Python使用while循环操作元组列表

Question

我正在尝试编写一个 python 函数，该函数使用 while 循环来遍历包含日期和在每个日期行走的步数的元组的输入列表。我的函数的两个输入是列表和一些必需的步骤。该函数需要遍历列表并计算达到所需步数所需的天数。下面是我到目前为止的代码，它适用于初始部分，但是我还需要添加它，以便如果在元组的输入列表中未达到所需的步骤数，则该函数返回“无”。另一件事是我只能有一个返回语句，因为问题说明了，这也是我坚持下去的部分原因。

def days_to_reach_n_steps(step_records, n):
  """DOCSTRING"""
  total_steps = 0
  counter = 0
  while total_steps < n:
      total_steps = total_steps + step_records[counter][1]
      counter = counter + 1
  return(counter)

我正在测试我的函数的示例，这个特定示例应该返回 None，因为列表中的步骤从未达到或超过 50。

step_records = [('2010-01-01',3), ('2010-01-02',2), ('2010-01-03',1)] days = days_to_reach_n_steps(step_records, 50) print(days)

Answer 1

在我看来，最简单的解决方案是这里的for 循环，因为从根本上说您想要的是迭代。您还可以使用enumerate 跟踪当天：

sample_steps = [("2010-01-1", 1),
                ("2010-01-2", 3),
                ("2010-01-3", 5),
                ("2010-01-4", 7),
                ("2010-01-5", 9),
                ("2010-01-6", 11)]

def days_to_reach_n_steps(step_records, n):
    total_steps = 0
    for counter, (date, steps) in enumerate(step_records, 1):
        total_steps += steps
        if total_steps >= n:
            return counter, date

我选择了奇数作为步骤的顺序，因为它们的累积和是平方（更容易用肉眼检查）：

for boundary in range(1, 7):
    for steps in range(boundary ** 2 - 1, boundary ** 2 + 2):
        result = days_to_reach_n_steps(sample_steps, steps)
        if result:
            days, date = result
            print("{} steps in {} days (arrived at {})".format(steps, days, date))
        else:
            print("{} was unreached".format(steps))

这将返回：

0 steps in 1 days (arrived at 2010-01-1)
1 steps in 1 days (arrived at 2010-01-1)
2 steps in 2 days (arrived at 2010-01-2)
3 steps in 2 days (arrived at 2010-01-2)
4 steps in 2 days (arrived at 2010-01-2)
5 steps in 3 days (arrived at 2010-01-3)
8 steps in 3 days (arrived at 2010-01-3)
9 steps in 3 days (arrived at 2010-01-3)
10 steps in 4 days (arrived at 2010-01-4)
15 steps in 4 days (arrived at 2010-01-4)
16 steps in 4 days (arrived at 2010-01-4)
17 steps in 5 days (arrived at 2010-01-5)
24 steps in 5 days (arrived at 2010-01-5)
25 steps in 5 days (arrived at 2010-01-5)
26 steps in 6 days (arrived at 2010-01-6)
35 steps in 6 days (arrived at 2010-01-6)
36 steps in 6 days (arrived at 2010-01-6)
37 was unreached

请注意，days_to_reach_n_steps 只有一个return 语句，但仍设法将return None 用于37。这是因为一个不返回任何内容的函数隐含地返回None。但是，这与您对 0 的规范不太匹配。如果您希望 0 异常，我建议您这样做：

for counter, (date, steps) in enumerate([("start", 0)] + step_records):

答案的第一行会变成

0 steps in 0 days (arrived at start)

这维护了算法的其余部分，因此您无需编写边缘情况。

如果它必须是一个while循环，你可以稍微滑稽地把for循环改写成这样：

def days_to_reach_n_steps(step_records, n):
    total_steps = 0
    counter = 0
    step_records = [("start", 0)] + step_records
    while counter < len(step_records):
        date, steps = step_records[counter]
        total_steps += steps
        if total_steps >= n:
            return counter, date
        counter += 1

这与 for 循环方法的第二次迭代完全相同（$ diff <(python while.py) <(python code.py) 干净地退出）。

要使其类似于 for 循环的第一次迭代，请删除对 step_records 的重新分配并返回 counter + 1。

请注意，这并不是一个很好的 while 循环应用程序——也许是为了练习使用 while 循环，但我真的不赞成强制使用丑陋的代码——Python 已经有简单的习惯用法来迭代列表和保持索引。见the Zen of Python。

Answer 2

在IndexError 退出并在未到达时将c 设置为None：

def daystoreach(steprecords, n):
total = 0
x = 0
while total < n:
    try:
        total += sr[x][1]
        x += 1
    except IndexError:
        break
if total < n:
    x = None
return x