recursive factorial function

Question

How can I combine these two functions into one recursive function to have this result:

factorial(6)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

This is the current code for my factorial function:

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))

and the output that this code produces is the following:

fact(6)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

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.

Don't. It looks fine the way it is. Combining them will just make things more difficult. — FrustratedWithFormsDesigner, Dec 21 '10 at 18:08
@ FrustratedWithFormsDesigner: last year exam ... hahah .... I wish I could take you guys with me to write my exam for me but it's not possible :P — user531225, Dec 21 '10 at 18:14
The asker had possibly graduated since the question was set. Anyway, I hope the teacher who wanted them to implement the *factorial recursively* told them that the efficiency of the recursive solution is so terrible that it should never be allowed. :) — pepr, Apr 25 '19 at 07:49

score 31 · Accepted Answer · edited Jun 30 '21 at 16:42

31

We can combine the two functions to this single recursive function:

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

edited Jun 30 '21 at 16:42

Alan Bagel

818
5
24

answered Dec 21 '10 at 18:13

pythonFoo

2,194
3
15
17

score 26 · Answer 2 · answered May 06 '14 at 17:52

26

2 lines of code:

def fac(n):
    return 1 if (n < 1) else n * fac(n-1)

Test it:

print fac(4)

Result:

answered May 06 '14 at 17:52

martynas

12,120
3
55
60

score 7 · Answer 3 · answered Dec 21 '10 at 18:12

7

def factorial(n):
    result = 1 if n <= 1 else n * factorial(n - 1)
    print '%d! = %d' % (n, result)
    return result

answered Dec 21 '10 at 18:12

Will McCutchen

13,047
3
44
43

score 5 · Answer 4 · answered Dec 17 '12 at 22:29

5

a short one:

def fac(n):
    if n == 0:
        return 1
    else:
        return n * fac(n-1)
print fac(0)

answered Dec 17 '12 at 22:29

Ilyas

59
1
1

score 4 · Answer 5 · edited Jan 29 '12 at 00:17

4

I've no experience with Python, but something like this?

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

edited Jan 29 '12 at 00:17

Tadeck

132,510
28
152
198

answered Dec 21 '10 at 18:11

Mchl

61,444
9
118
120

I'm not 100% sure that this is correct, but since OP said it's for an exam, I won't go into any further details... – FrustratedWithFormsDesigner Dec 21 '10 at 18:12

score 4 · Answer 6 · answered Dec 21 '10 at 18:11

try this:

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.

score 3 · Answer 7 · answered Dec 21 '10 at 18:14

3

Is this homework by any chance?

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 PEP227 a read for more details. The short of it is that Python lets you define functions within functions.

answered Dec 21 '10 at 18:14

D.Shawley

58,213
10
98
113

@D.Shawley: This is quite inefficient solution, as you calculate factorial(1) `n` times, factorial(2) `n-1` times, factorial(3) `n-2` times and so on... – Tadeck Jan 29 '12 at 00:20

score 2 · Answer 8 · answered Mar 04 '16 at 10:13

2

fac = lambda x: 1 if x == 0 else x * fac(x - 1)

answered Mar 04 '16 at 10:13

T-kin-ter

99
9

score 1 · Answer 9 · edited Jun 25 '21 at 17:48

1

One more

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))

edited Jun 25 '21 at 17:48

zganger

3,572
1
10
13

answered Jul 21 '11 at 20:49

MattyW

1,519
3
16
33

1

Mathematically 0! evaluates to 1. So the first part of your conditional should be changed. – mcocdawc May 09 '16 at 22:22

score 0 · Answer 10 · edited Jun 30 '21 at 11:14

0

I don't really know the factorial of negative numbers, but this will work with all n >= 0:

def factorial(n):
    if n >= 0:
        if n == 1 or n == 0:
            return 1
        else:
            n = n * factorial(n-1)
            return n

    return 'error'

edited Jun 30 '21 at 11:14

Alan Bagel

818
5
24

answered Jun 28 '17 at 14:50

Salah Hamza

89
2
6

score 0 · Answer 11 · edited Jun 30 '21 at 09:15

0

Can use these 4 lines of code...

   def factorial(n):
        f = lambda n: n * f(n - 1) if n > 1 else 1

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

edited Jun 30 '21 at 09:15

0 _

10,524
11
77
109

answered Jun 04 '18 at 21:27

Meir Keller

511
1
7
21

2

This doesn't work. The second reference to `factorial()` should instead be `f()` – cdlane Jun 25 '21 at 19:00

Szczerski · Answer 12 · 2019-03-26T11:24:41.250

0

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

def factorial(n):
    while  n >= 1:
        return n * factorial(n - 1)
    return 1

Although the option that TrebledJ wrote in the comments about using if is better. Because while loop performs more operations (SETUP_LOOP, POP_BLOCK) than if. The function is slower.

def factorial(n):
    if  n >= 1:
        return n * factorial(n - 1)
    return 1

timeit -n 10000 -r 10

while 836 µs ± 11.8 µs per loop
if 787 µs ± 7.22 µs per loop

edited Mar 26 '19 at 11:24

answered Nov 13 '18 at 13:09

Szczerski

839
11
11

Although correct, the `while` is redundant as `return` will kick in on the first iteration (and no other iterations will be performed). Changing `while` to `if` is much better. – TrebledJ Mar 26 '19 at 10:00
What I meant by redundant was the communicative aspect... other coders seeing the function will see `while` and think: "Okay, it's factorial by looping"; then one line later they see `return` and realise it's actually factorial by recursion. (Usually, recursion is a substitute for loops.) And... ah, I see a benchmark. A small difference in performance between while and if, but your new content seems well researched. :-) – TrebledJ Mar 26 '19 at 11:35

score 0 · Answer 13 · answered Sep 01 '22 at 15:37

0

In Python 3.8 you can try factorial function itself.

import math
n=int(input())
print(math.factorial(n))

For example:

Input: 5

Output:120

answered Sep 01 '22 at 15:37

score -1 · Answer 14 · edited Apr 25 '19 at 07:39

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

public int recursivefactorial(int number)
{
    if(number==1)
        return 1;
    else
        return recursivefactorial(number-1)*number;
}

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

score -2 · Answer 15 · answered Apr 23 '18 at 11:37

-2

One more =)

#FAC calculation

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


print (fakulteta(12))

answered Apr 23 '18 at 11:37

XraySensei

167
7

1

you will have a recursion error as you don't handle 0 and below – imbatman Oct 16 '18 at 10:31

recursive factorial function

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