Python defaultdict 和 lambda [英] Python defaultdict and lambda

问题描述

在别人的代码中，我读到了以下两行:

In someone else's code I read the following two lines:

x = defaultdict(lambda: 0)
y = defaultdict(lambda: defaultdict(lambda: 0))

由于 defaultdict 的参数是默认工厂，我认为第一行的意思是当我为不存在的键 k 调用 x[k] 时(例如像 v=x[k] 这样的语句)，键值pair (k,0) 将自动添加到字典中，就好像语句 x[k]=0 首先被执行一样.我说得对吗?

As the argument of defaultdict is a default factory, I think the first line means that when I call x[k] for a nonexistent key k (such as a statement like v=x[k]), the key-value pair (k,0) will be automatically added to the dictionary, as if the statement x[k]=0 is first executed. Am I correct?

你呢?似乎默认工厂会创建一个默认值为 0 的 defaultdict.但这具体意味着什么?我试图在 Python shell 中玩弄它，但无法弄清楚它到底是什么.

And what about y? It seems that the default factory will create a defaultdict with default 0. But what does that mean concretely? I tried to play around with it in Python shell, but couldn't figure out what it is exactly.

推荐答案

我认为第一行的意思是当我为不存在的键 k 调用 x[k](例如像 v=x[k])，键值对(k,0)会自动添加到字典中，就好像语句x[k]=0是首先执行.

I think the first line means that when I call x[k] for a nonexistent key k (such as a statement like v=x[k]), the key-value pair (k,0) will be automatically added to the dictionary, as if the statement x[k]=0 is first executed.

没错.这个写得比较地道

That's right. This is more idiomatically written

x = defaultdict(int)

在y的情况下，当你做y["ham"]["spam"]时，关键"ham"如果 y 不存在，则插入.与它相关联的值成为一个 defaultdict，其中 "spam" 自动插入一个值为 0.

In the case of y, when you do y["ham"]["spam"], the key "ham" is inserted in y if it does not exist. The value associated with it becomes a defaultdict in which "spam" is automatically inserted with a value of 0.

即，y 是一种两层"defaultdict.如果 "ham" 不在 y，那么评估 y["ham"]["spam"] 就像在做

I.e., y is a kind of "two-tiered" defaultdict. If "ham" not in y, then evaluating y["ham"]["spam"] is like doing

y["ham"] = {}
y["ham"]["spam"] = 0

就普通dict而言.

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