This answer mentions Json.NET but stops short of telling you how you can use Json.NET to serialize a dictionary: return JsonConvert.SerializeObject( myDictionary ); As opposed to JavaScriptSerializer, myDictionary does not have to be a dictionary of type <string, string> for JsonConvert to work.
xmltodict (full disclosure: I wrote it) does exactly that: xmltodict.parse(“”” <?xml version=”1.0″ ?> <person> <name>john</name> <age>20</age> </person>”””) # {u’person’: {u’age’: u’20’, u’name’: u’john’}}
The semantics of LinkedHashMap are still those of a Map, rather than that of a LinkedList. It retains insertion order, yes, but that’s an implementation detail, rather than an aspect of its interface. The quickest way to get the “first” entry is still entrySet().iterator().next(). Getting the “last” entry is possible, but will entail iterating over … Read more
There’s no such thing a the “first n” keys because a dict doesn’t remember which keys were inserted first. You can get any n key-value pairs though: n_items = take(n, d.iteritems()) This uses the implementation of take from the itertools recipes: from itertools import islice def take(n, iterable): “Return first n items of the iterable … Read more
If its an OrderedDict() you can easily access the elements by indexing by getting the tuples of (key,value) pairs as follows >>> import collections >>> d = collections.OrderedDict() >>> d[‘foo’] = ‘python’ >>> d[‘bar’] = ‘spam’ >>> d.items() [(‘foo’, ‘python’), (‘bar’, ‘spam’)] >>> d.items()[0] (‘foo’, ‘python’) >>> d.items()[1] (‘bar’, ‘spam’) Note for Python 3.X dict.items … Read more
Just use the indexer – it will overwrite if it’s already there, but it doesn’t have to be there first: Dictionary<string, object> currentViews = new Dictionary<string, object>(); currentViews[“Customers”] = “view1”; currentViews[“Customers”] = “view2”; currentViews[“Employees”] = “view1”; currentViews[“Reports”] = “view1”; Basically use Add if the existence of the key indicates a bug (so you want it … Read more
You can use set operations on the keys: diff = set(dictb.keys()) – set(dicta.keys()) Here is a class to find all the possibilities: what was added, what was removed, which key-value pairs are the same, and which key-value pairs are changed. class DictDiffer(object): “”” Calculate the difference between two dictionaries as: (1) items added (2) items … Read more
This data is JSON! You can deserialize it using the built-in json module if you’re on Python 2.6+, otherwise you can use the excellent third-party simplejson module. import json # or `import simplejson as json` if on Python < 2.6 json_string = u'{ “id”:”123456789″, … }’ obj = json.loads(json_string) # obj now contains a dict … Read more
Try: Iterator<Map.Entry<String,String>> iter = testMap.entrySet().iterator(); while (iter.hasNext()) { Map.Entry<String,String> entry = iter.next(); if(“Sample”.equalsIgnoreCase(entry.getValue())){ iter.remove(); } } With Java 1.8 and onwards you can do the above in just one line: testMap.entrySet().removeIf(entry -> “Sample”.equalsIgnoreCase(entry.getValue()));
You do not need to call d.keys(), so if key not in d: d[key] = value is enough. There is no clearer, more readable method. You could update again with dict.get(), which would return an existing value if the key is already present: d[key] = d.get(key, value) but I strongly recommend against this; this is … Read more