Python闭包和单元格（闭包值） [英] Python closures and cells (closed-over values)

问题描述

什么是Python机制，使

What is the Python mechanism that makes it so that

[lambda: x for x in range(5)][2]()

是4？

将x的副本绑定到每个lamba表达式的常见技巧是使上述表达式等于2？

What is the usual trick for binding a copy of x to each lamba expression so that the above expression will equal 2?

我的最终解决方案：

for template, model in zip(model_templates, model_classes):
    def create_known_parameters(known_parms):
        return lambda self: [getattr(self, p.name)
                             for p in known_parms]
    model.known_parameters = create_known_parameters(template.known_parms)

推荐答案

我通常使用 functools.partial

[ partial(lambda x: x, x) for x in range(5) ]

或者，您自己可以：

[ (lambda x: (lambda: x))(x) for x in range(5) ]

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