Python 递归阶乘函数
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recursive factorial function
提问by user531225
how can I combine these two functions in to one recursive function to have this result:
我怎样才能将这两个函数组合成一个递归函数来得到这个结果：
factorial(6)
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
these are the codes
这些是代码
def factorial( n ):
if n <1: # base case
return 1
else:
return n * factorial( n - 1 ) # recursive call
def fact(n):
for i in range(1, n+1 ):
print "%2d! = %d" % ( i, factorial( i ) )
fact(6)
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
as you see execution of these two gives a correct answer, I just want to make it to one recursive function.
当你看到这两个的执行给出了正确的答案时，我只想把它变成一个递归函数。
采纳答案by pythonFoo
def factorial( n ):
if n <1: # base case
return 1
else:
returnNumber = n * factorial( n - 1 ) # recursive call
print(str(n) + '! = ' + str(returnNumber))
return returnNumber
回答by Mchl
I've no experience with Python, but something like this?
我没有使用 Python 的经验，但类似这样的事情？
def factorial( n ):
if n <1: # base case
return 1
else:
f = n * factorial( n - 1 ) # recursive call
print "%2d! = %d" % ( n, f )
return f
回答by Vinay Pandey
try this:
尝试这个：
def factorial( n ):
if n <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
One thing I noticed is that you are returning '1' for n<1, that means your function will return 1 even for negative numbers. You may want to fix that.
我注意到的一件事是，对于 n<1，您将返回 '1'，这意味着即使对于负数，您的函数也会返回 1。你可能想解决这个问题。
回答by Will McCutchen
def factorial(n):
result = 1 if n <= 1 else n * factorial(n - 1)
print '%d! = %d' % (n, result)
return result
回答by D.Shawley
Is this homework by any chance?
这是作业吗？
def traced_factorial(n):
def factorial(n):
if n <= 1:
return 1
return n * factorial(n - 1)
for i in range(1, n + 1):
print '%2d! = %d' %(i, factorial(i))
Give PEP227a read for more details. The short of it is that Python lets you define functions within functions.
给PEP227读更多的细节。简而言之，Python 允许您在函数内定义函数。
回答by MattyW
One more
多一个
def fact(x):
if x == 0:
return 0
elif x == 1:
return 1
else:
return x * fact(x-1)
for x in range(0,10):
print '%d! = %d' %(x, fact(x))
回答by Ilyas
a short one:
一个简短的：
def fac(n):
if n == 0:
return 1
else:
return n * fac(n-1)
print fac(0)
回答by martynas
2 lines of code:
2行代码：
def fac(n):
return 1 if (n < 1) else n * fac(n-1)
Test it:
测试一下：
print fac(4)
Result:
结果：
24
回答by T-kin-ter
fac = lambda x: 1 if x == 0 else x * fac(x - 1)
回答by Salah Hamza
I don't really know the factorial of negative numbers, but this will work with all n >= 0:
我真的不知道负数的阶乘，但这适用于所有 n >= 0：
def factorial(n):
if n >= 0:
if n == 1 or n==0:
return 1
else:
n = n * factorial(n-1)
return n
else:
return 'error'
