blocks|key|202222|text|Python+3:使用functools.reduce|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|202223|>>>+from+functools+import+reduce
>>>+reduce(lambda+x,+y:+x*y,+[1,2,3,4,5,6])
720|code-block|syntax|javascript|202224|Python+2:使用reduce|202225|>>>+reduce(lambda+x,+y:+x*y,+[1,2,3,4,5,6])
720|202226|要与2和3兼容，请使用pip+install+six，然后：|202227|>>>+from+six.moves+import+reduce
>>>+reduce(lambda+x,+y:+x*y,+[1,2,3,4,5,6])
720|202228|entityMap^0|B|G|0|0|B|6|0|0|B|F|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Python 3: use <code>functools.reduce</code>:

<pre><code>&gt;&gt;&gt; from functools import reduce
&gt;&gt;&gt; reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
</code></pre>

Python 2: use <code>reduce</code>:

<pre><code>&gt;&gt;&gt; reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
</code></pre>

For compatible with 2 and 3 use <code>pip install six</code>, then:

<pre><code>&gt;&gt;&gt; from six.moves import reduce
&gt;&gt;&gt; reduce(lambda x, y: x*y, [1,2,3,4,5,6])
720
</code></pre>

blocks|key|3051971|text|您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3051972|import+operator
import+functools
functools.reduce(operator.mul,+[1,2,3,4,5,6],+1)|code-block|syntax|javascript|3051973|有关说明，请参阅reduce和operator.mul文档。|offset|length|style|CODE|3051974|您需要在Python+3%2B中使用import+functools行。|3051975|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/functions.html#reduce|1|http://docs.python.org/2/library/operator.html#operator.mul^0|0|0|8|6|F|C|8|6|0|F|C|1|0|G|G|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@$I|15|J|16|1|17]|$I|18|J|19|1|1A]]|A|$]]|$1|M|3|N|5|6|7|1B|8|@$I|1C|J|1D|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

You can use:

<pre><code>import operator
import functools
functools.reduce(operator.mul, [1,2,3,4,5,6], 1)
</code></pre>

See <a href="http://docs.python.org/2/library/functions.html#reduce" rel="noreferrer"><code>reduce</code></a> and <a href="http://docs.python.org/2/library/operator.html#operator.mul" rel="noreferrer"><code>operator.mul</code></a> documentations for an explanation.

You need the <code>import functools</code> line in Python 3+.

blocks|key|2839000|text|我将使用numpy.prod来执行该任务。见下文。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2839001|import+numpy+as+np
mylist+=+[1,+2,+3,+4,+5,+6]+
result+=+np.prod(np.array(mylist))++|code-block|syntax|javascript|2839002|entityMap^0|4|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I would use the <code>numpy.prod</code> to perform the task. See below.

<pre><code>import numpy as np
mylist = [1, 2, 3, 4, 5, 6] 
result = np.prod(np.array(mylist)) 
</code></pre>

blocks|key|206606|text|如果希望避免导入任何内容并避免Python中更复杂的区域，可以使用简单的for循环|type|unstyled|depth|inlineStyleRanges|entityRanges|data|206607|product+=+1++#+Don't+use+0+here,+otherwise,+you'll+get+zero+
+++++++++++++#+because+anything+times+zero+will+be+zero.
list+=+[1,+2,+3]
for+x+in+list:
++++product+*=+x|code-block|syntax|javascript|206608|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 want to avoid importing anything and avoid more complex areas of Python, you can use a simple for loop

<pre><code>product = 1 # Don't use 0 here, otherwise, you'll get zero 
 # because anything times zero will be zero.
list = [1, 2, 3]
for x in list:
 product *= x
</code></pre>

blocks|key|205141|text|启动Python+3.8后，标准库中的math模块中包含了一个.prod函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|205142|205143|+math.prod(iterable,+*,+start=1)|blockquote|205144|205145|205146|该方法返回start值(默认值:+1)乘以可迭代数字的乘积：|205147|import+math
math.prod([1,+2,+3,+4,+5,+6])

>>>+720|code-block|syntax|javascript|205148|如果iterable为空，这将生成1+(或start值，如果提供的话)。|205149|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/math.html#math.prod^0|2|A|J|4|V|5|V|5|0|0|0|1|V|0|0|0|5|5|0|0|H|1|L|5|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|N|3|O|5|P|7|1L|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

Starting <code>Python 3.8</code>, a <a href="https://docs.python.org/3/library/math.html#math.prod" rel="noreferrer"><code>.prod</code></a> function has been included to the <code>math</code> module in the standard library:

<blockquote>
 <code>math.prod(iterable, *, start=1)</code>
</blockquote>

The method returns the product of a <code>start</code> value (default: 1) times an iterable of numbers:

<pre class="lang-py prettyprint-override"><code>import math
math.prod([1, 2, 3, 4, 5, 6])

&gt;&gt;&gt; 720
</code></pre>

If the iterable is empty, this will produce <code>1</code> (or the <code>start</code> value, if provided).

blocks|key|206678|text|这是来自我的机器的一些性能测量。适用于在长时间运行的循环中对小输入执行此操作的情况：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|206679|import+functools,+operator,+timeit
import+numpy+as+np

def+multiply_numpy(iterable):
++++return+np.prod(np.array(iterable))

def+multiply_functools(iterable):
++++return+functools.reduce(operator.mul,+iterable)

def+multiply_manual(iterable):
++++prod+=+1
++++for+x+in+iterable:
++++++++prod+*=+x

++++return+prod

sizesToTest+=+[5,+10,+100,+1000,+10000,+100000]

for+size+in+sizesToTest:
++++data+=+[1]+*+size

++++timerNumpy+=+timeit.Timer(lambda:+multiply_numpy(data))
++++timerFunctools+=+timeit.Timer(lambda:+multiply_functools(data))
++++timerManual+=+timeit.Timer(lambda:+multiply_manual(data))

++++repeats+=+int(5e6+/+size)
++++resultNumpy+=+timerNumpy.timeit(repeats)
++++resultFunctools+=+timerFunctools.timeit(repeats)
++++resultManual+=+timerManual.timeit(repeats)
++++print(f'Input+size:+{size:>7d}+Repeats:+{repeats:>8d}++++Numpy:+{resultNumpy:.3f},+Functools:+{resultFunctools:.3f},+Manual:+{resultManual:.3f}')|code-block|syntax|javascript|206680|结果：|206681|Input+size:+++++++5+Repeats:++1000000++++Numpy:+4.670,+Functools:+0.586,+Manual:+0.459
Input+size:++++++10+Repeats:+++500000++++Numpy:+2.443,+Functools:+0.401,+Manual:+0.321
Input+size:+++++100+Repeats:++++50000++++Numpy:+0.505,+Functools:+0.220,+Manual:+0.197
Input+size:++++1000+Repeats:+++++5000++++Numpy:+0.303,+Functools:+0.207,+Manual:+0.185
Input+size:+++10000+Repeats:++++++500++++Numpy:+0.265,+Functools:+0.194,+Manual:+0.187
Input+size:++100000+Repeats:+++++++50++++Numpy:+0.266,+Functools:+0.198,+Manual:+0.185|206682|您可以看到，在较小的输入上，Numpy相当慢，因为它在执行乘法之前分配了一个数组。另外，要注意Numpy中的溢出。|206683|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here's some performance measurements from my machine. Relevant in case this is performed for small inputs in a long-running loop:

<pre><code>import functools, operator, timeit
import numpy as np

def multiply_numpy(iterable):
 return np.prod(np.array(iterable))

def multiply_functools(iterable):
 return functools.reduce(operator.mul, iterable)

def multiply_manual(iterable):
 prod = 1
 for x in iterable:
 prod *= x

 return prod

sizesToTest = [5, 10, 100, 1000, 10000, 100000]

for size in sizesToTest:
 data = [1] * size

 timerNumpy = timeit.Timer(lambda: multiply_numpy(data))
 timerFunctools = timeit.Timer(lambda: multiply_functools(data))
 timerManual = timeit.Timer(lambda: multiply_manual(data))

 repeats = int(5e6 / size)
 resultNumpy = timerNumpy.timeit(repeats)
 resultFunctools = timerFunctools.timeit(repeats)
 resultManual = timerManual.timeit(repeats)
 print(f'Input size: {size:&gt;7d} Repeats: {repeats:&gt;8d} Numpy: {resultNumpy:.3f}, Functools: {resultFunctools:.3f}, Manual: {resultManual:.3f}')
</code></pre>

Results:

<pre><code>Input size: 5 Repeats: 1000000 Numpy: 4.670, Functools: 0.586, Manual: 0.459
Input size: 10 Repeats: 500000 Numpy: 2.443, Functools: 0.401, Manual: 0.321
Input size: 100 Repeats: 50000 Numpy: 0.505, Functools: 0.220, Manual: 0.197
Input size: 1000 Repeats: 5000 Numpy: 0.303, Functools: 0.207, Manual: 0.185
Input size: 10000 Repeats: 500 Numpy: 0.265, Functools: 0.194, Manual: 0.187
Input size: 100000 Repeats: 50 Numpy: 0.266, Functools: 0.198, Manual: 0.185
</code></pre>

You can see that Numpy is quite a bit slower on smaller inputs, since it allocates an array before multiplication is performed. Also, watch out for the overflow in Numpy.

blocks|key|202240|text|我个人喜欢这个函数，它可以将泛型列表的所有元素相乘：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|202241|def+multiply(n):
++++total+=+1
++++for+i+in+range(0,+len(n)):
++++++++total+*=+n[i]
++++print+total|code-block|syntax|javascript|202242|它很紧凑，使用了简单的东西(一个变量和一个for循环)，而且对我来说很直观(看起来就像我思考问题的方式，只需取一个，乘以它，然后乘以下一个，依此类推！)|202243|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I personally like this for a function that multiplies all elements of a generic list together:

<pre><code>def multiply(n):
 total = 1
 for i in range(0, len(n)):
 total *= n[i]
 print total
</code></pre>

It's compact, uses simple things (a variable and a for loop), and feels intuitive to me (it looks like how I'd think of the problem, just take one, multiply it, then multiply by the next, and so on!)

blocks|key|204728|text|Numpy有一个prod()函数，它返回一个列表的乘积，或者在本例中，因为它是numpy，所以它是一个数组在给定轴上的乘积：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|204729|import+numpy
a+=+[1,2,3,4,5,6]
b+=+numpy.prod(a)|code-block|syntax|javascript|204730|...or或者你可以直接导入numpy.prod()|204731|from+numpy+import+prod
a+=+[1,2,3,4,5,6]
b+=+prod(a)|204732|entityMap^0|0|5|8|6|0|0|E|C|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

<code>Numpy</code> has the <code>prod()</code> function that returns the product of a list, or in this case since it's numpy, it's the product of an array over a given axis:

<pre><code>import numpy
a = [1,2,3,4,5,6]
b = numpy.prod(a)
</code></pre>

...or else you can just import <code>numpy.prod()</code>:

<pre><code>from numpy import prod
a = [1,2,3,4,5,6]
b = prod(a)
</code></pre>

blocks|key|2839112|text|简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2839113|import+numpy+as+np
np.exp(np.log(your_array).sum())|code-block|syntax|javascript|2839114|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 simple way is:

<pre><code>import numpy as np
np.exp(np.log(your_array).sum())
</code></pre>

blocks|key|202264|text|nums+=+str(tuple([1,2,3]))
mul_nums+=+nums.replace(',','*')
print(eval(mul_nums))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|202265|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>nums = str(tuple([1,2,3]))
mul_nums = nums.replace(',','*')
print(eval(mul_nums))
</code></pre>

blocks|key|206632|text|今天我找到了这个问题，但我注意到它没有列表中有None的情况。所以，完整的解决方案是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|206633|from+functools+import+reduce

a+=+[None,+1,+2,+3,+None,+4]
print(reduce(lambda+x,+y:+(x+if+x+else+1)+*+(y+if+y+else+1),+a))|code-block|syntax|javascript|206634|在添加的情况下，我们有：|206635|print(reduce(lambda+x,+y:+(x+if+x+else+0)+%2B+(y+if+y+else+0),+a))|206636|entityMap^0|N|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

Found this question today but I noticed that it does not have the case where there are <code>None</code>'s in the list. So, the complete solution would be:

<pre><code>from functools import reduce

a = [None, 1, 2, 3, None, 4]
print(reduce(lambda x, y: (x if x else 1) * (y if y else 1), a))
</code></pre>

In the case of addition, we have: 

<pre><code>print(reduce(lambda x, y: (x if x else 0) + (y if y else 0), a))
</code></pre>

blocks|key|2839004|text|我希望这样做的方式如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2839005|++++def+product_list(p):
++++++++++total+=1+#critical+step+works+for+all+list
++++++++++for+i+in+p:
+++++++++++++total=total*i+#+this+will+ensure+that+each+elements+are+multiplied+by+itself
++++++++++return+total
+++print+product_list([2,3,4,2])+#should+print+48|code-block|syntax|javascript|2839006|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|$]]

I would like this in following way:

<pre><code> def product_list(p):
 total =1 #critical step works for all list
 for i in p:
 total=total*i # this will ensure that each elements are multiplied by itself
 return total
 print product_list([2,3,4,2]) #should print 48
</code></pre>

blocks|key|3052255|text|这是我的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3052256|def+product_list(list_of_numbers):
++++xxx+=+1
++++for+x+in+list_of_numbers:
++++++++xxx+=+xxx*x
++++return+xxx

print(product_list([1,2,3,4]))|code-block|syntax|javascript|3052257|结果：('1*1*2*3*4'，24)|3052258|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is my code:

<pre><code>def product_list(list_of_numbers):
 xxx = 1
 for x in list_of_numbers:
 xxx = xxx*x
 return xxx

print(product_list([1,2,3,4]))
</code></pre>

result : ('1*1*2*3*4', 24)

blocks|key|206705|text|使用递归怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|206706|def+multiply(lst):
++++if+len(lst)+>+1:
++++++++return+multiply(lst[:-1])*+lst[-1]
++++else:
++++++++return+lst[0]|code-block|syntax|javascript|206707|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|$]]

How about using recursion? 

<pre><code>def multiply(lst):
 if len(lst) &gt; 1:
 return multiply(lst[:-1])* lst[-1]
 else:
 return lst[0]
</code></pre>

blocks|key|202358|text|我的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|202359|def+multiply(numbers):
++++a+=+1
++++for+num+in+numbers:
++++++++a+*=+num
++++return+a|code-block|syntax|javascript|202360|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|$]]

My solution:
<pre><code>def multiply(numbers):
 a = 1
 for num in numbers:
 a *= num
 return a
</code></pre>

blocks|key|206729|text|‘理解循环逻辑使用的唯一简单方法’|type|unstyled|depth|inlineStyleRanges|entityRanges|data|206730|in的Lap=2,5,7,7,9+x=1+:+x=i*x+print(x)|206731|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

'''the only simple method to understand the logic
use for loop'''

Lap=[2,5,7,7,9] 
x=1 
for i in Lap: 
 x=i*x 
print(x)

blocks|key|2839182|text|这很简单，不导入任何东西。这是我的代码。这将定义一个函数，该函数将一个列表中的所有项目相乘并返回它们的乘积。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2839183|def+myfunc(lst):
++++multi=1
++++for+product+in+lst:
++++++++multi*=product
++++return+product|code-block|syntax|javascript|2839184|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|$]]

It is very simple do not import anything. This is my code.
This will define a function that multiplies all the items in a list and returns their product.

<pre><code>def myfunc(lst):
 multi=1
 for product in lst:
 multi*=product
 return product
</code></pre>

I need to write a function that takes
a list of numbers and multiplies them together. Example:
<code>[1,2,3,4,5,6]</code> will give me <code>1*2*3*4*5*6</code>. I could really use your help.

How can I multiply all items in a list together with Python?

Python

我需要编写一个函数，该函数接受数字的列表，然后将它们相乘。例如：[1,2,3,4,5,6]会给我1*2*3*4*5*6。我真的需要你的帮助。

问如何使用Python将列表中的所有项相乘？
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回答 17

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问如何使用Python将列表中的所有项相乘？
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