python - 延迟函数的解包结果

Question

在使用延迟转换我的程序时，我偶然发现了一种不适用于延迟的常用编程模式。例子：

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

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

提高：TypeError: Delayed objects of unspecified length are not iterable 虽然以下解决方法没有。但是看起来笨拙了很多

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？

Answer 1

使用延迟文档字符串nout=中描述的关键字

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

x, y = f(1)

文档字符串

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.