根据相邻的差异值将列表分成组

Question

我对列表中的元素分组有以下问题。数字图像转换后，我将孔中心分开并将它们收集在 values 列表中，然后在计算相邻元素之间的差异后，我得到了 diff_ar .现在我想获取属于一个 group/cluster 的元素的索引。我假设一个部分中元素之间的最大差异应小于 3。此外，只有当内部至少有 7 个元素时才能创建组。结果，我期望元组列表包含索引开始和索引结束每个 od 检测到的组（本例中为 2）。

Image for better issue statement

values = [73.0, 143.0, 323.0, 324.0, 325.0, 325.0, 325.0, 325.0, 325.5,
          325.5, 326.0, 326.0, 326.0, 326.0, 406.0, 406.5, 432.5, 433.0,
          433.5, 434.5, 435.0, 435.0, 436.0, 436.5, 437.5, 438.0]

diff_ar = [70.0, 180.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.5, 0.0, 0.5, 0.0,
           0.0, 0.0, 80.0, 0.5, 26.0, 0.5, 0.5, 1.0, 0.5, 0.0, 1.0, 0.5,
           1.0, 0.5]

expected_output = [(2,12),(16,24)]

Answer 1

第一个解决方案：使用more_itertools.split_when.

import more_itertools

values = [73.0, 143.0, 323.0, 324.0, 325.0, 325.0, 325.0, 325.0, 325.5,
          325.5, 326.0, 326.0, 326.0, 326.0, 406.0, 406.5, 432.5, 433.0,
          433.5, 434.5, 435.0, 435.0, 436.0, 436.5, 437.5, 438.0]

threshold = 3
min_items = 7
groups = [(g[0][0], g[-1][0]) for g in more_itertools.split_when(enumerate(values), lambda x,y: abs(x[1]-y[1])>=threshold) if len(g) >= min_items]

print(groups)
# [(2, 13), (16, 25)]

第二种方案：自己写循环。

def split_values(values, threshold, min_items):
    result = []
    prev = values[0]
    last_cut = 0
    for i,x in enumerate(values[1:], start=1):
        if abs(x - prev) >= threshold:
            if i - last_cut >= min_items:
                result.append((last_cut, i-1))
            last_cut = i
        prev = x
    if len(values) - last_cut >= min_items:
        result.append((last_cut, len(values)-1))
    return result

values = [73.0, 143.0, 323.0, 324.0, 325.0, 325.0, 325.0, 325.0, 325.5,
          325.5, 326.0, 326.0, 326.0, 326.0, 406.0, 406.5, 432.5, 433.0,
          433.5, 434.5, 435.0, 435.0, 436.0, 436.5, 437.5, 438.0]

print(split_values(values, 3, 7))
# [(2, 13), (16, 25)]

根据相邻的差异值将列表分成组

Divide list into groups based on adjoining difference values

python

sorting

grouping

list