Question

假设您有一个列表[1,2,3,4,5,0,0,0,7,8, ...]。给定当前索引，什么是获取下一个非零索引的最佳方法？例如，如果index为2，则下一个非零的值为3；否则为0。如果是4，则为8。

Answer 1

a = [1,2,3,4,5,0,0,0,7,8]

def nextZeroIndex(array, start):
    nextItems = array[start+1:]
    for i, x in enumerate(nextItems):
        if x != 0:
            return 1 + start + i

print(nextZeroIndex(a, 2)) # 3
print(nextZeroIndex(a, 5)) # 8
print(nextZeroIndex(a, 4)) # 8

Answer 2

def get_next_index(lst, index):
    try:
        return next(i + index + 1 for i, n in enumerate(lst[index + 1:]) if n != 0)
    except StopIteration:
        return None

lst = [1, 2, 3, 4, 5, 0, 0, 0, 7, 8]
get_next_index(lst, 2)  # 3
get_next_index(lst, 9)  # None
get_next_index(lst, 100)  # None

从指定的索引开始遍历列表，直到找到一个不同于零的数字，如果索引超出范围或发现非零，则返回None

Answer 3

您还可以使用itertools实现此目的：

from itertools import compress,count
start = 4
next = list(compress(count(),l[start+1:]))[0]+start+1
print(next)

output:

8

Answer 4

您可以拥有一个用于存储所有“有效”索引（非零值）的类，当您要查找从给定索引开始的非零值的下一个索引时，只需搜索下一个有效索引，该索引大于您的起始索引

from bisect import bisect_right

class NextIndexNonZero:
    def __init__(self, your_numbers: list):

        # filter out the indices for zero values/numbers
        valid_indices = filter(lambda x: x[1] != 0, enumerate(your_numbers))
        self.valid_indices = list(map(lambda x: x[0], valid_indices))

        self.max_index = self.valid_indices[-1]


    def find(self, start_index: int) -> int:

        if start_index >= self.max_index:
            return f'There is no index non-zero greater than {start_index}'

        next_index = bisect_right(self.valid_indices, start_index)
        return self.valid_indices[next_index]

a = [1, 0, 0, 0, 1, 0, 0 ]
n = NextIndexNonZero(a)
print(n.find(2))
print(n.find(4))

输出：

4
There is no index non-zero greater than 4

获取Python列表中下一个非零值的索引

4 个答案: