George Sakkis wrote: > It's not the *most* efficient way because value is looked up twice if > it is contained in the dictionary; if you absolutely need it to be as > efficient as possible and can't figure it out on your own, ask again > and someone will help you out. > How do you *know* it is not the *most* efficient way? Have you tried timing different ways of approaching this problem and found that looking up the value twice is slow?
I've tried timing dictionary lookups in the past, and the three obvious solutions roughly come out as follows: try/except is fastest when the value is in the dictionary, but it is a *lot* slower if the exception gets thrown. If missing values are a very rare occurrence this might be a good way to do it, but usually the code doesn't read as well so its best to avoid. [0.26/4.11] Test with the 'in' operator and then retrieving the value is the fastest solution when the value isn't in the dictionary (it only does a single lookup then), and is still fast when it is. [0.36/0.2] Using the get method of the dictionary with a default value to be retrieved if the key is not present is slower than using the 'in' operator in all cases (it does beat try/except when an exception is thrown) [0.49/0.54] The numbers above are the times produced in each case for a key present/key missing using a simple test with timeit.py. Part of the reason, BTW, that calling d.get(key,default) is slow is that is also requires two dictionary lookups: one to find the get method and then another to access the key in the dictionary, plus it has other overheads (a method call) which test&get avoids. These figures could of course be invalidated if the actual use is too far from the simple string lookup I tried. For example if the key has a slow hash function saving the second lookup would be worthwhile. -- http://mail.python.org/mailman/listinfo/python-list