Python defaultdict和lambda [英] Python defaultdict and lambda

问题描述

在别人的代码中，我读了以下两行：

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，我认为第一行意味着当我为不存在的密钥k（例如像v = x [k]的语句）调用x [k]时，键值对（k，0）将被自动添加到字典，好像首先执行语句x [k] = 0。我是否正确？

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?

那么y呢？看来默认工厂将创建一个默认为0的defaultdict。但是具体是什么意思？

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

我认为第一行意味着当我为 x [k] 调用不存在的键 k 例如 v = x [k] 的语句），键值对（k，0）将自动添加到字典中，如同第一次执行语句 x [k] = 0 。

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。这是更惯用的写法

That's right. This is more idiomatically written

x = defaultdict(int)

在 y 的情况下，当您执行 y [ham] [ spam] 时，ham中的键插入 y 存在。与其关联的值变为 defaultdict ，其中spam自动插入值 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.

Ie， y 是一种 defaultdict 。如果ham不在y ，则评估 y [ham] [spam]

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

按照普通 dict 。

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