Python初学者如何实现闭包中的变量绑定技巧？

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Python 在闭包（closures）中如何绑定变量：

看这个例子：

pythondef create_multipliers(): return [lambda x: i * x for i in range(5)]

for multiplier in create_multipliers(): print(multiplier(2))

期望得到以下结果：

python

02

46

8

搞不清楚在闭包（closures）中Python是怎样绑定变量的

看这个例子：

>>> def create_multipliers(): ... return [lambda x : i * x for i in range(5)] >>> for multiplier in create_multipliers(): ... print multiplier(2) ...

期望得到下面的输出：

0

2

4

6

8

但是实际上得到的是：

8

8

8

8

8

实例扩展：

# coding=utf-8 __author__ = 'xiaofu' # 解释参考 docs.python-guide.org/en/latest/writing/gotchas/#late-binding-closures def closure_test1(): """ 每个closure的输出都是同一个i值 :return: """ closures = [] for i in range(4): def closure(): print("id of i: {}, value: {} ".format(id(i), i)) closures.append(closure) # Python's closures are late binding. # This means that the values of variables used in closures are looked up at the time the inner function is called. for c in closures: c() def closure_test2(): def make_closure(i): def closure(): print("id of i: {}, value: {} ".format(id(i), i)) return closure closures = [] for i in range(4): closures.append(make_closure(i)) for c in closures: c() if __name__ == '__main__': closure_test1() closure_test2()

输出:

id of i: 10437280, value: 3 id of i: 10437280, value: 3 id of i: 10437280, value: 3 id of i: 10437280, value: 3 id of i: 10437184, value: 0 id of i: 10437216, value: 1 id of i: 10437248, value: 2 id of i: 10437280, value: 3

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