具有可变繁殖力的斐波那契凡人兔

Question

我正在尝试修改 Fibonacci 凡人兔子的 python 代码，以便根据它们的年龄改变兔子的繁殖力。举个例子吧。

我的兔子3个月后成熟，6个月后死亡。在它们 4 个月的生育期中，它们会根据年龄产生不同数量的后代。 3个月大时生2对小兔，4个月大时生3对小兔，以此类推直到第6个月。每对兔子由一雌一雄组成。最后我会计算对的数量而不是个人的数量。从出生到死亡的生育力值：

fecundity = [0, 0, 2, 3, 3, 1]

我使用的 python 代码 (https://github.com/jschendel/Rosalind/blob/master/011_FIBD.py) 是：

n = 12
m = 6
#n = months to run
#m = how many months the rabbits live

# Populate the initial rabbits.
Rabbits = [1]+[0]*(m-1)

# Calculate the new rabbits (bunnies), in a given month.
# Start at use range(1,n) since our initial population is 0 month old.
for month in range(1, n):
    Bunnies = 0
    # Get the number of Rabbits able to old enough to give birth.
    for j in range(1,m):
        Bunnies += Rabbits[(month-j-1)%m]
    # Bunnies replace the old rabbits who died.
    Rabbits[(month)%m] = Bunnies

# Total rabbits is the sum of the living rabbits.
Total_Rabbits = sum(Rabbits)

我不确定如何实现繁殖力的变化。感谢您的帮助！

谢谢，瓦伦蒂娜

Answer 1

定义你的生育力数组在一只兔子死亡时停止:

fecundity = [0, 0, 2, 3, 3, 1]

表示您的兔子在 7 个月大时死亡。之后，你只要写一个递归函数来计算特定步骤中new_borns的个数。我将步骤初始化为 1 对用于步骤 0，0 对用于步骤 < 0。您当然可以更改它以适合您的情况。（我所说的一步是一个时间单位，这里是月份）。这是函数：

def new_borns(step):
    if step < 0:
        return 0
    if step == 0:
        return 1
    nb_newborns = 0
    # We create a loop on living pairs
    for old_step in range(1, len(fecondity) +1):
        nb_newborns += (fecundity[old_step -1]) * new_borns(step - old_step)
    return nb_newborns

特定步骤的总人口是前面步骤的 new_borns 的总和，仍然活着（即，对于您的繁殖力数组的长度）。

def FIBD(step):
    population = 0
    for i in range(len(fecundity)):
        population += new_borns(step - i)
    return population

要知道你在第7步有多少对，只需调用FIBD(7)兔子可以活的月数就是繁殖力数组的长度。

当然，这个递归函数非常非常慢而且很糟糕。您需要一个缓存系统来避免多次计算同一步骤。这是要写入的完整文件。

#!/usr/bin/env python

fecundity = [0, 0, 2, 3, 3, 1]
new_borns_cache = [1]

def new_borns(step):
    if step < 0:
        return 0
    try :
        return new_borns_cache[step]
    except IndexError:
        if step == 0:
            return 1
        sum = 0
        for old_step in range(1, len(fecundity) +1):
            sum += (fecundity[old_step -1]) * new_borns(step - old_step)
        return sum

def fibd(step):
    population = 0
    for i in range(len(fecundity)):
        population += new_borns(step - i)
    return population

要使用它，只需导入，然后调用 fibd(7)

Answer 2

我自己得出答案，我真的修改了我之前发布的代码。我觉得现在简单多了

import numpy as np

m = 15
n = 18
fecundity = np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 2, 1, 1, 1])
Bunnies = np.array([0]*m)
Rabbits = np.array([1]+[0]*(m-1))

for month in range(0, 18):
    # every month I shift the list of 1 since they're getting older
    Rabbits = np.roll(Rabbits, 1)
    # I set the newborns as 0
    Rabbits[0] = 0
    # I calculate the newborns
    Bunnies = Rabbits * fecundity
    # and then I assign them to the rabbits 0 month old
    Rabbits[0] = sum(Bunnies)

# the sum of Rabbits is the number of final pairs of Rabbits after n months
Total_Rabbits = sum(Rabbits)
# 26 Rabbits

具有可变繁殖力的斐波那契凡人兔

Fibonacci mortal rabbits with variable fecundity

python

fibonacci

rosalind