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1.5 Adding an Entry to a DictionaryCredit: Alex Martelli 1.5.1 ProblemWorking with a dictionary D, you need to use the entry D[k] if it's already present, or add a new D[k] if k isn't yet a key in D. 1.5.2 SolutionThis is what the setdefault method of dictionary objects is for. Say we're building a word-to-page numbers index. A key piece of code might be: theIndex = {} def addword(word, pagenumber): if theIndex.has_key(word): theIndex[word].append(pagenumber) else: theIndex[word] = [pagenumber] Good Pythonic instincts suggest substituting this "look before you leap" pattern with an "easier to get permission" pattern (see Recipe 5.4 for a detailed discussion of these phrases): def addword(word, pagenumber): try: theIndex[word].append(pagenumber) except AttributeError: theIndex[word] = [pagenumber] This is just a minor simplification, but it satisfies the pattern of "use the entry if it is already present; otherwise, add a new entry." Here's how using setdefault simplifies this further: def addword(word, pagenumber): theIndex.setdefault(word, []).append(pagenumber) 1.5.3 DiscussionThe setdefault method of a dictionary is a handy shortcut for this task that is especially useful when the new entry you want to add is mutable. Basically, dict.setdefault(k, v) is much like dict.get(k, v), except that if k is not a key in the dictionary, the setdefault method assigns dict[k]=v as a side effect, in addition to returning v. (get would just return v, without affecting dict in any way.) Therefore, setdefault is appropriate any time you have get-like needs but also want to produce this specific side effect on the dictionary. setdefault is particularly useful in a dictionary with values that are lists, as detailed in Recipe 1.6. The single most typical usage form for setdefault is: somedict.setdefault(somekey, []).append(somevalue) Note that setdefault is normally not very useful if the values are immutable. If you just want to count words, for example, something like the following is no use: theIndex.setdefault(word, 1) In this case, you want: theIndex[word] = 1 + theIndex.get(word, 0) since you will be rebinding the dictionary entry at theIndex[word] anyway (because numbers are immutable). 1.5.4 See AlsoRecipe 5.4; the Library Reference section on mapping types. |
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