python - 根据间隙大小更改 numpy 数组中的间隙

Question

我需要过滤掉零之间的短非零系列。例如，这个数组:

t = np.array([1, 3, 1, 0, 0, 1, 8, 3, 0, 8, 2, 4, 7, 0,0,4,1])

应该变成:

array([1, 3, 1, 0, 0, 0, 0, 0, 0, 8, 2, 4, 7, 0, 0, 4, 1])

我找到了非零序列的第一个索引，并计算了它们之间的非零数。我写了以下内容，它有效，但看起来很糟糕。我试过工作人员但有错误。
如何用pythonicly重写它？

minseq = 4  # length of minimal non zero seq
p = np.where(fhr>0, 1, 0).astype(int)
s = np.array([1]+ list(np.diff(p)))
sind = np.where(s==1)[0][1:]
print(sind)
    
for i in range(len(sind) - 1):
    s1 = sind[i]
    e1 = sind[i+1]
    
    subfhr = np.where(fhr[s1:e1] > 0, 1, 0).sum()
    
    if (subfhr < minseq):
        
        print(s1, e1, subfhr)
        fhr[s1:e1] = 0

出去:

[ 5  9 15]
5 9 3
array([1, 3, 1, 0, 0, 0, 0, 0, 0, 8, 2, 4, 7, 0, 0, 4, 1])

Answer 1

您可以使用基于图像处理的 binary_closing ——

from scipy.ndimage.morphology import binary_closing

def remove_small_nnz(a, W):
    K = np.ones(W, dtype=int)
    m = a==0
    p = binary_closing(m,K)
    a[~m & p] = 0
    return a

sample 运行 -

In [97]: a
Out[97]: array([1, 3, 1, 0, 0, 1, 8, 3, 0, 8, 2, 4, 7, 0, 0, 4, 1])

In [98]: remove_small_nnz(a, W=3)
Out[98]: array([1, 3, 1, 0, 0, 1, 8, 3, 0, 8, 2, 4, 7, 0, 0, 4, 1])

In [99]: remove_small_nnz(a, W=4)
Out[99]: array([1, 3, 1, 0, 0, 0, 0, 0, 0, 8, 2, 4, 7, 0, 0, 4, 1])

In [100]: remove_small_nnz(a, W=5)
Out[100]: array([1, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1])