Python 默认值和 lambda

在别人的代码中，我读到了以下两行:

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

由于 default 的参数是默认工厂，我认为第一行意味着当我为一个不存在的键 k 调用 x [ k ]时(比如像 v = x [ k ]这样的语句) ，键值对(k，0)将自动添加到字典中，就像语句 x [ k ] = 0首先被执行一样。我说的对吗？

那你呢？似乎默认工厂将创建一个默认值为0的默认设置。但具体是什么意思呢？我试着在 Python shell 中使用它，但是不能确切地找出它是什么。

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最佳答案

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.

That's right. This is more idiomatically written

x = defaultdict(int)

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.

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

in terms of ordinary dict.

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defaultdict takes a zero-argument callable to its constructor, which is called when the key is not found, as you correctly explained.

lambda: 0 will of course always return zero, but the preferred method to do that is defaultdict(int), which will do the same thing.

As for the second part, the author would like to create a new defaultdict(int), or a nested dictionary, whenever a key is not found in the top-level dictionary.

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You are correct for what the first one does. As for y, it will create a defaultdict with default 0 when a key doesn't exist in y, so you can think of this as a nested dictionary. Consider the following example:

y = defaultdict(lambda: defaultdict(lambda: 0))
print y['k1']['k2']   # 0
print dict(y['k1'])   # {'k2': 0}

To create an equivalent nested dictionary structure without defaultdict you would need to create an inner dict for y['k1'] and then set y['k1']['k2'] to 0, but defaultdict does all of this behind the scenes when it encounters keys it hasn't seen:

y = {}
y['k1'] = {}
y['k1']['k2'] = 0

The following function may help for playing around with this on an interpreter to better your understanding:

def to_dict(d):
if isinstance(d, defaultdict):
return dict((k, to_dict(v)) for k, v in d.items())
return d

This will return the dict equivalent of a nested defaultdict, which is a lot easier to read, for example:

>>> y = defaultdict(lambda: defaultdict(lambda: 0))
>>> y['a']['b'] = 5
>>> y
defaultdict(<function <lambda> at 0xb7ea93e4>, {'a': defaultdict(<function <lambda> at 0xb7ea9374>, {'b': 5})})
>>> to_dict(y)
{'a': {'b': 5}}
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y = defaultdict(lambda:defaultdict(lambda:0))

will be helpful if you try this y['a']['b'] += 1

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All answers are good enough still I am giving the answer to add more info:

"defaultdict requires an argument that is callable. That return result of that callable object is the default value that the dictionary returns when you try to access the dictionary with a key that does not exist."

Here's an example

SAMPLE= {'Age':28, 'Salary':2000}
SAMPLE = defaultdict(lambda:0,SAMPLE)


>>> SAMPLE
defaultdict(<function <lambda> at 0x0000000002BF7C88>, {'Salary': 2000, 'Age': 28})


>>> SAMPLE['Age']----> This will return 28
>>> SAMPLE['Phone']----> This will return 0   # you got 0 as output for a non existing key inside SAMPLE

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