python获取列表中数字开始和结束位置的索引

Question

我有一个query，它是一个数字列表。我想获取出现数字1 的索引范围。范围从1 出现时开始，到它没有出现的索引处结束。我举了一个例子来说明这一点。

query= [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]

answer = [[5,8],[9,10],[12,14]]

注意：我不是在 Python 的列表中寻找某个值的第一个和最后一个索引。我正在寻找它们开始和结束的所有地方。

更新：从下面的一些建议答案看来，Itertools 对这些东西非常方便。

Answer 1

您可以使用itertools.dropwhile 来执行此操作。

>>> query = [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]
>>> n = 1
>>>
>>> from itertools import dropwhile
>>> itr = enumerate(query)
>>> [[i, next(dropwhile(lambda t: t[1]==n, itr), [len(query)])[0]] for i,x in itr if x==n]
[[5, 8], [9, 10], [12, 14]]

Answer 2

您也可以为此使用itertools.groupby。使用enumerate 获取索引，然后groupby 实际值，然后按值过滤，最后从组中获取第一个和最后一个索引。

>>> from itertools import groupby
>>> query = [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]
>>> [(g[0][0], g[-1][0]+1) for g in (list(g) for k, g in
...   groupby(enumerate(query), key=lambda t: t[1]) if k == 1)]
...
[(5, 8), (9, 10), (12, 14)]

Answer 3

query= [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]
first = 0 # Track the first index in the current group
ingroup = False # Track whether we are currently in a group of ones
answer = []
for i, e in enumerate(query):
    if e:
        if not ingroup:
            first = i
    else:
        if ingroup:
            answer.append([first, i])
    ingroup = e
if ingroup:
    answer.append([first, len(query)])

>>> answer
[[5, 8], [9, 10], [12, 14]]

我想你可能想要这样的东西。

Answer 4

你可以只使用基本的 for 循环和 if 语句你在哪里检查 其中“0”系列变为“1”系列，反之亦然

query= [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]

r_0 = []
r_1 = []

for i in range(len(query)-1):
    if query[i] == 0 and query[i+1] == 1:
        r_0.append(i+1) # [5, 9, 12]
    if query[i] == 1 and query[i + 1] == 0:
        r_1.append(i + 1) # [8, 10, 14]

print (list(zip(r_0,r_1)))

输出：

[(5, 8), (9, 10), (12, 14)]

Answer 5

希望这会有所帮助。这是一个没有循环的解决方案

from itertools import chain

query = [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]

result = list(zip(
    filter(lambda i: query[i] == 1 and (i == 0 or query[i-1] != 1), range(len(query))),
    chain(filter(lambda i: query[i] != 1 and query[i-1] == 1, range(1, len(query))), [len(query)-1])
))
print(result)

输出是：

[(2, 3), (5, 8), (9, 10), (12, 14)]

Answer 6

想分享一个递归的方法

query= [0,0,0,0,0,1,1,1,0,1,0,0,1,1,0]

def findOccurrences(of, at, index=0, occurrences=None):
    if occurrences == None: occurrences = [] # python has a weird behavior over lists as the default
                                           # parameter, unfortunately this neets to be done

    try:
        last = start = query.index(of, index)

        for i in at[start:]:
            if i == of:
                last += 1
            else:
                break

        occurrences.append([start, last])
        return findOccurrences(of, at, last, occurrences)
    except:
        pass
    return occurrences


print(findOccurrences(1, query))
print(findOccurrences(1, query, 0)) # Offseting
print(findOccurrences(0, query, 9)) # Offseting with defaul list