How do I print a fibonacci sequence to the nth number in Python?

问题

I have a homework assignment that I'm stumped on. I'm trying to write a program that outputs the fibonacci sequence up the nth number. Here's what I have so far:

def fib():
   n = int(input("Please Enter a number: "))

   if n == 1:
      return(1)
   elif n == 0:   
      return(0)            
   else:                      
      return (n-1) + (n-2)


mylist = range[0:n]
print(mylist)

I'm thinking I could use separate functions but I can't figure out how to pass the argument that calculates the fibonacci sequence. Then the next step would be to to print out the sequence of numbers up to that number.

回答1:

Non-recursive solution

def fib(n):
    cur = 1
    old = 1
    i = 1
    while (i < n):
        cur, old, i = cur+old, cur, i+1
    return cur

for i in range(10):
    print(fib(i))

Generator solution:

def fib(n):
    old = 0
    cur = 1
    i = 1
    yield cur
    while (i < n):
        cur, old, i = cur+old, cur, i+1
        yield cur

for f in fib(10):
    print(f)

Note that generator solution outperforms the non-recursive (and non-recursive outperforms recursive, if memoization is not applied to recursive solution)

One more time, recursive with memoization:

def memoize(func):
    memo = dict()
    def decorated(n):
        if n not in memo:
            memo[n] = func(n)
        return memo[n]

    return decorated

@memoize
def fib(n):
    #added for demonstration purposes only - not really needed
    global call_count
    call_count = call_count + 1
    #end demonstration purposes

    if n<=1:
        return 1
    else:
        return fib(n-1) + fib(n-2)

call_count = 0 #added for demonstration purposes only - not really needed
for i in range(100):
    print(fib(i))
print(call_count) #prints 100

This time each fibbonacci number calculated exactly once, and than stored. So, this solution would outperform all the rest. However, the decorator implementation is just quick and dirty, don't let it into production. (see this amazing question on SO for details about python decorators :)

So, having fib defined, the program would be something like (sorry, just looping is boring, here's some more cool python stuff: list comprehensions)

fib_n = int(input("Fib number?"))
fibs = [fib(i) for i in range(fib_n)]
print " ".join(fibs)

this prints all numbers in ONE line, separated by spaces. If you want each on it's own line - replace " " with "\n"

回答2:

def fibonacci(n):
  if n <= 1:
    return n
  else:
    return fibonacci(n-1) + fibonacci(n-2)

print(fibonacci(int(input())))

And since you want to print up to the nth number:

[print(fibonacci(n)) for n in range (int(input()))]

And for python2.7 change input to raw_input.

回答3:

Please note, in your call

You are not calling fib() recursively
You need a wrapper method so that the input is not requested every time the method is called recursively
You do not need to send in a list. Just the number n is good enough.

This method would only give you the nth number in the sequence. It does not print the sequence.

You need to return fib(n-1) + fib(n-2)

def f():
    n = int(input("Please Enter a number: "))
    print fib(n)

def fib(n):    
    if n == 0: 
        return 0
    elif n == 1: 
        return 1
    else: 
        return fib(n-1)+fib(n-2)

回答4:

def fib(n):
   if n == 1:
      return(1)
   elif n == 0:   
      return(0)            
   else:                      
      return fib(n-1) + fib(n-2)

my_num = int(input("Enter a number:"))
print fib(my_num)

Im not really sure what your question is... but the answer is probably something like this

回答5:

Separate functions would be best, as the recursive function would be far easier to deal with. On the other hand, you could code an iterative function that would take only one parameter

Recursively::

def fib(n):
    if n == 1:
        return (1);
    elif n == 0:
        return (0);
    else:
        return fib(n-1) + fib(n-2);

def callFib():
    n = int(raw_input('Enter n::\t'));
    mylist = fib(n);
    print mylist;

callFib();

Iteratively::

def fib():
    n = int(raw_input('Enter n::\t'));
    terms = [0,1];
    i=2;
    while i<=n:
        terms.append(terms[i-1] + terms[i-2]);
        i=i+1;
    print terms[n];

fib();

回答6:

for a recursive solution:

def fib(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n-1) + fib(n-2)
x=input('which fibonnaci number do you want?')
print fib(x)

explanation: if n is 0, then ofcourse the '0th' term is 0, and the 1st term is one. From here, you know that the next numbers will be the sum of the previous 2. That is what's inferred by the line after the else.

回答7:

This might be faster incase of long list

# Get nth Fibonacci number 
def nfib(nth):
  sq5 = 5**.5
  phi1 = (1+sq5)/2
  phi2 = -1 * (phi1 -1)
  resp = (phi1**(nth+1) - phi2**(nth+1))/sq5
  return long(resp)

for i in range(10):
  print i+1, ": ",  nfib(i)

OUTPUT

来源：https://stackoverflow.com/questions/15820601/how-do-i-print-a-fibonacci-sequence-to-the-nth-number-in-python

标签

python

parameter-passing

fibonacci