延迟功能的拆箱结果 [英] Unpacking result of delayed function

问题描述

在使用延迟转换程序的同时，我偶然发现了一种常用的编程模式，该模式不适用于延迟。示例：

While converting my program using delayed, I stumbled upon a commonly used programming pattern that doesn't work with delayed. Example:

from dask import delayed
@delayed
def myFunction():
    return 1,2

a, b = myFunction()
a.compute()

引发： TypeError：未指定长度的延迟对象是不可迭代的
而以下解决方法则不能。但是看起来很笨拙

Raises: TypeError: Delayed objects of unspecified length are not iterable While the following work around does not. But looks a lot more clumsy

from dask import delayed
@delayed
def myFunction():
    return 1,2

dummy = myFunction()
a, b = dummy[0], dummy[1]
a.compute()

这是预期的行为吗？

推荐答案

使用 nout = 关键字，如延迟的文档字符串

@delayed(nout=2)
def f(...):
    return a, b

x, y = f(1)

文档字符串

Docstring

nout : int, optional
    The number of outputs returned from calling the resulting ``Delayed``
    object. If provided, the ``Delayed`` output of the call can be iterated
    into ``nout`` objects, allowing for unpacking of results. By default
    iteration over ``Delayed`` objects will error. Note, that ``nout=1``
    expects ``obj``, to return a tuple of length 1, and consequently for
    `nout=0``, ``obj`` should return an empty tuple.

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