可以将可变数量的参数传递给函数吗？

Question

与在C或C++中使用varargs的方式类似:

fn(a, b)
fn(a, b, c, d, ...)

Answer 1

是.

这很简单,如果您忽略关键字参数,则有效:

def manyArgs(*arg):
  print "I was called with", len(arg), "arguments:", arg

>>> manyArgs(1)
I was called with 1 arguments: (1,)
>>> manyArgs(1, 2, 3)
I was called with 3 arguments: (1, 2, 3)

如您所见,Python将为您提供包含所有参数的单个元组.

对于关键字参数,您需要将它们作为单独的实际参数接受,如Skurmedel的答案中所示.

Answer 2

添加解除帖子:

您也可以发送多个键值参数.

def myfunc(**kwargs):
    # kwargs is a dictionary.
    for k,v in kwargs.iteritems():
         print "%s = %s" % (k, v)

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

你可以将两者结合起来:

def myfunc2(*args, **kwargs):
   for a in args:
       print a
   for k,v in kwargs.iteritems():
       print "%s = %s" % (k, v)

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123

它们必须以该顺序声明和调用,即函数签名需要是*args,**kwargs,并按此顺序调用.

Answer 3

如果可以的话,Skurmedel的代码是针对python 2的; 使其适应python 3,更改iteritems为items并添加括号print.这可能会阻止像我这样的初学者碰到: AttributeError: 'dict' object has no attribute 'iteritems'并在其他地方搜索(例如,错误"'dict'对象在尝试使用NetworkX的write_shp()时没有属性'iteritems')为什么会发生这种情况.

def myfunc(**kwargs):
for k,v in kwargs.items():
   print("%s = %s" % (k, v))

myfunc(abc=123, efh=456)
# abc = 123
# efh = 456

和:

def myfunc2(*args, **kwargs):
   for a in args:
       print(a)
   for k,v in kwargs.items():
       print("%s = %s" % (k, v))

myfunc2(1, 2, 3, banan=123)
# 1
# 2
# 3
# banan = 123

Answer 4

添加到其他优秀的帖子.

有时你不想指定参数的数量并想要为它们使用键(如果在方法中没有使用字典中传递的一个参数,编译器会抱怨).

def manyArgs1(args):
  print args.a, args.b #note args.c is not used here

def manyArgs2(args):
  print args.c #note args.b and .c are not used here

class Args: pass

args = Args()
args.a = 1
args.b = 2
args.c = 3

manyArgs1(args) #outputs 1 2
manyArgs2(args) #outputs 3

然后你可以做的事情

myfuns = [manyArgs1, manyArgs2]
for fun in myfuns:
  fun(args)