Fibonacci numbers, with an one-liner in Python 3?

I know there is nothing wrong with writing with proper function structure, but I would like to know how can I find nth fibonacci number with most Pythonic way with a one-line.

I wrote that code, but It didn't seem to me best way:

>>> fib = lambda n:reduce(lambda x, y: (x[0]+x[1], x[0]), [(1,1)]*(n-2))[0]
>>> f开发者_Go百科ib(8)
13

How could it be better and simplier?

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fib = lambda n:reduce(lambda x,n:[x[1],x[0]+x[1]], range(n),[0,1])[0]

(this maintains a tuple mapped from [a,b] to [b,a+b], initialized to [0,1], iterated N times, then takes the first tuple element)

>>> fib(1000)
43466557686937456435688527675040625802564660517371780402481729089536555417949051
89040387984007925516929592259308032263477520968962323987332247116164299644090653
3187938298969649928516003704476137795166849228875L

(note that in this numbering, fib(0) = 0, fib(1) = 1, fib(2) = 1, fib(3) = 2, etc.)

(also note: reduce is a builtin in Python 2.7 but not in Python 3; you'd need to execute from functools import reduce in Python 3.)

---

A rarely seen trick is that a lambda function can refer to itself recursively:

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

By the way, it's rarely seen because it's confusing, and in this case it is also inefficient. It's much better to write it on multiple lines:

def fibs():
    a = 0
    b = 1
    while True:
        yield a
        a, b = b, a + b

---

I recently learned about using matrix multiplication to generate Fibonacci numbers, which was pretty cool. You take a base matrix:

[1, 1]
[1, 0]

and multiply it by itself N times to get:

[F(N+1), F(N)]
[F(N), F(N-1)]

This morning, doodling in the steam on the shower wall, I realized that you could cut the running time in half by starting with the second matrix, and multiplying it by itself N/2 times, then using N to pick an index from the first row/column.

With a little squeezing, I got it down to one line:

import numpy

def mm_fib(n):
    return (numpy.matrix([[2,1],[1,1]])**(n//2))[0,(n+1)%2]

>>> [mm_fib(i) for i in range(20)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

---

This is a closed expression for the Fibonacci series that uses integer arithmetic, and is quite efficient.

fib = lambda n:pow(2<<n,n+1,(4<<2*n)-(2<<n)-1)%(2<<n)

>> fib(1000)
4346655768693745643568852767504062580256466051737178
0402481729089536555417949051890403879840079255169295
9225930803226347752096896232398733224711616429964409
06533187938298969649928516003704476137795166849228875L

It computes the result in O(log n) arithmetic operations, each acting on integers with O(n) bits. Given that the result (the nth Fibonacci number) is O(n) bits, the method is quite reasonable.

It's based on genefib4 from http://fare.tunes.org/files/fun/fibonacci.lisp , which in turn was based on an a less efficient closed-form integer expression of mine (see: http://paulhankin.github.io/Fibonacci/)

---

If we consider the "most Pythonic way" to be elegant and effective then:

def fib(nr):
    return int(((1 + math.sqrt(5)) / 2) ** nr / math.sqrt(5) + 0.5)

wins hands down. Why use a inefficient algorithm (and if you start using memoization we can forget about the oneliner) when you can solve the problem just fine in O(1) by approximation the result with the golden ratio? Though in reality I'd obviously write it in this form:

def fib(nr):
    ratio = (1 + math.sqrt(5)) / 2
    return int(ratio ** nr / math.sqrt(5) + 0.5)

More efficient and much easier to understand.

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This is a non-recursive (anonymous) memoizing one liner

fib = lambda x,y=[1,1]:([(y.append(y[-1]+y[-2]),y[-1])[1] for i in range(1+x-len(y))],y[x])[1]

---

fib = lambda n, x=0, y=1 : x if not n else fib(n-1, y, x+y)

run time O(n), fib(0) = 0, fib(1) = 1, fib(2) = 1 ...

---

I'm Python newcomer, but did some measure for learning purposes. I've collected some fibo algorithm and took some measure.

from datetime import datetime
import matplotlib.pyplot as plt
from functools import wraps
from functools import reduce
from functools import lru_cache
import numpy


def time_it(f):

    @wraps(f)
    def wrapper(*args, **kwargs):
        start_time = datetime.now()

        f(*args, **kwargs)

        end_time = datetime.now()
        elapsed = end_time - start_time
        elapsed = elapsed.microseconds

        return elapsed
    return wrapper


@time_it
def fibslow(n):
    if n <= 1:
        return n

    else:
        return fibslow(n-1) + fibslow(n-2)


@time_it
@lru_cache(maxsize=10)
def fibslow_2(n):
    if n <= 1:
        return n

    else:
        return fibslow_2(n-1) + fibslow_2(n-2)


@time_it
def fibfast(n):
    if n <= 1:
        return n

    a, b = 0, 1

    for i in range(1, n+1):
        a, b = b, a + b

    return a


@time_it
def fib_reduce(n):
    return reduce(lambda x, n: [x[1], x[0]+x[1]], range(n), [0, 1])[0]


@time_it
def mm_fib(n):
    return (numpy.matrix([[2, 1], [1, 1]])**(n//2))[0, (n+1) % 2]


@time_it
def fib_ia(n):
    return pow(2 << n, n+1, (4 << 2 * n) - (2 << n)-1) % (2 << n)


if __name__ == '__main__':

    X = range(1, 200)
#    fibslow_times = [fibslow(i) for i in X]
    fibslow_2_times = [fibslow_2(i) for i in X]
    fibfast_times = [fibfast(i) for i in X]
    fib_reduce_times = [fib_reduce(i) for i in X]
    fib_mm_times = [mm_fib(i) for i in X]
    fib_ia_times = [fib_ia(i) for i in X]

#    print(fibslow_times)
#    print(fibfast_times)
#    print(fib_reduce_times)

    plt.figure()
#    plt.plot(X, fibslow_times, label='Slow Fib')
    plt.plot(X, fibslow_2_times, label='Slow Fib w cache')
    plt.plot(X, fibfast_times, label='Fast Fib')
    plt.plot(X, fib_reduce_times, label='Reduce Fib')
    plt.plot(X, fib_mm_times, label='Numpy Fib')
    plt.plot(X, fib_ia_times, label='Fib ia')
    plt.xlabel('n')
    plt.ylabel('time (microseconds)')
    plt.legend()
    plt.show()

The result is usually the same.

Fiboslow_2 with recursion and cache, Fib integer arithmetic and Fibfast algorithms seems the best ones. Maybe my decorator not the best thing to measure performance, but for an overview it seemed good.

---

Another example, taking the cue from Mark Byers's answer:

fib = lambda n,a=0,b=1: a if n<=0 else fib(n-1,b,a+b)

---

I wanted to see if I could create an entire sequence, not just the final value.

The following will generate a list of length 100. It excludes the leading [0, 1] and works for both Python2 and Python3. No other lines besides the one!

(lambda i, x=[0,1]: [(x.append(x[y+1]+x[y]), x[y+1]+x[y])[1] for y in range(i)])(100)

Output

[1,
 2,
 3,
 ...
 218922995834555169026,
 354224848179261915075,
 573147844013817084101]

---

Here's an implementation that doesn't use recursion, and only memoizes the last two values instead of the whole sequence history.

nthfib() below is the direct solution to the original problem (as long as imports are allowed)

It's less elegant than using the Reduce methods above, but, although slightly different that what was asked for, it gains the ability to to be used more efficiently as an infinite generator if one needs to output the sequence up to the nth number as well (re-writing slightly as fibgen() below).

from itertools import imap, islice, repeat

nthfib = lambda n: next(islice((lambda x=[0, 1]: imap((lambda x: (lambda setx=x.__setitem__, x0_temp=x[0]: (x[1], setx(0, x[1]), setx(1, x0_temp+x[1]))[0])()), repeat(x)))(), n-1, None))    

>>> nthfib(1000)
43466557686937456435688527675040625802564660517371780402481729089536555417949051
89040387984007925516929592259308032263477520968962323987332247116164299644090653
3187938298969649928516003704476137795166849228875L

from itertools import imap, islice, repeat

fibgen = lambda:(lambda x=[0,1]: imap((lambda x: (lambda setx=x.__setitem__, x0_temp=x[0]: (x[1], setx(0, x[1]), setx(1, x0_temp+x[1]))[0])()), repeat(x)))()

>>> list(islice(fibgen(),12))
[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144]

---

def fib(n):
    x =[0,1]
    for i in range(n):
        x=[x[1],x[0]+x[1]]
    return x[0]

take the cue from Jason S, i think my version have a better understanding.

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Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), we can use and update a variable within a list comprehension:

fib = lambda n,x=(0,1):[x := (x[1], sum(x)) for i in range(n+1)][-1][0]

This:

• Initiates the duo n-1 and n-2 as a tuple x=(0, 1)
• As part of a list comprehension looping n times, x is updated via an assignment expression (x := (x[1], sum(x))) to the new n-1 and n-2 values
• Finally, we return from the last iteration, the first part of the x

---

To solve this problem I got inspired by a similar question here in Stackoverflow Single Statement Fibonacci, and I got this single line function that can output a list of fibonacci sequence. Though, this is a Python 2 script, not tested on Python 3:

(lambda n, fib=[0,1]: fib[:n]+[fib.append(fib[-1] + fib[-2]) or fib[-1] for i in range(n-len(fib))])(10)

assign this lambda function to a variable to reuse it:

fib = (lambda n, fib=[0,1]: fib[:n]+[fib.append(fib[-1] + fib[-2]) or fib[-1] for i in range(n-len(fib))])
fib(10)

output is a list of fibonacci sequence:

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

---

I don't know if this is the most pythonic method but this is the best i could come up with:->

Fibonacci = lambda x,y=[1,1]:[1]*x if (x<2) else ([y.append(y[q-1] + y[q-2]) for q in range(2,x)],y)[1]

The above code doesn't use recursion, just a list to store the values.

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My 2 cents

# One Liner
def nthfibonacci(n):
    return long(((((1+5**.5)/2)**n)-(((1-5**.5)/2)**n))/5**.5)

OR

# Steps
def nthfibonacci(nth):
    sq5 = 5**.5
    phi1 = (1+sq5)/2
    phi2 = -1 * (phi1 -1)
    n1 = phi1**(nth+1)
    n2 = phi2**(nth+1)
    return long((n1 - n2)/sq5)

---

Why not use a list comprehension?

from math import sqrt, floor
[floor(((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))) for n in range(100)]

Without math imports, but less pretty:

[int(((1+(5**0.5))**n-(1-(5**0.5))**n)/(2**n*(5**0.5))) for n in range(100)]

---

Similar:

    def fibonacci(n):
      f=[1]+[0]
      for i in range(n):
        f=[sum(f)] + f[:-1]
      print f[1]

---

A simple Fibonacci number generator using recursion

fib = lambda x: 1-x if x < 2 else  fib(x-1)+fib(x-2)
print fib(100)

This takes forever to calculate fib(100) in my computer.

There is also closed form of Fibonacci numbers.

fib = lambda n: int(1/sqrt(5)*((1+sqrt(5))**n-(1-sqrt(5))**n)/2**n)
print fib(50)

This works nearly up to 72 numbers due to precision problem.

---

Lambda with logical operators

fibonacci_oneline = lambda n = 10, out = []: [ out.append(i) or i if i <= 1 else out.append(out[-1] + out[-2]) or out[-1] for i in range(n)]

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here is how i do it ,however the function returns None for the list comprehension line part to allow me to insert a loop inside .. so basically what it does is appending new elements of the fib seq inside of a list which is over two elements

>>f=lambda list,x :print('The list must be of 2 or more') if len(list)<2 else [list.append(list[-1]+list[-2]) for i in range(x)]
>>a=[1,2]
>>f(a,7)

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You can generate once a list with some values and use as needed:

fib_fix = []

fib = lambda x: 1 if x <=2 else fib_fix[x-3] if x-2 <= len(fib_fix) else (fib_fix.append(fib(x-2) + fib(x-1)) or fib_fix[-1])

fib_x = lambda x: [fib(n) for n in range(1,x+1)]

fib_100 = fib_x(100)

than for example:

a = fib_fix[76]

---

import math

sqrt_five = math.sqrt(5)
phi = (1 + sqrt_five) / 2

fib = lambda n : int(round(pow(phi, n) / sqrt_five))

print([fib(i) for i in range(1, 26)])

single line lambda fibonacci but with some extra variables