blocks|key|447607|text|args[:-1]是一个元组，因此嵌套调用实际上是mySum((4,))，args[0]的嵌套返回返回一个元组。因此，最后一行被缩减为return+3+%2B+(4,)。要解决此问题，您需要在调用mySum时通过将最后一行更改为return+args[-1]+%2B+mySum(*args[:-1])来扩展元组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|447608|entityMap^0|0|9|P|B|11|7|1V|F|34|Z|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

<code>args[:-1]</code> is a tuple, so the nested call is actually <code>mySum((4,))</code>, and the nested return of <code>args[0]</code> returns a tuple. So you end up with the last line being reduced to <code>return 3 + (4,)</code>. To fix this you need to expand the tuple when calling mySum by changing the last line to <code>return args[-1] + mySum(*args[:-1])</code>.

blocks|key|229602|text|这一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229603|return+args[-1]+%2B+mySum(args[:-1])|code-block|syntax|javascript|229604|args[:-1]返回参数元组的一部分。我假设你的目标是使用那片参数递归调用你的函数。不幸的是，您当前的代码只是使用单个对象调用您的函数-切片本身。|offset|length|style|CODE|229605|您想要做的是调用这些展开的参数。|229606|return+args[-1]+%2B+mySum(*args[:-1])
++++++++++++++++++++++++%5E----+note+the+asterisk|229607|这种技术称为"unpacking+argument+lists“，而星号有时(非正式地)称为"splat”运算符。|229608|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/tutorial/controlflow.html#unpacking-argument-lists^0|0|0|0|9|0|0|0|7|O|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|17|8|@]|9|@$I|18|J|19|1|1A]]|A|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

This line:

<pre><code>return args[-1] + mySum(args[:-1])
</code></pre>

<code>args[:-1]</code> returns a slice of the arguments tuple. I assume your goal is to recursively call your function using that slice of the arguments. Unfortunately, your current code simply calls your function using a single object - the slice itself.

What you want to do instead is to call with those args unrolled.

<pre><code>return args[-1] + mySum(*args[:-1])
 ^---- note the asterisk
</code></pre>

This technique is called "<a href="http://docs.python.org/2/tutorial/controlflow.html#unpacking-argument-lists">unpacking argument lists</a>," and the asterisk is sometimes (informally) called the "splat" operator.

blocks|key|229648|text|如果您不想以递归方式执行此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229649|def+mySum(*args):
++++sum+=+0
++++for+i+in+args:
++++++++sum+=+sum+%2B+i
++++return+sum|code-block|syntax|javascript|229650|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you don't want to do it recursively:
<pre class="lang-py prettyprint-override"><code>def mySum(*args):
 sum = 0
 for i in args:
 sum = sum + i
 return sum
</code></pre>

blocks|key|229662|text|Option1：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229663|def+mySum(*args):
++++return+sum(args)

mySum(1,2,3)+#+6
mySum(1,2)+#+3|code-block|syntax|javascript|229664|选项2：|229665|mySum2+=+lambda+*args:+sum(args)
mySum2(1,2,3)+#+6
mySum2(1,2)+#+3|229666|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Option1:

<pre><code>def mySum(*args):
 return sum(args)

mySum(1,2,3) # 6
mySum(1,2) # 3
</code></pre>

Option 2:

<pre><code>mySum2 = lambda *args: sum(args)
mySum2(1,2,3) # 6
mySum2(1,2) # 3
</code></pre>

blocks|key|447755|text|任意参数作为元组(带有一个星号*)传递给函数(您可以将其更改为代码中所示的列表)，并通过使用for循环自己编码来计算其元素的和；如果不想使用python的sum()方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447756|def+summing(*arg):
++++li+=+list(*arg)
++++x+=+0
++++for+i+in+range((len(li)-1)):
++++++++x+=+li[i]%2Bx

++++return+x

#creating+a+list+and+pass+it+as+arbitrary+argument+to+the+function
#to+colculate+the+sum+of+it's+elements

li+=+[4,+5+,3,+24,+6,+67,+1]
print+summing(li)|code-block|syntax|javascript|447757|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The arbitrary arguments are passed as tuple (with one asterisk*) to the function, (you can change it to a list as shown in the code) and calculate the sum of its elements, by coding yourself using a for loop; if don't want to use the sum() method of python. 

<pre><code>def summing(*arg):
 li = list(*arg)
 x = 0
 for i in range((len(li)-1)):
 x = li[i]+x

 return x

#creating a list and pass it as arbitrary argument to the function
#to colculate the sum of it's elements

li = [4, 5 ,3, 24, 6, 67, 1]
print summing(li)
</code></pre>

I'd like to learn how to pass an arbitrary number of args in a python function, so I wrote a simple sum function in a recursive way as follows:

<pre><code>def mySum(*args):
 if len(args) == 1:
 return args[0]
 else:
 return args[-1] + mySum(args[:-1])
</code></pre>

but when I tested <code>mySum(3, 4)</code>, I got this error: 

<pre><code>TypeError: unsupported operand type(s) for +: 'int' and 'tuple'
</code></pre>

Does anyone have an idea about this and gimme some clue to correct it?

arbitrary number of arguments in a python function

我想学习如何在python函数中传递任意数量的参数，所以我以递归的方式编写了一个简单的sum函数，如下所示：def mySum(*args):  if len(args) == 1:    return args[0]  else:    return args[-1] + mySum(args[:-1])但是当我测试...

问python函数中任意数量的参数
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