如何迭代函数参数

Question

我有一个Python函数接受几个字符串参数def foo(a, b, c):并将它们连接在一个字符串中.我想迭代所有函数参数来检查它们不是None.怎么做？有没有快速的方法将无效转换为""？

谢谢.

Answer 1

locals() 如果你在你的职能中首先称呼它可能是你的朋友.

例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': {...}}

例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

答案:

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

更新:

"示例1"强调,如果参数的顺序很重要,我们可能会做一些额外的工作,因为(或)dict返回的顺序是无序的.上述功能也不能非常优雅地处理数字.所以这里有几个改进:locals()vars()

>>> 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'

Answer 2

def func(*args):
    ' '.join(i if i is not None else '' for i in args)

如果你加入一个空字符串,你可以这样做 ''.join(i for i in args if i is not None)