blocks|key|51855|text|我们可以将这两个函数组合成一个递归函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|51856|def+factorial(n):
+++if+n+<+1:+++#+base+case
+++++++return+1
+++else:
+++++++returnNumber+=+n+*+factorial(n+-+1)++#+recursive+call
+++++++print(str(n)+%2B+'!+=+'+%2B+str(returnNumber))
+++++++return+returnNumber|code-block|syntax|javascript|51857|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|$]]

We can combine the two functions to this single recursive function:
<pre><code>def factorial(n):
 if n &lt; 1: # base case
 return 1
 else:
 returnNumber = n * factorial(n - 1) # recursive call
 print(str(n) + '! = ' + str(returnNumber))
 return returnNumber
</code></pre>

blocks|key|2535514|text|两行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2535515|def+fac(n):
++++return+1+if+(n+<+1)+else+n+*+fac(n-1)|code-block|syntax|javascript|2535516|测试它：|2535517|print+fac(4)|2535518|结果：|2535519|24|2535520|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

2 lines of code:

<pre><code>def fac(n):
 return 1 if (n &lt; 1) else n * fac(n-1)
</code></pre>

Test it:

<pre><code>print fac(4)
</code></pre>

Result:

<pre><code>24
</code></pre>

blocks|key|51461|text|def+factorial(n):
++++result+=+1+if+n+<=+1+else+n+*+factorial(n+-+1)
++++print+'%25d!+=+%25d'+%25+(n,+result)
++++return+result|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|51462|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>def factorial(n):
 result = 1 if n &lt;= 1 else n * factorial(n - 1)
 print '%d! = %d' % (n, result)
 return result
</code></pre>

blocks|key|2745246|text|一个简短的例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2745247|def+fac(n):
++++if+n+==+0:
++++++++return+1
++++else:
++++++++return+n+*+fac(n-1)
print+fac(0)|code-block|syntax|javascript|2745248|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|$]]

a short one:

<pre><code>def fac(n):
 if n == 0:
 return 1
 else:
 return n * fac(n-1)
print fac(0)
</code></pre>

blocks|key|52732|text|我没有使用Python的经验，但是像这样的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|52733|def+factorial(+n+):
+++if+n+<1:+++#+base+case
+++++++return+1
+++else:
+++++++f+=+n+*+factorial(+n+-+1+)++#+recursive+call
+++++++print+"%252d!+=+%25d"+%25+(+n,+f+)
+++++++return+f|code-block|syntax|javascript|52734|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've no experience with Python, but something like this?

<pre><code>def factorial( n ):
 if n &lt;1: # base case
 return 1
 else:
 f = n * factorial( n - 1 ) # recursive call
 print "%2d! = %d" % ( n, f )
 return f
</code></pre>

blocks|key|51842|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|51843|def+factorial(+n+):
+++if+n+<1:+++#+base+case
+++++++print+"%252d!+=+%25d"+%25+(n,+n)
+++++++return+1
+++else:
+++++++temp+=+factorial(+n+-+1+)
+++++++print+"%252d!+=+%25d"+%25+(n,+n*temp)
+++++++return+n+*+temp++#+recursive+call|code-block|syntax|javascript|51844|我注意到的一件事是，你为n<1返回了'1‘，这意味着即使是负数，你的函数也会返回1。你可能想解决这个问题。|51845|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|$]]

try this:

<pre><code>def factorial( n ):
 if n &lt;1: # base case
 print "%2d! = %d" % (n, n)
 return 1
 else:
 temp = factorial( n - 1 )
 print "%2d! = %d" % (n, n*temp)
 return n * temp # recursive call
</code></pre>

One thing I noticed is that you are returning '1' for n&lt;1, that means your function will return 1 even for negative numbers. You may want to fix that.

blocks|key|2535486|text|这是不是碰巧有家庭作业？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2535487|def+traced_factorial(n):
++def+factorial(n):
++++if+n+<=+1:
++++++return+1
++++return+n+*+factorial(n+-+1)
++for+i+in+range(1,+n+%2B+1):
++++print+'%252d!+=+%25d'+%25(i,+factorial(i))|code-block|syntax|javascript|2535488|请阅读PEP227以了解更多详细信息。它的不足之处在于，Python允许在函数中定义函数。|offset|length|2535489|entityMap|0|LINK|mutability|MUTABLE|url|http://www.python.org/dev/peps/pep-0227/^0|0|0|3|6|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Is this homework by any chance?

<pre><code>def traced_factorial(n):
 def factorial(n):
 if n &lt;= 1:
 return 1
 return n * factorial(n - 1)
 for i in range(1, n + 1):
 print '%2d! = %d' %(i, factorial(i))
</code></pre>

Give <a href="http://www.python.org/dev/peps/pep-0227/" rel="nofollow">PEP227</a> a read for more details. The short of it is that Python lets you define functions within functions.

blocks|key|51489|text|fac+=+lambda+x:+1+if+x+==+0+else+x+*+fac(x+-+1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|51490|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>fac = lambda x: 1 if x == 0 else x * fac(x - 1)
</code></pre>

blocks|key|52127|text|再来一个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|52128|def+fact(x):
++++if+x+in+{0,+1}:
++++++++return+1
++++else:
++++++++return+x+*+fact(x-1)

for+x+in+range(0,10):
++++print+'%25d!+=+%25d'+%25(x,+fact(x))|code-block|syntax|javascript|52129|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|$]]

One more
<pre><code>def fact(x):
 if x in {0, 1}:
 return 1
 else:
 return x * fact(x-1)

for x in range(0,10):
 print '%d! = %d' %(x, fact(x))
</code></pre>

blocks|key|2745304|text|我真的不知道负数的阶乘，但这将适用于所有n+>=+0：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2745305|def+factorial(n):
++++if+n+>=+0:
++++++++if+n+==+1+or+n+==+0:
++++++++++++return+1
++++++++else:
++++++++++++n+=+n+*+factorial(n-1)
++++++++++++return+n

++++return+'error'|code-block|syntax|javascript|2745306|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 don't really know the factorial of negative numbers, but this will work with all n &gt;= 0:
<pre><code>def factorial(n):
 if n &gt;= 0:
 if n == 1 or n == 0:
 return 1
 else:
 n = n * factorial(n-1)
 return n

 return 'error'
</code></pre>

blocks|key|2535567|text|可以使用这4行代码...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2535568|+++def+factorial(n):
++++++++f+=+lambda+n:+n+*+f(n+-+1)+if+n+>+1+else+1

++++++++for+x+in+range(n):
++++++++++++print('{}!+=+{}'.format(x+%2B+1,+factorial(x+%2B+1)))|code-block|syntax|javascript|2535569|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|$]]

Can use these 4 lines of code...
<pre><code> def factorial(n):
 f = lambda n: n * f(n - 1) if n &gt; 1 else 1

 for x in range(n):
 print('{}! = {}'.format(x + 1, factorial(x + 1)))
</code></pre>

blocks|key|2745370|text|并且第一次使用递归和while循环来计算阶乘。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2745371|def+factorial(n):
++++while++n+>=+1:
++++++++return+n+*+factorial(n+-+1)
++++return+1|code-block|syntax|javascript|2745372|尽管TrebledJ在评论中写的关于使用if的选项更好。因为while循环比if执行更多的操作(SETUP_LOOP,+POP_BLOCK)。该函数速度较慢。|offset|length|style|CODE|2745373|def+factorial(n):
++++if++n+>=+1:
++++++++return+n+*+factorial(n+-+1)
++++return+1|2745374|timeit+-n+10000+-r+10|2745375|每个loop|2745376|if的|unordered-list-item|2745377|while+836µs±11.8µs每个环路787µs±7.22µs+|2745378|2745379|2745380|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/10239789/trebledj^0|0|0|K|2|U|5|12|2|1C|L|2|8|0|0|0|0|0|0|2|0|0|5|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|19|8|@$I|1A|J|1B|K|L]|$I|1C|J|1D|K|L]|$I|1E|J|1F|K|L]|$I|1G|J|1H|K|L]]|9|@$I|1I|J|1J|1|1K]]|A|$]]|$1|M|3|N|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1M|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1N|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1O|8|@$I|1P|J|1Q|K|L]]|9|@]|A|$]]|$1|V|3|W|5|U|7|1R|8|@$I|1S|J|1T|K|L]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|10|$11|$5|12|13|14|A|$15|16]]]]

And for the first time calculate the factorial using recursive and the while loop.

<pre><code>def factorial(n):
 while n &gt;= 1:
 return n * factorial(n - 1)
 return 1
</code></pre>

Although the option that <a href="https://stackoverflow.com/users/10239789/trebledj">TrebledJ</a> wrote in the comments about using <code>if</code> is better. Because <code>while</code> loop performs more operations (<code>SETUP_LOOP, POP_BLOCK</code>) than <code>if</code>. The function is slower. 

<pre><code>def factorial(n):
 if n &gt;= 1:
 return n * factorial(n - 1)
 return 1
</code></pre>

timeit -n 10000 -r 10 

<ul>
<li><code>while</code> 836 µs ± 11.8 µs per loop</li>
<li><code>if</code> 787 µs ± 7.22 µs per loop</li>
</ul>

blocks|key|2745399|text|在递归函数中总是有某种类型的循环和一些停止代码来停止循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2745400|public+int+recursivefactorial(int+number)
{
++++if(number==1)
++++++++return+1;
++++else
++++++++return+recursivefactorial(number-1)*number;
}|code-block|syntax|javascript|2745401|正如您所看到的，满足if条件会导致实际结束“循环”的代码，这是递归函数最重要的部分。相反，条件的else部分导致再次调用recursivefactorial函数，这实际上是一种循环。|offset|length|style|CODE|2745402|entityMap^0|0|0|A|2|1C|4|1O|I|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

There is always some kind of a loop in recursive functions and some stoping codes which stop the loop:

<pre><code>public int recursivefactorial(int number)
{
 if(number==1)
 return 1;
 else
 return recursivefactorial(number-1)*number;
}
</code></pre>

As you can see the fulfilling the <code>if</code> condition leads to the code that actually ends the "loop" and this is the most important part of a recursive function. In contrast, the <code>else</code> part of the condition leads to calling <code>recursivefactorial</code> function once again which is effectively a kind of loop.

blocks|key|51508|text|再加一个=)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|51509|#FAC+calculation

def+fakulteta(x):
++++if+x!=1:
+++++++++return+x*fakulteta(x-1)
++++return+1


print+(fakulteta(12))|code-block|syntax|javascript|51510|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|$]]

One more =)

<pre><code>#FAC calculation

def fakulteta(x):
 if x!=1:
 return x*fakulteta(x-1)
 return 1


print (fakulteta(12))
</code></pre>

How can I combine these two functions into one recursive function to have this result:
<pre><code>factorial(6)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
</code></pre>
This is the current code for my factorial function:
<pre><code>def factorial(n):
 if n &lt; 1: # base case
 return 1
 else:
 return n * factorial(n - 1) # recursive call


def fact(n):
 for i in range(1, n+1 ):
 print &quot;%2d! = %d&quot; % (i, factorial(i))
</code></pre>
and the output that this code produces is the following:
<pre><code>fact(6)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
</code></pre>
As you see, the execution of these two functions gives me correct answers, but I just wanted to simplify the two functions to a single recursive function.

recursive factorial function

Python

如何将这两个函数组合成一个递归函数才能得到这样的结果：factorial(6)1! = 12! = 23! = 64! = 245! = 1206! = 720这是我阶乘函数的当前代码：def factorial(n): if n < 1: # base case return 1 else: return n * factorial(n - 1) # rec

问递归阶乘函数
EN

回答 14

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问递归阶乘函数EN

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问递归阶乘函数
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