在 python 工具中柯里化 merge_with

Question

我希望能够咖喱 merge_with:

merge_with 如我所料

>>> from cytoolz import curry, merge_with
>>> d1 = {"a" : 1, "b" : 2}
>>> d2 = {"a" : 2, "b" : 3}
>>> merge_with(sum, d1, d2)
{'a': 3, 'b': 5}

在一个简单的函数上，curry 像我预期的那样工作：

>>> def f(a, b):
...     return a * b
... 
>>> curry(f)(2)(3)
6

但我无法"manually"制作merge_with的咖喱版本：

>>> curry(merge_with)(sum)(d1, d2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'dict' object is not callable
>>> curry(merge_with)(sum)(d1)(d2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'dict' object is not callable

预柯里化版本有效：

>>> from cytoolz.curried import merge_with as cmerge
>>> cmerge(sum)(d1, d2)
{'a': 3, 'b': 5}

我的错误在哪里？

Answer 1

这是因为merge_with将dicts作为位置参数：

merge_with(func, *dicts, **kwargs)

所以 f 是唯一必须的参数，对于空 *args 你得到一个空字典：

>>> curry(merge_with)(sum)  # same as merge_with(sum)
{}

所以：

curry(f)(2)(3)

等同于

>>> {}(2)(3)
Traceback (most recent call last):
...
TypeError: 'dict' object is not callable

您必须显式定义助手

def merge_with_(f):
    def _(*dicts, **kwargs):
        return merge_with(f, *dicts, **kwargs)
    return _

想怎么用就怎么用：

>>> merge_with_(sum)(d1, d2)
{'a': 3, 'b': 5}

或：

def merge_with_(f, d1, d2, *args, **kwargs):
    return merge_with(f, d1, d2, *args, **kwargs)

哪个都可以

>>> curry(merge_with_)(sum)(d1, d2)
{'a': 3, 'b': 5}

和：

>>> curry(merge_with_)(sum)(d1)(d2)
{'a': 3, 'b': 5}

在 python 工具中柯里化 merge_with

Currying merge_with in python toolz

python

currying

toolz