获取 python 中列表中下一个非零值的索引

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，获得下一个非零索引的最佳方法是什么？ 例如，如果 index 是 2，那么下一个非零的一个是 3； 如果为 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

使用过滤器查找列表中的第一个非零项

i+list(filter(lambda x: x[1]!=0, enumerate(arr[i:])))[0][0]
# for i=5, the result is 8

---

常规循环

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

def firstNonZero(arr,i):
  for k,v in enumerate(arr[i:]):
      if v:
        break
  else :
      return None
  return k+i

print(firstNonZero(arr,5)) # 8
print(firstNonZero(arr,10)) # None

Answer 3

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

从指定索引开始遍历列表，直到找到一个不为零的数字，如果索引超出范围或找到非零返回None

Answer 4

您也可以使用 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 5

你可以有一个 class 存储所有“有效”索引（非零值），当你想从给定索引开始查找非零值的下一个索引时，你只需搜索下一个有效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))

output：

4
There is no index non-zero greater than 4