Question

实现for_each函数的正确方法是什么，以便它可以使用任意数量的参数，列表或元组作为参数？

def do_something(arg):
    print("done", arg)


def for_each(func, *args):
    if len(args) == 1:  # How to do this, since this gives an
        args = args[0]  # error if there's only one parameter besides func?
    for arg in args:
        func(arg)


for_each(do_something, 1, 2)
for_each(do_something, ['foo', 'bar'])
for_each(do_something, (3, 4, 5))

输出：

done 1
done 2
done foo
done bar
done 3
done 4
done 5

实现此目标的正确方法是什么？因为这样调用会破裂：

for_each(do_something, 1)

Traceback (most recent call last):
  File "main.py", line 12, in <module>
    for_each(do_something, 1)
  File "main.py", line 8, in for_each
    for arg in args:
TypeError: 'int' object is not iterable

Answer 1

您要检查您的第一个元素是列表还是元组，如下所示：

（您需要检查实例，以防用户仅使用一个int实例，否则代码将失败）

def for_each(func, *args):
  if len(args) == 1 and (isinstance(args[0], tuple) or isinstance(args[0], list)): # You can even check on Iterable by importing collections.abc.Iterable
    args = args[0]
  for arg in args:
    func(arg)

但是，您可以走得更远，让用户输入多个Iterable，而不仅仅是元组或列表，例如：

from collections.abc import Iterable
from itertools import chain


def do_something(arg):
    print("done", arg)

def for_each(func, *args):
  if all(map(lambda x: isinstance(x, Iterable), args)) and not any(map(lambda x: isinstance(x, str), args)):
    args = chain(*args)
  for arg in args:
    func(arg)


for_each(do_something, [0, 1], [0, 2])

Answer 2

好，现在我明白了您的问题：

不要这样做：

if len(args) == 1:  # is there really no cleaner
    args = args[0]  # way to do this part?

只要您不需要返回，只想调用一个函数，这看起来就更漂亮了。

def for_each(func, *args):
    for arg in args: 
        if any([isinstance(args, myiter) for myiter in [tuple, list, set]]): 
            [for_each(func, arg) for arg in args]
        else:
            func(args)



for_each(do_something, 1, 2) 
for_each(do_something, ['foo', 'bar']) 
for_each(do_something, (3, 4, 5))                                                                                                                                                              
done 1
done 2
done foo
done bar
done 3
done 4
done 5
for_each(do_something, 1)                                                                                                                                                                      
done 1

制作可以带有列表或任意数量参数的函数的最佳方法？

2 个答案: