从数组中提取单独的非零块

Question

有一个这样的数组，例如：

[1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1]

在 Python 中获取组织在列表中的非零元素的最快方法是什么，其中每个元素都包含连续非零值块的索引？

这里的结果将是一个包含许多数组的列表：

([0, 1, 2, 3], [9, 10, 11], [14, 15], [20, 21])

Answer 1

>>> L = [1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1]
>>> import itertools
>>> import operator
>>> [[i for i,value in it] for key,it in itertools.groupby(enumerate(L), key=operator.itemgetter(1)) if key != 0]

[[0, 1, 2, 3], [9, 10, 11], [14, 15], [20, 21]]

Answer 2

看看scipy.ndimage.measurements.label:

import numpy as np
from scipy.ndimage.measurements import label

x = np.asarray([1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1])
labelled, numfeats = label(x)
indices = [np.nonzero(labelled == k) for k in np.unique(labelled)[1:]]

indices 完全符合您的要求。请注意，根据您的最终目标，labelled 可能还会为您提供有用的（额外）信息。

Answer 3

我在 Finding the consecutive zeros in a numpy array 的回答的一个微不足道的变化给出了函数 find_runs:

def find_runs(value, a):
    # Create an array that is 1 where a is `value`, and pad each end with an extra 0.
    isvalue = np.concatenate(([0], np.equal(a, value).view(np.int8), [0]))
    absdiff = np.abs(np.diff(isvalue))
    # Runs start and end where absdiff is 1.
    ranges = np.where(absdiff == 1)[0].reshape(-1, 2)
    return ranges

例如，

In [43]: x
Out[43]: array([1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1])

In [44]: find_runs(1, x)
Out[44]: 
array([[ 0,  4],
       [ 9, 12],
       [14, 16],
       [20, 22]])

In [45]: [range(*run) for run in find_runs(1, x)]
Out[45]: [[0, 1, 2, 3], [9, 10, 11], [14, 15], [20, 21]]

如果您的示例中的值 1 不具有代表性，并且您确实希望运行任何非零值（如问题文本所建议的），您可以更改 np.equal(a, value)到 (a != 0) 并适当地更改参数和注释。例如

def find_nonzero_runs(a):
    # Create an array that is 1 where a is nonzero, and pad each end with an extra 0.
    isnonzero = np.concatenate(([0], (np.asarray(a) != 0).view(np.int8), [0]))
    absdiff = np.abs(np.diff(isnonzero))
    # Runs start and end where absdiff is 1.
    ranges = np.where(absdiff == 1)[0].reshape(-1, 2)
    return ranges

例如，

In [63]: y
Out[63]: 
array([-1,  2, 99, 99,  0,  0,  0,  0,  0, 12, 13, 14,  0,  0,  1,  1,  0,
        0,  0,  0, 42, 42])

In [64]: find_nonzero_runs(y)
Out[64]: 
array([[ 0,  4],
       [ 9, 12],
       [14, 16],
       [20, 22]])

Answer 4

你可以使用np.split，一旦你知道非零长度的区间和A中的相应索引。假设 A 作为输入数组，实现看起来像这样 -

# Append A on either sides with zeros
A_ext = np.diff(np.hstack(([0],A,[0])))

# Find interval of non-zeros lengths
interval_lens = np.where(A_ext==-1)[0] - np.where(A_ext==1)[0]

# Indices of non-zeros places in A
idx = np.arange(A.size)[A!=0]

# Finally split indices based on the interval lengths
out = np.split(idx,interval_lens.cumsum())[:-1]

样本输入、输出-

In [53]: A
Out[53]: array([1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1])

In [54]: out
Out[54]: [array([0, 1, 2, 3]), array([ 9, 10, 11]), array([14, 15]), array([20, 21])]

从数组中提取单独的非零块

Extract separate non-zero blocks from array

python

numpy

extract