Numpy:找到掩码边缘的索引

问题描述

我正在尝试查找掩码段的 indeces.例如:

I'm trying to find indeces of masked segments. For example:

mask = [1, 0, 0, 1, 1, 1, 0, 0]
segments = [(0, 0), (3, 5)]

当前的解决方案看起来像这样(并且非常很慢，因为我的掩码包含数百万个数字):

Current solution looks like this (and it's very slow, because my mask contains millions of numbers):

segments = []
start = 0
for i in range(len(mask) - 1):
    e1 = mask[i]
    e2 = mask[i + 1]
    if e1 == 0 and e2 == 1:
        start = i + 1
    elif e1 == 1 and e2 == 0:
        segments.append((start, i))

有没有办法用 numpy 有效地做到这一点?

Is there any way to do this efficiently with numpy?

我唯一能用谷歌搜索的是 numpy.ma.notmasked_edges，但它看起来不像我需要的.

The only thing that i've managed to google is numpy.ma.notmasked_edges, but it doesn't look like what i need.

推荐答案

这是一种方法 -

def start_stop(a, trigger_val):
    # "Enclose" mask with sentients to catch shifts later on
    mask = np.r_[False,np.equal(a, trigger_val),False]

    # Get the shifting indices
    idx = np.flatnonzero(mask[1:] != mask[:-1])

    # Get the start and end indices with slicing along the shifting ones
    return zip(idx[::2], idx[1::2]-1)

样品运行 -

In [216]: mask = [1, 0, 0, 1, 1, 1, 0, 0]

In [217]: start_stop(mask, trigger_val=1)
Out[217]: [(0, 0), (3, 5)]

使用它来获取0s -

In [218]: start_stop(mask, trigger_val=0)
Out[218]: [(1, 2), (6, 7)]

100000x 上的时序放大数据大小 -

Timings on 100000x scaled up datasize -

In [226]: mask = [1, 0, 0, 1, 1, 1, 0, 0]

In [227]: mask = np.repeat(mask,100000)

# Original soln
In [230]: %%timeit
     ...: segments = []
     ...: start = 0
     ...: for i in range(len(mask) - 1):
     ...:     e1 = mask[i]
     ...:     e2 = mask[i + 1]
     ...:     if e1 == 0 and e2 == 1:
     ...:         start = i + 1
     ...:     elif e1 == 1 and e2 == 0:
     ...:         segments.append((start, i))
1 loop, best of 3: 401 ms per loop

# @Yakym Pirozhenko's soln
In [231]: %%timeit
     ...: slices = np.ma.clump_masked(np.ma.masked_where(mask, mask))
     ...: result = [(s.start, s.stop - 1) for s in slices]
100 loops, best of 3: 4.8 ms per loop

In [232]: %timeit start_stop(mask, trigger_val=1)
1000 loops, best of 3: 1.41 ms per loop

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