Ruby 阶乘函数

There is no factorial function in the standard library.

It's not in the standard library but you can extend the Integer class.

class Integer
def factorial_recursive
self <= 1 ? 1 : self * (self - 1).factorial
end
def factorial_iterative
f = 1; for i in 1..self; f *= i; end; f
end
alias :factorial :factorial_iterative
end

N.B. Iterative factorial is a better choice for obvious performance reasons.

You could also use Math.gamma function which boils down to factorial for integer parameters.

I just wrote my own:

def fact(n)
if n<= 1
1
else
n * fact( n - 1 )
end
end

Also, you can define a falling factorial:

def fall_fact(n,k)
if k <= 0
1
else
n*fall_fact(n - 1, k - 1)
end
end

Shamelessly cribbed from http://rosettacode.org/wiki/Factorial#Ruby, my personal favorite is

class Integer
def fact
(1..self).reduce(:*) || 1
end
end


>> 400.fact
=> 64034522846623895262347970319503005850702583026002959458684445942802397169186831436278478647463264676294350575035856810848298162883517435228961988646802997937341654150838162426461942352307046244325015114448670890662773914918117331955996440709549671345290477020322434911210797593280795101545372667251627877890009349763765710326350331533965349868386831339352024373788157786791506311858702618270169819740062983025308591298346162272304558339520759611505302236086810433297255194852674432232438669948422404232599805551610635942376961399231917134063858996537970147827206606320217379472010321356624613809077942304597360699567595836096158715129913822286578579549361617654480453222007825818400848436415591229454275384803558374518022675900061399560145595206127211192918105032491008000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

This implementation also happens to be the fastest among the variants listed in Rosetta Code.

update #1

Added || 1 to handle the zero case.

update #2

With thanks and appreciation to Mark Thomas, here's a version that is a bit more efficient, elegant and obscure:

class Integer
def fact
(2..self).reduce(1,:*)
end
end

Like this is better

(1..n).inject(:*) || 1

class Integer
def factorial
return self < 0 ? false : self==0 ? 1 : self.downto(1).inject(:*)
#Not sure what other libraries say, but my understanding is that factorial of
#anything less than 0 does not exist.
end
end

Using Math.gamma.floor is an easy way to produce an approximation and then round it back down to the correct integer result. Should work for all Integers, include an input check if necessary.

I would do

(1..n).inject(1, :*)

Just another way to do it, although it really isn't necessary.

class Factorial
attr_reader :num
def initialize(num)
@num = num
end


def find_factorial
(1..num).inject(:*) || 1
end
end


number = Factorial.new(8).find_factorial
puts number

And yet another way (=

def factorial(number)
number = number.to_i
number_range = (number).downto(1).to_a
factorial = number_range.inject(:*)
puts "The factorial of #{number} is #{factorial}"
end
factorial(#number)

You will probably find a Ruby feature request useful. It contains a nontrivial patch that includes a demo Bash script. The speed difference between a naive loop and the solution presented in the batch can be literally 100x (hundred fold). Written all in pure Ruby.

Just call this function

def factorial(n=0)
(1..n).inject(:*)
end

examples

factorial(3)
factorial(11)

Here is my version seems to be clear to me even though it's not as clean.

def factorial(num)
step = 0
(num - 1).times do (step += 1 ;num *= step) end
return num
end

This was my irb testing line that showed each step.

num = 8;step = 0;(num - 1).times do (step += 1 ;num *= step; puts num) end;num

class Integer
def !
(1..self).inject(:*)
end
end

examples

!3  # => 6
!4  # => 24

In math, factorial of n is just the gamma function of n+1
(see: http://en.wikipedia.org/wiki/Gamma_function)

Ruby has Math.gamma() so just use Math.gamma(n+1) and cast it back to an integer if desired.

Just one more way to do it:

# fact(n) => Computes the Factorial of "n" = n!


def fact(n) (1..n).inject(1) {|r,i| r*i }end


fact(6) => 720

Why would the standard library require a factorial method, when there is a built-in iterator for this exact purpose? It is called upto.

No you do not need to use recursion, like all these other answers show.

def fact(n)
n == 0 ? 1 : n * fact(n - 1)
end

Rather, the built-in iterator upto can be used to calculate factorials:

factorial = 1
1.upto(10) {|x| factorial *= x }
factorial
=> 3628800

With high respect to all who participated and spent their time to help us, I would like to share my benchmarks of the solutions listed here. Params:

iterations = 1000

n = 6

                                     user     system      total        real
Math.gamma(n+1)                   0.000383   0.000106   0.000489 (  0.000487)
(1..n).inject(:*) || 1            0.003986   0.000000   0.003986 (  0.003987)
(1..n).reduce(1, :*)              0.003926   0.000000   0.003926 (  0.004023)
1.upto(n) {|x| factorial *= x }   0.003748   0.011734   0.015482 (  0.022795)

For n = 10

  user     system      total        real
0.000378   0.000102   0.000480 (  0.000477)
0.004469   0.000007   0.004476 (  0.004491)
0.004532   0.000024   0.004556 (  0.005119)
0.027720   0.011211   0.038931 (  0.058309)

In Ruby standard library function for factorial is not available. We can make a simple function of factorial in ruby in this way.

def factorial_number(n)
if n <= 1
1
else
n * factorial_number(n-1)
end
end


puts factorial_number(6) #Output is 720   => (6*5*4*3*2*1)
puts factorial_number(8) #Output is 40320 => (8*7*6*5*4*3*2*1)