Python 获取numpy数组中具有最小长度的相同项序列的范围

考虑一个数组，其中的条目完全由-1或1组成。如何获取包含1且最小长度为

t

（例如

t=3

）的所有片的范围

例如：

>>>a=np.array([-1,-1,1,1,1,1,1,-1,1,-1,-1,1,1,1,1], dtype=int)
>>> a
array([-1, -1,  1,  1,  1,  1,  1, -1,  1, -1, -1,  1,  1,  1,  1])

然后，

t=3

的期望输出将是

[（2,7）、（11,15）]

一种使用和的方法-

---

我不太了解numpy，但使用简单的函数不是更好吗

def slices(a, t):
    start = None
    i = 0 # index into array
    slices = [] 
    for val in a:
        if a[i] == 1: # start of sequence
            if start is None:
                start = i
        else: # -1 end of sequence
            if start is not None:
                if i - start >= t: # check sequence for minimum size
                    slices.append((start, i))
                start = None
        i += 1

    # if sequence of 1's doesn't end with -1 within array
    if start is not None:
        if i - start >= t:
            slices.append((start, i))

   return slices

---

@科琳娜很有趣，所以没问题！

def slices(a, t):
    start = None
    i = 0 # index into array
    slices = [] 
    for val in a:
        if a[i] == 1: # start of sequence
            if start is None:
                start = i
        else: # -1 end of sequence
            if start is not None:
                if i - start >= t: # check sequence for minimum size
                    slices.append((start, i))
                start = None
        i += 1

    # if sequence of 1's doesn't end with -1 within array
    if start is not None:
        if i - start >= t:
            slices.append((start, i))

   return slices