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copied!<p>One good place to start is by getting iPython (<code>easy_install ipython</code>) then poking around with tab completion, <code>?</code> and <code>dir</code>:</p> <pre><code>In [2]: dir {} ------> dir({}) Out[2]: ['__class__', ... 'keys', 'pop', 'popitem', 'setdefault', 'update', 'values'] In [3]: {}.update? Type: dict Base Class: <type 'dict'> String Form: {} Namespace: Interactive Length: 0 Docstring: dict() -> new empty dictionary. dict(mapping) -> new dictionary initialized from a mapping object's (key, value) pairs. dict(seq) -> new dictionary initialized as if via: d = {} for k, v in seq: d[k] = v dict(**kwargs) -> new dictionary initialized with the name=value pairs in the keyword argument list. For example: dict(one=1, two=2) </code></pre> <p>(just for example)</p> <p>Anyway, your problem with checking keys common between the two dictionaries: there are a few things to consider (maybe look at the <code>set</code> class?), but here's how I'd do it:</p> <pre><code>common_keys = [k for k in dict1 if k in dict2] </code></pre> <p>(ie, "each key <code>k</code> in dict1 if that key is also in dict2") (note, too, that testing for dictionary membership is an O(1) operation, so this will run in O(|dict1|))</p> <p><strong>edit</strong>: alright, so this doesn't solve the problem of merging the two dicts into one with lists... But Lott's answer is good for that, or you could use the <code>setdefault</code> method:</p> <pre><code>new = {} for (k, v) in dict1.items(): new.setdefault(k, []).append(v) for (k, v) in dict2.items(): new.setdefault(k, []).append(v) </code></pre>

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