Question

我有一个像这样的列表

mixed_list = [None, 1, 3, None, 5, 6, 7, 8, 10, None, None, 11, 12, None, None]

我需要找到此列表中的最后一个数字，其中连续的数字大于n，在这种情况下为2。所以...

输入：mixed_list 输出：10

我知道我必须编写一个循环，但是在此之后不知道如何开始。有人可以告诉我如何入门吗？

Answer 1

首先尝试reverse列表，然后从其开始iterating

标记找到的first number，并在其后开始连续计数

If在计数器大于None之前得到n，然后重置计数器，标记next number after None并继续迭代。

Else，您标记的号码是答案：）

下面的代码：

def find_the_number(the_list, n):
    counter = 0
    possible_answer = None
    for i in reversed(the_list):
        if i is not None:
            if counter == 0:
                possible_answer = i
            counter += 1
        else:
            counter = 0
        if counter > n:
            return possible_answer


mixed_list = [None, 1, 3, None, 5, 6, 7, 8, 10, None, None, 11, 12, None, None]
cons_number = 2
print(find_the_number(mixed_list, cons_number))

Answer 2

>>> next((v[-1] for v in reversed(list(zip(*[mixed_list[i:] for i in range(n+1)]))) if all(v)), None)
>>> 10

说明

zip(*[mixed_list[i:] for i in range(n+1)将返回n + 1个连续的数字作为元组

>>> list(zip(*[mixed_list[i:] for i in range(n+1)]))
[(None, 1, 3), (1, 3, None), (3, None, 5), (None, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 10), (8, 10, None), (10, None, None), (None, None, 11), (None, 11, 12), (11, 12, None), (12, None, None)]

然后您将其反转

>>> list(reversed(list(zip(*[mixed_list[i:] for i in range(n+1)]))))
[(12, None, None), (11, 12, None), (None, 11, 12), (None, None, 11), (10, None, None), (8, 10, None), (7, 8, 10), (6, 7, 8), (5, 6, 7), (None, 5, 6), (3, None, 5), (1, 3, None), (None, 1, 3)]

然后，仅在元组包含所有数字时才对其进行过滤，并仅从元组中返回第一个数字

>>> [v[-1] for v in reversed(list(zip(*[mixed_list[i:] for i in range(n+1)]))) if all(v)]
[10, 8, 7]

您所要做的就是从返回的列表中获取第一个电话号码：）

Answer 3

类似的事情应该起作用。警告：未经测试，只需在此处输入

minimum_consecutives = 2
mixed_list = [None, 1, 3, None, 5, 6, 7, 8, 10, None, None, 11, 12, None, None]
consecutive_non_nulls = 0
last_item = None
for item in mixed_list:
   if item is not None:
       consecutive_non_nulls = consecutive_non_nulls + 1
   else:
       if consecutive_non_nulls > minimum_consecutives:
           break;
       consecutive_non_nulls = 0
   last_item = item
print(last_item)

Answer 4

我会在倒排的列表上使用itertools.groupby。

from itertools import groupby

n = 2
lst = [None, 1, 3, None, 5, 6, 7, 8, 10, None, None, 11, 12, None, None]
groups = groupby(reversed(lst), lambda x: isinstance(x, int))
result = next((grplist[0] for p, grp in groups
                          for grplist in [list(grp)] 
                          if p and len(grplist) > n), None)

Answer 5

我接受了答案，但意识到他想念我，但由于Igor S，我修改了他的代码，这项工作奏效了。

def get_last_number2(list, n=2):
    counter = 0
    possible_answer = None

    for i in list:
      if i is not None:
         possible_answer = i
         counter +=1
      else:
         if counter>n:
            return possible_answer
         counter = 0

在列表中找到连续数字大于“ n”的最后一个数字

5 个答案: