Arrays in JavaScript don’t use strings as keys. You will probably find that the value is there, but the key is an integer. If you make Dict into an object, this will work: var dict = {}; var addPair = function (myKey, myValue) { dict[myKey] = myValue; }; var giveValue = function (myKey) { return … Read more
A default constructed std::string ins inserted into the std::map with key ‘b’ and a reference to that is returned. It is often useful to consult the documentation, which defines the behavior of operator[] as: Returns a reference to the object that is associated with a particular key. If the map does not already contain such … Read more
No, you can’t do that. For your usecase, use a tuple instead: key = (2013, 1) Since you don’t need to do date manipulations on the value a tuple more than suffices.
According to PEP 589 you can use alternative syntax to create a TypedDict as follows: Movie = TypedDict(‘Movie’, {‘name’: str, ‘year’: int}) So, in your case, you could write: from typing import TypedDict RandomAlphabet = TypedDict(‘RandomAlphabet’, {‘A(2)’: str}) or for the second example: RandomAlphabet = TypedDict(‘RandomAlphabet’, {‘return’: str}) PEP 589 warns, though: This syntax doesn’t … Read more
SVG would probably be easier to implement. I would look in to Raphael JS Here is a relevant example of what raphael JS can do and how little code it takes SVG world map definitions can be found numerous places, including Wikipedia, they shouldn’t be too hard to search for.
You can use this: for key in sorted(D.iterkeys()): .. code .. In Python 3.x, use D.keys() (which is the same as D.iterkeys() in Python 2.x).
In your example the map key is a pointer to the struct, not the struct itself. Pointers can be compared for equality even when the items they point to can’t be compared. This comparison is not based on the contents of the item, but only on its memory address.
It sounds like you probably want something like: Dictionary<string, Func<string[], int>> functions = …; This is assuming the function returns an int (you haven’t specified). So you’d call it like this: int result = functions[name](parameters); Or to validate the name: Func<string[], int> function; if (functions.TryGetValue(name, out function)) { int result = function(parameters); … } else … Read more
Update: check for existence using TryGetValue to do only one lookup in the case where you have the list: List<int> list; if (!dictionary.TryGetValue(“foo”, out list)) { list = new List<int>(); dictionary.Add(“foo”, list); } list.Add(2); Original: Check for existence and add once, then key into the dictionary to get the list and add to the list … Read more
This would work: random.choice([k for k in d for x in d[k]])