Question

我希望在快速调用图中使下游函数接受多个参数，但是当我调用get时，dask说我缺少位置参数。

>>> import dask.threaded
>>> dsk = {'in':2, 'f1': (lambda x:(x**2, x**3), 'in'), 'out':(lambda x2,x3:(x2,x3), 'f1')}
>>> dask.threaded.get(dsk, 'out')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/nathan/GLOBAL/lib/python3.6/site-packages/dask/threaded.py", line 76, in get
    pack_exception=pack_exception, **kwargs)
  File "/home/nathan/GLOBAL/lib/python3.6/site-packages/dask/local.py", line 462, in get_async
    raise_exception(exc, tb)
  File "/home/nathan/GLOBAL/lib/python3.6/site-packages/dask/compatibility.py", line 112, in reraise
    raise exc
  File "/home/nathan/GLOBAL/lib/python3.6/site-packages/dask/local.py", line 230, in execute_task
    result = _execute_task(task, data)
  File "/home/nathan/GLOBAL/lib/python3.6/site-packages/dask/core.py", line 119, in _execute_task
    return func(*args2)
TypeError: <lambda>() missing 1 required positional argument: 'x3'

如何让一个函数在不使用*args的情况下接受多个参数？

Answer 1

输出(x**2, x**3)是单个tuple，不会自动解压缩；就像如果您将单个tuple传递给具有多个参数的函数一样，则不会。

但是，如果您不喜欢*args，则可以做的一件事是显式地将元组解包：

In [21]: def out(x):
    ...:     x2, x3 = x
    ...:     return (x2, 3*x3)
    ...:

In [22]: dsk = {'in': 2, 'f1': (lambda x:(x**2, x**3), 'in'), 'out': (out, 'f1')}

In [23]: dask.threaded.get(dsk, 'out')
Out[23]: (4, 24)

或者，如果您不介意只按索引访问元素：

In [27]: dsk = {'in': 2, 'f1': (lambda x: (x**2, x**3), 'in'), 'out': (lambda x: (x[0], 3*x[1]), 'f1')}

In [29]: dask.threaded.get(dsk, 'out')
Out[29]: (4, 24)

快速调用图：缺少必需的位置参数

1 个答案: