Python使用while循環操作元組列表

Question

我正在嘗試編寫一個python函數，它使用while循環來遍歷包含日期和步數在每個日期上行走的元組的輸入列表。我的功能的兩個輸入是列表和一些必需的步驟。該功能需要通過列表並計算達到所需步驟所需的時間。下面是我的代碼到目前爲止，它適用於初始部分，但我也需要添加到它，以便如果元組的輸入列表中沒有達到所需的步數，該函數返回'None'。另一件事是，我只能有一個返回語句，因爲這個問題說明了這一點，這也是我爲什麼被卡住的原因。

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) 

的什麼，我測試我的函數，這個特殊的例子應該返回None在列表中的步驟，不要老是達到或超過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

這似乎是一個例子對我來說，最簡單的解決方案是for循環，因爲基本上你想要的是迭代。您還可以使用enumerate保持一天的軌跡：

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 

我選擇了奇數的步驟順序作爲其累加的平方（使得它更容易用肉眼檢查）：

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)) 

這將返回此：

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 

注意days_to_reach_n_steps只有一個return聲明，但仍設法returnNone爲37。這是因爲隱含的函數不返回任何東西會返回None。然而，這並不完全匹配您的規範0我建議這樣做，如果你想0是例外：

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

答案的第一行會變成

0 steps in 0 days (arrived at start) 

這保持算法的其餘部分，所以你不需要編程邊界情況。

如果它是一個while循環，你可以稍微開玩笑改寫爲循環成這樣：

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 

這工作完全一樣的for循環方法的第二次迭代（$ diff <(python while.py) <(python code.py)乾淨地退出） 。

爲了使其適應類似於for循環的第一次迭代，請將重新分配移除到step_records並返回counter + 1。

請注意，這不是一個真正的while循環的應用程序 - 也許其目的是通過while循環獲得練習，但我並不真的贊成強制醜陋的代碼 - Python已經有了用於遍歷列表的簡單成語並保持指數。見the Zen of Python。

Answer 2

在IndexError退出並設置c到None時未達到：

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