PyCkBk-1-5 - Woodpecker Wiki for CPUG

1.5 Adding an Entry to a Dictionary

增加一个入口到字典里

Credit: Alex Martelli

1.5.1 Problem

问题

Working 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. 用一个字典D工作，如果D[k]存在，使用它。如果不存在增加一个新的D[k]

1.5.2 Solution

解决

This 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: 这个是字典对象的setdefault方法所做的。假定，我们构造一个词到页（word-to-page）的数字索引。关键的代码片断是这样的：

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): === 良好的python本能建议用“轻松获得许可”（easier to get permission）模式来替换掉“三思而后行”（look before you leap）模式（这些过程的详细讨论请看Recipe 5.4）

【译注：“轻松获得许可”，即我们使用异常。这样就不用去检查变量。“三思而后行”就是说，每次我们都需要检查变量然后在根据检查的结果来做不同的处理】

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: 这仅仅是一个次要的简化。但是它满足了“如果入口存在使用它，否则，增加新的入口”的模式。这里是怎么样使用setdefault来更进一步简化 def addword(word, pagenumber): theIndex.setdefault(word, []).append(pagenumber)

1.5.3 Discussion

讨论

The 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. 对于这个任务，字典的Setdefault方法是一个便利的捷径。尤其对你想增加的新入口是可变的这种情况非常有用。如果k不是字典里的键值，setdefault方法有dict[k]=v的作用，并且返回一个v（get方法只返回v，不以任何方式影响字典）。除此之外，基本上，dict.setdefault(k, v)更象dict.get(k, v)。

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: 当一个字典的值是list，setdefault是非常有用的，细节在Recipe 1.6。大多数典型的使用形式是： 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: 注意，当值是不可变的时候，setdefault不是非常有用。如果你仅仅想计算单词，例如，象下面这些是不能使用的：

theIndex.setdefault(word, 1)

In this case, you want: 在这种情况下，你想：

theIndex[word] = 1 + theIndex.get(word, 0)

since you will be rebinding the dictionary entry at theIndex[word] anyway (because numbers are immutable). 因为你无论如何将重新绑定字典入口在theIndex[word]上。（因为数字是不可变的）

1.5.4 See Also

参考 Recipe 5.4; the Library Reference section on mapping types.