Note: I did search on SO, but could only find questions related to Printing nth Fibonacci number. e.g Fibonacci numbers, with an one-liner in Python 3?
I have the following working solutions:
A. Using lambda() + reduce():
>>> fib = lambda n: reduce(lambda x, _: x+[x[-1]+x[-2]], range(n-2), [0, 1]) >>> fib(10) >>> [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
Note: Not able to use x.append(x[-1]+x[-2]). Gives AttributeError (Don't know why)
B. Using lambda() + map(): (Have to use a variable for result)
>>> result = [0,1] >>> fib = lambda n: map(lambda _: result.append(result[-1] + result[-2]), xrange(n-2)) >>> fib(10) ## Ignore its output ## >>> result ## Stores the result ## >> [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
lambda() argument and treat reduce() as a simple iterator . The second problem is that fib(1) and fib(2) produce the same result but they shouldn't.
x.append(...) return None because is a operation in place of list, while reduce expect your the function to return a value that is going to be used in the next iteration, for that you can do this (x.append(...),x)[1]result until something consumes the result of map(). (E.g. try wrapping your map() in an any()) The second problem is that fib(1) and fib(2) produce the same result but they shouldn't.[:n] to the list [0,1] -> [0,1][:n]
fib = lambda n: reduce(lambda x, _: x+[x[-1]+x[-2]], range(n-1), [0, 1])[:-1] also works.Fibonacci using reduce() and lambda()
from functools import reduce
def fibonacci(count):
sequence = (0, 1)
for _ in range(2, count):
sequence += (reduce(lambda a, b: a + b, sequence[-2:]), )
return sequence[:count]
print(fibonacci(10))
OUTPUT
(0, 1, 1, 2, 3, 5, 8, 13, 21, 34)
Fibonacci using map() and lambda()
def fibonacci(count):
sequence = [0, 1]
any(map(lambda _: sequence.append(sum(sequence[-2:])), range(2, count)))
return sequence[:count]
print(fibonacci(10))
OUTPUT
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
fibonacci = lambda number: number if number <= 1 else fibonacci(number - 1) + fibonacci(number - 2);
listOfFibonacciNumbers = list(map(fibonacci, range(0, 20, 1)));
print(listOfFibonacciNumbers);
Output:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]
In place of 20, you can give your number. I hope this helps:)
yo can try this for Fibonacci using reduce() and lambda()
def Fib(count):
first =[0,1]
for i in range(0,count-2):
first.append(reduce(lambda x,y : x+y,first[-2:]))
print(first)
Fib(10)
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
Here's what might help you!
Implemented using reduce() and lambda
from functools import reduce
fibonacci_seq = [0, 1]
n = 10
reduce(lambda a, b: len(fibonacci_seq) < n and (fibonacci_seq.append(a+b) or a+b), fibonacci_seq)
(fibonacci_seq.append(a+b) or a+b) : as the <any_list>.append(ele) returns None, I'm using it to append the next element in the series to the fibonacci_seq. oring it with a+b allows to return a+b as the result to the reduce() function to operate on it with the next element in the sequence.
len(fibonacci_seq) < n : The list stops growing once the len(list) reaches n.
OUTPUT
print(fibonacci_seq)
>>> [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
The above method returns the right sequence for n = 2 onwards.
try this code which prints the first 10 Fibonacci numbers using LAMBDA
fib = lambda n: n if n<=1 else fib(n-1)+fib(n-2)
for i in range(10):print(fib(i))
