问题：是否可以将可变数量的参数传递给函数？
与在C或C ++中使用varargs的方式类似：
fn(a, b)
fn(a, b, c, d, ...)
In a similar way to using varargs in C or C++:
fn(a, b)
fn(a, b, c, d, ...)
回答 0
是。您可以将其*args用作非关键字参数。然后，您将可以传递任意数量的参数。
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会将所有参数作为一个元组解压缩参数。
对于关键字参数，您需要将其作为单独的实际参数接受，如Skurmedel的answer所示。
Yes. You can use *args as a non-keyword argument. You will then be able to pass any number of arguments.
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)
As you can see, Python will unpack the arguments as a single tuple with all the arguments.
For keyword arguments you need to accept those as a separate actual argument, as shown in Skurmedel’s answer.
回答 1
添加到发布帖子：
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
您可以将两者混合使用：
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
必须同时声明和调用它们，也就是说，函数签名必须为* args，** kwargs，并以该顺序调用。
Adding to unwinds post:
You can send multiple key-value args too.
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
And you can mix the two:
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
They must be both declared and called in that order, that is the function signature needs to be *args, **kwargs, and called in that order.
回答 2
如果可以的话，Skurmedel的代码适用于python 2；使其适应python 3，将更iteritems改为items并添加括号print。这可能会阻止像我这样的初学者进入：
AttributeError: 'dict' object has no attribute 'iteritems'并在其他位置搜索（例如，尝试使用NetworkX的write_shp（）时错误“’dict’对象没有属性’iteritems”））。
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
If I may, Skurmedel’s code is for python 2; to adapt it to python 3, change iteritems to items and add parenthesis to print. That could prevent beginners like me to bump into:
AttributeError: 'dict' object has no attribute 'iteritems' and search elsewhere (e.g. Error “ ‘dict’ object has no attribute ‘iteritems’ ” when trying to use NetworkX’s write_shp()) why this is happening.
def myfunc(**kwargs):
for k,v in kwargs.items():
print("%s = %s" % (k, v))
myfunc(abc=123, efh=456)
# abc = 123
# efh = 456
and:
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
回答 3
添加到其他优秀职位。
有时，您不想指定参数的数目，而不想为它们使用键（如果在方法中未使用在字典中传递的一个参数，则编译器会抱怨）。
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
然后你可以做类似的事情
myfuns = [manyArgs1, manyArgs2]
for fun in myfuns:
fun(args)
Adding to the other excellent posts.
Sometimes you don’t want to specify the number of arguments and want to use keys for them (the compiler will complain if one argument passed in a dictionary is not used in the method).
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
Then you can do things like
myfuns = [manyArgs1, manyArgs2]
for fun in myfuns:
fun(args)
回答 4
def f(dic):
if 'a' in dic:
print dic['a'],
pass
else: print 'None',
if 'b' in dic:
print dic['b'],
pass
else: print 'None',
if 'c' in dic:
print dic['c'],
pass
else: print 'None',
print
pass
f({})
f({'a':20,
'c':30})
f({'a':20,
'c':30,
'b':'red'})
____________
上面的代码将输出
None None None
20 None 30
20 red 30
这就像通过字典传递变量参数一样好
def f(dic):
if 'a' in dic:
print dic['a'],
pass
else: print 'None',
if 'b' in dic:
print dic['b'],
pass
else: print 'None',
if 'c' in dic:
print dic['c'],
pass
else: print 'None',
print
pass
f({})
f({'a':20,
'c':30})
f({'a':20,
'c':30,
'b':'red'})
____________
the above code will output
None None None
20 None 30
20 red 30
This is as good as passing variable arguments by means of a dictionary
回答 5
除了已经提到的好答案之外，另一种解决方法还取决于您可以按位置传递可选的命名参数这一事实。例如，
def f(x,y=None):
print(x)
if y is not None:
print(y)
Yield
In [11]: f(1,2)
1
2
In [12]: f(1)
1
Another way to go about it, besides the nice answers already mentioned, depends upon the fact that you can pass optional named arguments by position. For example,
def f(x,y=None):
print(x)
if y is not None:
print(y)
Yields
In [11]: f(1,2)
1
2
In [12]: f(1)
1
