Note: This answer was written before the implementation of the dict type changed, in Python 3.6. Most of the implementation details in this answer still apply, but the listing order of keys in dictionaries is no longer determined by hash values. The set implementation remains unchanged. The order is not arbitrary, but depends on the … Read more
Lets say your data is – data = {‘a’: [ [1, 2] ], ‘b’: [ [3, 4] ],’c’:[ [5,6]] } You can use the data.items() method to get the dictionary elements. Note, in django templates we do NOT put (). Also some users mentioned values[0] does not work, if that is the case then try … Read more
You could use row_factory, as in the example in the docs: import sqlite3 def dict_factory(cursor, row): d = {} for idx, col in enumerate(cursor.description): d[col[0]] = row[idx] return d con = sqlite3.connect(“:memory:”) con.row_factory = dict_factory cur = con.cursor() cur.execute(“select 1 as a”) print cur.fetchone()[“a”] or follow the advice that’s given right after this example in … Read more
As said by Niklas, you need recursion, i.e. you want to define a function to print your dict, and if the value is a dict, you want to call your print function using this new dict. Something like : def myprint(d): for k, v in d.items(): if isinstance(v, dict): myprint(v) else: print(“{0} : {1}”.format(k, v))
If lakes is your DataFrame, you can do something like area_dict = dict(zip(lakes.id, lakes.value))
A general answer involves using a Manager object. Adapted from the docs: from multiprocessing import Process, Manager def f(d): d[1] += ‘1’ d[‘2’] += 2 if __name__ == ‘__main__’: manager = Manager() d = manager.dict() d[1] = ‘1’ d[‘2’] = 2 p1 = Process(target=f, args=(d,)) p2 = Process(target=f, args=(d,)) p1.start() p2.start() p1.join() p2.join() print d … Read more
As @Falanwe points out in a comment, doing something like this is incorrect: int LastCount = mydict.Keys.ElementAt(mydict.Count -1); You should not depend on the order of keys in a Dictionary. If you need ordering, you should use an OrderedDictionary, as suggested in this answer. The other answers on this page are interesting as well.
Don’t use an array if you want named keys, use a plain object. var a = {}; a[“key1”] = “value1”; a[“key2”] = “value2”; Then: if (“key1” in a) { // something } else { // something else }
Notice that your error message only contains five K, V pairs, 10 arguments total. This is by design; the ImmutableMap class provides six different of() methods, accepting between zero and five key-value pairings. There is not an of(…) overload accepting a varags parameter because K and V can be different types. You want an ImmutableMap.Builder: … Read more
If I can rephrase your question, what you want is a dictionary with the years as keys and an array for each year containing a list of values associated with that year, right? Here’s how I’d do it: years_dict = dict() for line in list: if line[0] in years_dict: # append the new number to … Read more