如何迭代函数参数 [英] How to iterate over function arguments

问题描述

我有一个 Python 函数，它接受多个字符串参数 def foo(a, b, c): 并将它们连接成一个字符串.我想遍历所有函数参数以检查它们不是 None.怎么做?有没有一种快速的方法可以将 None 转换为 ""?

谢谢.

解决方案

locals() 如果您在函数中首先调用它，则可能是您的朋友.

示例 1:

<预><代码>>>>定义乐趣(a，b，c):... d = 当地人()... e = d... 打印 e...打印当地人()...>>>乐趣(1, 2, 3){'a':1，'c':3，'b':2}{'a':1，'c':3，'b':2，'e':{...}，'d':{...}}

示例 2:

<预><代码>>>>定义无(a，b，c，d):... 参数 = locals()... 打印 '以下参数不是 None:', ', '.join(k for k, v in arguments.items() if v is not None)...>>>nones("Something", None, 'N', False)以下参数不是 None:a、c、d

答案:

<预><代码>>>>def foo(a, b, c):...返回 ''.join(v for v in locals().values() 如果 v 不是 None)...>>>foo('Cleese', 'Palin', 无)'克利斯佩林'

更新:

示例 1"强调，如果参数的顺序很重要，因为 locals()(或 dict)返回的 dictcode>vars()) 是无序的.上面的函数也没有非常优雅地处理数字.所以这里有一些改进:

<预><代码>>>>def foo(a, b, c):... 参数 = locals()...返回 ''.join(str(arguments[k]) for k in sorted(arguments.keys()) 如果 arguments[k] 不是 None)...>>>foo(None, 'Antioch', 3)'Antioch3'

I have a Python function accepting several string arguments def foo(a, b, c): and concatenating them in a string. I want to iterate over all function arguments to check they are not None. How it can be done? Is there a quick way to convert None to ""?

Thanks.

解决方案

locals() may be your friend here if you call it first thing in your function.

Example 1:

>>> def fun(a, b, c):
...     d = locals()
...     e = d
...     print e
...     print locals()
... 
>>> fun(1, 2, 3)
{'a': 1, 'c': 3, 'b': 2}
{'a': 1, 'c': 3, 'b': 2, 'e': {...}, 'd': {...}}

Example 2:

>>> def nones(a, b, c, d):
...     arguments = locals()
...     print 'The following arguments are not None: ', ', '.join(k for k, v in arguments.items() if v is not None)
... 
>>> nones("Something", None, 'N', False)
The following arguments are not None:  a, c, d

Answer:

>>> def foo(a, b, c):
...     return ''.join(v for v in locals().values() if v is not None)
... 
>>> foo('Cleese', 'Palin', None)
'CleesePalin'

Update:

'Example 1' highlights that we may have some extra work to do if the order of your arguments is important as the dict returned by locals() (or vars()) is unordered. The function above also doesn't deal with numbers very gracefully. So here are a couple of refinements:

>>> def foo(a, b, c):
...     arguments = locals()
...     return ''.join(str(arguments[k]) for k in sorted(arguments.keys()) if arguments[k] is not None)
... 
>>> foo(None, 'Antioch', 3)
'Antioch3'

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