numpy 2D数组:获取所有连接的条目的索引并共享相同的值

时间:2018-04-11 09:44:41

标签: python arrays numpy indices

我有一个2D numpy数组,其中填充了从0到N的整数值,如何获得所有直接连接的条目的索引并共享相同的值。

补充:大多数条目为零,可以忽略!

示例输入数组:

[ 0 0 0 0 0 ]
[ 1 1 0 1 1 ]
[ 0 1 0 1 1 ]
[ 1 0 0 0 0 ]
[ 2 2 2 2 2 ]

希望的产出指数:

1: [ [1 0] [1 1] [2 1] [3 0] ] # first 1 cluster
   [ [1 3] [1 4] [2 3] [2 4] ] # second 1 cluster

2: [ [4 0] [4 1] [4 2] [4 3] [4 4] ] # only 2 cluster

输出数组的格式化并不重要,我只需要可以解决单个索引的分离值集群

我首先想到的是:

N = numberClusters
x = myArray

for c in range(N):
   for i in np.where(x==c):
         # fill output array with i

但这会错过具有相同值的聚类的分离

1 个答案:

答案 0 :(得分:1)

您可以使用skimage.measure.label(如果需要,可以使用pip install scikit-image安装):

import numpy as np
from skimage import measure

# Setup some data
np.random.seed(42)
img = np.random.choice([0, 1, 2], (5, 5), [0.7, 0.2, 0.1])
# [[2 0 2 2 0]
#  [0 2 1 2 2]
#  [2 2 0 2 1]
#  [0 1 1 1 1]
#  [0 0 1 1 0]]

# Label each region, considering only directly adjacent pixels connected
img_labeled = measure.label(img, connectivity=1)
# [[1 0 2 2 0]
#  [0 3 4 2 2]
#  [3 3 0 2 5]
#  [0 5 5 5 5]
#  [0 0 5 5 0]]

# Get the indices for each region, excluding zeros
idx = [np.where(img_labeled == label)
       for label in np.unique(img_labeled)
       if label]
# [(array([0]), array([0])),
#  (array([0, 0, 1, 1, 2]), array([2, 3, 3, 4, 3])),
#  (array([1, 2, 2]), array([1, 0, 1])),
#  (array([1]), array([2])),
#  (array([2, 3, 3, 3, 3, 4, 4]), array([4, 1, 2, 3, 4, 2, 3]))]

# Get the bounding boxes of each region (ignoring zeros)
bboxes = [area.bbox for area in measure.regionprops(img_labeled)]
# [(0, 0, 1, 1),
#  (0, 2, 3, 5),
#  (1, 0, 3, 2),
#  (1, 2, 2, 3),
#  (2, 1, 5, 5)]

可以使用非常有用的函数skimage.measure.regionprops找到边界框,其中包含有关区域的大量信息。对于边界框,它返回(min_row, min_col, max_row, max_col)元组,其中属于边界框的像素处于半开区间[min_row; max_row)[min_col; max_col)