CODEWITHSUNDEEP
pythonpython-3.xpython-itertoolsaccumulate
Dennis
Dennis

Reputation: 81

itertools accumulate to build a recursive list

I want to build a list from a seed in such a way: seed, seed*weight - seed*weight**2, and so on, where the previous number is the seed for the next.

so with seed of .5 and weight of 3 we would get

.5, -4, 24.0 and so on

This is what I thought would work:

from itertools import accumulate, repeat 

relationship = lambda seed, weight: seed*weight - seed*weight**2
seed = .5 
weight = 3
inputs = list(repeat(seed,10))
first = [x for x in accumulate(inputs, relationship)]

Is the problem that the inputs list creates a list of .5s, instead of populating the list forward with numbers that I want?

Upvotes: 0

Views: 428

Answers (1)

Blckknght
Blckknght

Reputation: 104762

Your accumulate call doesn't use the weight variable you've declared at all. Rather, it uses the first value from inputs as the first seed (which is as intended), and all the other values as weights (not what is intended).

You could make it work with a different inputs list:

inputs = [seed] + [weight]*10
results = list(accumulate(inputs, relationships))

But I'm not sure accumulate is really the best way to solve this problem. I'd think it would be more natural to solve the problem in another way, such as with an infinite generator function:

def gen(seed, weight):
    while True:
        seed = seed * weight - seed * weight**2
        yield seed

Since the series is a simple geometric progression (multiplying by weight - weight**2 each step), you could also write a function to calculate each value from a closed-form equation:

def calc(seed, weight, n):
    factor = weight - weight**2
    return seed * factor**(n+1)

Upvotes: 1

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