[key1] = my_dict[key1].keys()[0] Hope this helps solved Python – selectively add up values while iterating through dictionary of dictionaries [closed]
First, you start your question with expenses but ends with incomes and the code also has incomes in the parameters so I’ll go with the income. Second, the error says the answer. “expense_types(income_types)” is a list and “expenses(incomes)” is a dictionary. You’re trying to find a list (not hashable) in a dictionary. So to make … Read more
I’ve added comments to the answer to explain the flow, please add more details in your question if this is not clear enough 🙂 # Open file for reading file_house_price = open(“house_price.txt”, “r”) # read all lines from the file into a variable data_house_price = file_house_price.readlines() # as we’re done with the file, we can … Read more
This will add a numbered (name, job) pair for each name: dict = {} i = 0 # or 1, if you prefer. but it’s not proper to start a list from 1 when programming. while i < len(names): dict[i] = (names[i], Jobs[i]) i = i+1 as a for loop: for i in id_: dict[i] … Read more
This should give the basic idea – Use the item you want to count as the key, and the count itself as the value. If it is already in the Dictionary, increment it. If not, add it: Const TEST_DATA = “Apple,Orange,Banana,Banana,Orange,Pear” Sub Example() Dim counter As New Scripting.Dictionary Dim testData() As String testData = Split(TEST_DATA, … Read more
How can I extract a string from a text field that matches with a “value” in my Dictionary and return the dict’s key into a new column in the Output? [closed] solved How can I extract a string from a text field that matches with a “value” in my Dictionary and return the dict’s key … Read more
If your object isn’t a dict but could be converted to a dict easily (e.g. list of pairs), you could use: def make_a_dict(some_object): if isinstance(some_object, dict): return some_object else: try: return dict(some_object) except: return {‘return’: some_object} print(make_a_dict({‘a’: ‘b’})) # {‘a’: ‘b’} print(make_a_dict([(1,2),(3,4)])) # {1: 2, 3: 4} print(make_a_dict(2)) # {‘return’: 2} solved how to make … Read more
The only thing missing is initialization of the property – it is null otherwise (meaning that calling Add on it will throw a NullReferenceException). DatabaseLogger.ILogger logger = new DatabaseLogger.Logger(); logger.customProperties = new Dictionary<string, string>(); logger.customProperties.Add(“companyName”, “Company”); logger.customProperties.Add(“application”, “application”); Though this might be better done in the Logger constructor: public Logger() { customProperties = new Dictionary<string, … Read more
ids = [int(row[‘id’]) for row in b[‘cricket’][‘Arts & Entertainment’]] will give you a list of the numbers, [6003760631113, 6003350271605, 6003106646578, 6003252371486, 6003370637135] And for your new edits; ids = [int(row[‘id’]) for row in b[‘cricket’][‘Arts & Entertainment’] + b[‘News, Media & Publications’]] 3 solved Parse a nested dictionary [closed]
You can slightly improve your code by removing queue, you don’t need this. The rest of the code looks algorithmically optimal. #include <map> #include <iostream> using namespace std; int main() { int a,b; map<int, int> AB; map<int, int>::iterator it; std::ios::sync_with_stdio(false); while (1) { cin >> a; if (a == -1)break; cin >> b; AB.insert(pair<int, int>(a, … Read more