Question

我正在尝试修改Fibonacci凡人兔的python代码，以便根据他们的年龄改变兔子的繁殖力。让我们举个例子。

我的兔子在3个月后成熟，6个月后死亡。在他们4个月的繁殖期间，他们根据年龄产生不同数量的后代。当他们3个月大时生产2对兔子，4个月生产3对兔子等，直到第6个月。每对兔子由雌性和雄性形成。最后，我会计算对的数量，而不是个人的数量。从出生到死亡的繁殖价值：

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

我正在使用的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)

我不确定如何实施繁殖力的变化。任何帮助表示赞赏！

谢谢你，瓦伦蒂娜

Answer 1

定义你的生殖力阵列在兔子死亡时停止：

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

意味着您的兔子在7个月大时死亡。之后，您只需编写一个递归函数来计算特定步骤中new_borns的数量。我将步骤初始化为步骤0的1对，步骤0的步骤为0对。你当然可以改变它以适合你的情况。（我称之为步骤是一个单位的时间，这里是月份）。这是功能：

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

特定步骤的总人口数是前一步骤的新生儿总数，仍然存在（即，您的生育数组的长度）。

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

要了解您在步骤7中拥有多少对，只需致电FIBD(7)兔子可以居住的月数是繁殖阵列的长度。

当然，这种递归函数非常慢而且很糟糕。您需要一个缓存系统，以避免计算同一步骤的多次。这是要写的完整文件。

#!/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

我自己找到了答案，我真的修改了之前发布的代码。我认为现在它更简单了

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

Fibonacci凡人兔具有可变的繁殖力

2 个答案: