Question

我正在尝试编写一个函数，计算一种细菌为其初始种群达到最终种群所需的天数，因为每天它的种群数量增加一倍。我的代码是;

def num_doublings(initial_population, final_population):
   time = 0
   while initial_population < final_population:
       time+= 1
       initial_population = initial_population * 2
       print(time)

这会打印每一天，但我希望它只打印while循环停止所需的迭代总次数。使用num_doublings(2, 8)调用函数时，我的代码输出是

1
2
3

但我只需输出3.如何解决此问题？

Answer 1

你可以试试这个：

def num_doublings(initial_population, final_population):
    time = 0
    while initial_population <= final_population:
        time+= 1
        initial_population = initial_population * 2

    return time

该函数将返回时间变量，在您给出的示例中为3

Answer 2

正如Martijn Pieters正确观察到的那样，高性能解决方案根本不使用循环：

from math import log, ceil

def num_doublings(initial, final):
    return ceil(log(final/initial, 2))

Answer 3

您只需要在while循环外拉print行。像这样 -

def num_doublings(initial_population, final_population):
   time = 1
   while initial_population < final_population:
       time+= 1
       initial_population = initial_population * 2
   print(time)

请注意，我初始化time =1

Answer 4

是范围问题，只需在modifyStock(id: number, quantity: number) { if (id == null || quantity == null) { return Observable.throw('Error'); } const url = `/item/${id}/modifyStock`; const body = { quantity: quantity }; return this.http.put(url, body); }循环结束时打印数据：

while

Answer 5

取消缩进最后一行。

def num_doublings(initial_population, final_population):
   time = 0
   while initial_population < final_population:
       time+= 1
       initial_population = initial_population * 2
   print(time)

Python - While循环简单

5 个答案: