numpy数组：第一次出现N个小于阈值的连续值

Question

我有一个一维numpy数组-例如，

a = np.array([1, 4, 5, 7, 1, 2, 2, 4, 10])

我想获得N个后续值都低于某个值x的第一个数字的索引。

在这种情况下，对于N=3和x=3，我将搜索第一个数字，该数字后面的三个条目都小于3。这将是a[4]。

只需通过循环遍历所有值即可轻松实现这一点for，但是我想知道是否有更干净，更有效的方法来实现此目的。

Answer 1

方法1：

这是向量化NumPy的方式-

def start_valid_island(a, thresh, window_size):
    m = a<thresh
    me = np.r_[False,m,False]
    idx = np.flatnonzero(me[:-1]!=me[1:])
    lens = idx[1::2]-idx[::2]
    return idx[::2][(lens >= window_size).argmax()]

样品运行-

In [44]: a
Out[44]: array([ 1,  4,  5,  7,  1,  2,  2,  4, 10])

In [45]: start_valid_island(a, thresh=3, window_size=3)
Out[45]: 4

In [46]: a[:3] = 1

In [47]: start_valid_island(a, thresh=3, window_size=3)
Out[47]: 0

方法2：

与SciPy's binary-erosion-

from scipy.ndimage.morphology import binary_erosion

def start_valid_island_v2(a, thresh, window_size):
    m = a<thresh
    k = np.ones(window_size,dtype=bool)
    return binary_erosion(m,k,origin=-(window_size//2)).argmax()

方法3：

为了完成这套游戏，下面是一个基于短篇幅并使用numba-

from numba import njit

@njit
def start_valid_island_v3(a, thresh, window_size):
    n = len(a)
    out = None
    for i in range(n-window_size+1):
        found = True
        for j in range(window_size):
            if a[i+j]>=thresh:
                found = False
                break
        if found:
            out = i
            break
    return out

时间-

In [142]: np.random.seed(0)
     ...: a = np.random.randint(0,10,(100000000))

In [145]: %timeit start_valid_island(a, thresh=3, window_size=3)
1 loop, best of 3: 810 ms per loop

In [146]: %timeit start_valid_island_v2(a, thresh=3, window_size=3)
1 loop, best of 3: 1.27 s per loop

In [147]: %timeit start_valid_island_v3(a, thresh=3, window_size=3)
1000000 loops, best of 3: 608 ns per loop