为什么不能通过参数切换函数生成器的行为? [英] Why can't you toggle a function generator's behavior by an argument?

问题描述

请考虑以下两个功能:

def foo():
    x = 0
    while True:
        yield x
        x += 1

def wrap_foo(limit=10, gen=True):
    fg = foo()
    count = 0
    if gen:
        while count < limit:
            yield next(fg)
            count += 1
    else:
        return [next(fg) for _ in range(limit)]=

foo()是一个生成器，而wrap_foo()只是限制了生成多少数据.我正在尝试让包装器充当gen=True的生成器，或充当将所有生成的数据直接通过kwarg gen=False放入内存的常规函数​​.

foo() is a generator, and wrap_foo() just puts a limit on how much data gets generated. I was experimenting with having the wrapper behave as a generator with gen=True, or as a regular function that puts all generated data into memory directly with the kwarg gen=False.

常规生成器行为符合我的预期:

The regular generator behavior works as I'd expect:

In [1352]: [_ for _ in wrap_foo(gen=True)]
Out[1352]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

但是，使用gen=False不会生成任何内容.

However, with gen=False, nothing gets generated.

In [1351]: [num for num in wrap_foo(gen=False)]
Out[1351]: []

Python似乎根据yield语句的存在将函数预分类为生成器(如果注释了yield，则该示例可以很好地工作).

It seems like Python pre-classifies the function as a generator based on the presence of the yield statement (latter example works perfectly if yield is commented out).

这是为什么?我想了解这里的作用机制.我正在运行3.6

Why is this? I would like to understand the mechanisms at play here. I'm running 3.6

推荐答案

Python似乎根据yield语句的存在将函数预分类为生成器

It seems like Python pre-classifies the function as a generator based on the presence of the yield statement

是的，这正是发生的情况. wrap_foo在函数定义时被确定为生成器.您可以考虑改用生成器表达式:

Yes, that's exactly what happens. wrap_foo is determined to be a generator at function definition time. You could consider using generator expressions instead:

def wrap_foo(limit=10, gen=True):
    fg = foo()
    if gen:
        return (next(fg) for _ in range(limit))
    else:
        return [next(fg) for _ in range(limit)]

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