[Solved] c++ can not get value from map

Your void Write(const std::string& inString) function of OutputMemoryStream should not store additional byte of buffer for null terminator because std::string will not contain null terminator but if you use c_str(), a null terminator will be included in the return from this method. Don’t get confused with the internal structure of the memory. std::string stores the … Read more

[Solved] Feching user’s current location tips [closed]

So you really have two options. The first, and most likely to be accurate, is to ask the user through their browser via javascript, what their location is. HTML5 allows you to do this through geolocation: https://developer.mozilla.org/en-US/docs/Web/API/Geolocation/Using_geolocation The basics are as follows: navigator.geolocation.getCurrentPosition(function(position) { do_something(position.coords.latitude, position.coords.longitude); }); Obviously, this requires the user to agree to … Read more

[Solved] Convert string to Ordered Dictionary [closed]

You can try something like this : input_st = “Field1:’abc’,Field2:’HH:MM:SS’,Field3:’d,b,c'” output_st = {item.split(“:”)[0]:”:”.join(item.split(“:”)[1:]).replace(“‘”, “”) for item in input_st.split(“‘,”)} outputs : {‘Field1’: ‘abc’, ‘Field2’: ‘HH:MM:SS’, ‘Field3’: ‘d,b,c’} Kind of ugly but it does the job. 4 solved Convert string to Ordered Dictionary [closed]

[Solved] Is there a faster solution to solve this kind of job? [closed]

I’m not sure I understand your “job description”, but I think you want this: def find_matches(tuple_of_dicts, key_to_find): return [d for d in tuple_of_dicts if key_to_find in d] So: >>> tuple_of_dicts = ({18: None}, {10: None}, {16: None, 18: None, 5: None, 6: None, 7: None, 10: None}, {16: None}, {7: None}, {10: None}, {18: None}, … Read more

[Solved] pymongo result to String

Gives this: {u’_id’: ObjectId(‘5238273074f8edc6a20c48fe’), u’Command’: u’ABCDEF’, u’processed’: False} But really, all I want is ABCDEF in a string. What you get is a dictionary, you simply need to fetch what you want from it print myDict[ u’Command’] 0 solved pymongo result to String

[Solved] Python: Sort dictionary by value (equal values)

dictionaries are unordered (well pre 3.6 at least), instead sort the items d = {3: ‘__init__’, 5: ‘other’, 7: ‘hey ‘, 11: ‘hey’} print(sorted(d.items(),key=lambda item:(item[0].strip(),item[1]))) # output => [(3, ‘__init__’), (7, ‘hey ‘), (11, ‘hey’), (5, ‘other’)] if you really want it as a dict (for reasons i cant fathom) and you are using 3.6+ … Read more

[Solved] How to retrieve an element inside a Map in Java?

Your ENUM values on the map seem like another map. Therefore I put it like this Map<Enum, Map<String, Object>> However, question is not clear! Here is one possible solution. import java.util.*; public class Main { public static void main(String[] args) { Map<Enum, Map<String, Object>> map = new HashMap<>(); Map<String, Object> value = new HashMap<>(); value.put(“String1”, … Read more

[Solved] How to count elements in a list and return a dict? [duplicate]

Using no external modules and only the count, although there would be some redundancy using dict comprehension: d = {itm:L.count(itm) for itm in set(L)} If you are allowed to use external modules and do not need to implement everything on your own, you could use Python’s defaultdict which is delivered through the collections module: #!/usr/bin/env … Read more

[Solved] Are Swift dictionaries automatically sorted?

Swift’s Dictionary is a hash-based data structure. Unless a specific ordering mechanism is in place, the order of items in hash-based structures depends on several factors: Hash values of objects used as keys – hashValue method is used to determine the bucket number for the item Size of the structure – Since hashValue can be … Read more

[Solved] how to add just the values in key s1 [closed]

If you want to add all the values of s1, you need to iterate over the values() of dictionary students. students={ 1: {“pup1”: “001”, “s1”: 10, “s2”: 20}, 2: {“pup2”: “124”, “s1”: 20, “s2”: 30}, 3: {“pup3”: “125”, “s1”: 30, “s2”: 40}} sum1 = 0 for data in students.values(): sum1 += data[‘s1’] avg = sum1 … Read more