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
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
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]
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
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
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
map does a delayed execution, so when you cast the result to a list, the map is executed, which amounts in computing sum(1), sum(2), sum(-3) and sum(4) Any of those will fail because sum takes something like a list (an iterable object) in input. You pass integers which are not iterable, hence the error message. … Read more
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
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
Assuming you have an input that looks like this:dic={x:[a,b],y:[c]}. Try this, res=sorted([(x,len(dic[x])) for x in dic],key=lambda x:x[0]) solved How to compare number of values in a dictionary [closed]
To access a value from dictionary you need to simply use dictionary[key], so in your function you just need to return this value. solved Accessing a Dictionary [closed]
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
try dic.Values.OrderBy(i=>i.CreatedDateTime) 0 solved How to sort dictionary by DateTime attribute of the value
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
According your question and your comment, I assume you have no problem with the text and input. So the point is how to split up mark and grade? You can use zip() to do this # This is an example for Josh, assuming you already have list call lines, # lines = [(mark1, grad1), (mark2, … Read more