如何在numpy中找到数组中连续元素的组？

Question

我必须从numpy数组中聚集连续元素.考虑以下示例

    a = [ 0, 47, 48, 49, 50, 97, 98, 99]

输出应该是一个元组列表,如下所示

   [(0),(47, 48, 49, 50),(97, 98, 99)]

这里差别只是一个.元素之间.如果差异也可以指定为限制或硬编码数字,那将是很好的.

非常感谢.

Answer 1

def consecutive(data, stepsize=1):
    return np.split(data, np.where(np.diff(data) != stepsize)[0]+1)

a = np.array([0, 47, 48, 49, 50, 97, 98, 99])
consecutive(a)

产量

[array([0]), array([47, 48, 49, 50]), array([97, 98, 99])]

Answer 2

这是一个可能有帮助的lil功能:

def group_consecutives(vals, step=1):
    """Return list of consecutive lists of numbers from vals (number list)."""
    run = []
    result = [run]
    expect = None
    for v in vals:
        if (v == expect) or (expect is None):
            run.append(v)
        else:
            run = [v]
            result.append(run)
        expect = v + step
    return result

>>> group_consecutives(a)
[[0], [47, 48, 49, 50], [97, 98, 99]]
>>> group_consecutives(a, step=47)
[[0, 47], [48], [49], [50, 97], [98], [99]]

Answer 3

(a[1:]-a[:-1])==1 将产生一个布尔数组,False表示运行中断.您也可以使用内置的numpy.grad.

Answer 4

测试一维数组

获取diff不为一的地方

diffs = numpy.diff(array) != 1

获取 diff 的索引，获取第一个维度并对所有维度加一，因为diff与前一个索引进行比较

indexes = numpy.nonzero(diffs)[0] + 1

使用给定索引进行拆分

groups = numpy.split(array, indexes)

Answer 5

这是我到目前为止所提出的:不确定是100%正确

import numpy as np
a = np.array([ 0, 47, 48, 49, 50, 97, 98, 99])
print np.split(a, np.cumsum( np.where(a[1:] - a[:-1] > 1) )+1)

收益:

>>>[array([0]), array([47, 48, 49, 50]), array([97, 98, 99])]