如何将 numpy 2d 数组裁剪为非零值? [英] How to crop a numpy 2d array to non-zero values?

问题描述

假设我有一个像这样的二维布尔 numpy 数组:

Let's say i have a 2d boolean numpy array like this:

import numpy as np
a = np.array([
    [0,0,0,0,0,0],
    [0,1,0,1,0,0],
    [0,1,1,0,0,0],
    [0,0,0,0,0,0],
], dtype=bool)

通常如何将其裁剪为包含所有 True 值的最小框(矩形、内核)?

How can i in general crop it to the smallest box (rectangle, kernel) that includes all True values?

所以在上面的例子中:

b = np.array([
    [1,0,1],
    [1,1,0],
], dtype=bool)

推荐答案

这是一个使用切片和 argmax 来获取边界的方案 -

Here's one with slicing and argmax to get the bounds -

def smallestbox(a):
    r = a.any(1)
    if r.any():
        m,n = a.shape
        c = a.any(0)
        out = a[r.argmax():m-r[::-1].argmax(), c.argmax():n-c[::-1].argmax()]
    else:
        out = np.empty((0,0),dtype=bool)
    return out

样品运行 -

In [142]: a
Out[142]: 
array([[False, False, False, False, False, False],
       [False,  True, False,  True, False, False],
       [False,  True,  True, False, False, False],
       [False, False, False, False, False, False]])

In [143]: smallestbox(a)
Out[143]: 
array([[ True, False,  True],
       [ True,  True, False]])

In [144]: a[:] = 0

In [145]: smallestbox(a)
Out[145]: array([], shape=(0, 0), dtype=bool)

In [146]: a[2,2] = 1

In [147]: smallestbox(a)
Out[147]: array([[ True]])

基准测试

其他方法 -

def argwhere_app(a): # @Jörn Hees's soln
    coords = np.argwhere(a)
    x_min, y_min = coords.min(axis=0)
    x_max, y_max = coords.max(axis=0)
    return a[x_min:x_max+1, y_min:y_max+1]

不同程度稀疏性的时间(大约 10%、50% 和 90%)-

Timings for varying degrees of sparsity (approx. 10%, 50% & 90%) -

In [370]: np.random.seed(0)
     ...: a = np.random.rand(5000,5000)>0.1

In [371]: %timeit argwhere_app(a)
     ...: %timeit smallestbox(a)
1 loop, best of 3: 310 ms per loop
100 loops, best of 3: 3.19 ms per loop

In [372]: np.random.seed(0)
     ...: a = np.random.rand(5000,5000)>0.5

In [373]: %timeit argwhere_app(a)
     ...: %timeit smallestbox(a)
1 loop, best of 3: 324 ms per loop
100 loops, best of 3: 3.21 ms per loop

In [374]: np.random.seed(0)
     ...: a = np.random.rand(5000,5000)>0.9

In [375]: %timeit argwhere_app(a)
     ...: %timeit smallestbox(a)
10 loops, best of 3: 106 ms per loop
100 loops, best of 3: 3.19 ms per loop

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