python - kwarg-splatting 一个 numpy 数组

Question

我怎样才能编写一个包装器类来实现这一点？

def foo(a, b):
    print a

data = np.empty(20, dtype=[('a', np.float32), ('b', np.float32)])

data = my_magic_ndarray_subclass(data)

foo(**data[0])

---

更多背景:

我有一对这样的函数，我想对其进行矢量化:

def start_the_work(some_arg):
    some_calculation = ...
    something_else = ...

    cost = some_calculation * something_else

    return cost, dict(
        some_calculation=some_calculation,
        some_other_calculation=some_other_calculation
    )

def finish_the_work(some_arg, some_calculation, some_other_calculation):
    ...

目的是用一堆不同的参数调用start_the_work，然后完成成本最低的项目。这两个函数使用了很多相同的计算，因此使用字典和 kwarg-splatting 来传递这些结果:

def run():
    best, best_cost, continuation = min(
        ((some_arg,) + start_the_work(some_arg)
         for some_arg in [1, 2, 3, 4]),
        key=lambda t: t[1]  # cost
    )
    return finish_the_work(best, **continuation)

我可以对它们进行矢量化的一种方法如下:

def start_the_work(some_arg):
    some_calculation = ...
    something_else = ...

    cost = some_calculation * something_else

    continuation = np.empty(cost.shape, dtype=[
        ('some_calculation', np.float32),
        ('some_other_calculation', np.float32)
    ])
    continuation['some_calculation'] = some_calculation
    continuation['some_other_calculation'] = some_other_calculation

    return cost, continuation

但是尽管看起来像一本字典，continuation 不能用 kwarg-splatted。

Answer 1

这可能不是您想要的，但是将数组包装在 pandas DataFrame 中允许这样的事情:

import pandas as pd

def foo(a, b):
    print(a)

data = np.empty(20, dtype=[('a', np.float32), ('b', np.float32)])

data = pd.DataFrame(data).T

foo(**data[0])
# 0.0

请注意，dataframe 是转置的，因为 pandas 的主索引是列而不是行。