Question

如果列表包含特定的初始元素

，我如何从列表中提取元素

这是列表数据= [10， 16,2 ，45,52,12,0,0,2,0,54,85,23,15,48,78,45 ，16,16,0,4， 16,3 ，16,2,0，....]

示例数据： -

1)   [10,16,2,45,52,12,0,0,2,0,78,45,16,16,0,4,16,3,16,2,0,....] (upto 200bytes)
2)   [16,2,24,14,45,25,87,89,23,0,0,5,10,16,0,4,16,3,16,2,0,....]
3)   [11,12,24,14,45,25, 16,2, 121,45,0,10,16,0,4,16,3,16,2,0,....]

首先我必须搜索 16,2 元素并提取 16,2 之后的所有元素直到它找到元素 16,3 并丢弃之后的每个重复帧

根据数据流，初始字节可以在列表的开头，也可以从流中的任何位置开始

Answer 1

您可以使用zip函数创建对并获取开始和结束对的索引，使用切片来提取预期元素：

>>> def extracter(lst,start,end):
...     pairs = zip(lst,lst[1:])
...     start_index = pairs.index(tuple(start))+2
...     end_index = pairs.index(tuple(end))
...     return lst[start_index:end_index]
...

演示：

>>> lst = [10, 16,2, 45,52,12,0,0,2,0,78,45,16,16,0,4, 16,3, 16,2,0]
>>> extracter(lst,(16,2),(16,3))
[45, 52, 12, 0, 0, 2, 0, 78, 45, 16, 16, 0, 4]

Answer 2

这应该是相当有效的，虽然有点冗长

from itertools import islice

def n_grams(a, n):
    z = (islice(a, i, None) for i in range(n))
    return zip(*z)

def filterStartStop(lst, start, stop):
    it = n_grams(lst, 3)
    for i in it:
        if i[0:2] == start:
            break
    for i in it:
        if i[1:3] == stop:
            break
        yield i[1]

a = range(10)
print(list(filterStartStop(a, (1,2), (7,8))))
>>[3, 4, 5, 6]

与@Kasramvd解决方案相比。

a = range(10**5)
start = (100, 101)
stop = (12345,12346)
%timeit list(filterStartStop(a, start, stop))
%timeit list(extracter(a, start, stop))

b = list(filterStartStop(a, start, stop))
c =  list(extracter(a, start, stop))
print (all(i == j for i,j in zip(b,c)))

>>100 loops, best of 3: 6.18 ms per loop
>>100 loops, best of 3: 19.4 ms per loop
>>True

Answer 3

我认为我找到了解决方案。

def findIn(l):
    foundBeginning=False
    foundEnding=False
    for i in range(len(l)):
        if l[i] == 16:
            if l[i+1] == 2:
                if foundBeginning == False and foundEnding == False:
                    begin = i+2
                    foundBeginning = True
            if l[i+1] == 3:
                if foundBeginning == True and foundEnding == False:
                    end = I
                    foundEnding = True
    return(l[begin:end])

如何在python 2.7中从列表中迭代和提取数据

3 个答案: