在numpy数组中查找连续的重复nan

这是一种方法-

def max_repeatedNaNs(a):    # Mask of NaNs    mask = np.concatenate(([False],np.isnan(a),[False]))    if ~mask.any():        return 0    else:        # Count of NaNs in each NaN group. Then, get max count as o/p.        c = np.flatnonzero(mask[1:] < mask[:-1]) -  np.flatnonzero(mask[1:] > mask[:-1])        return c.max()

这是一个改进的版本-

def max_repeatedNaNs_v2(a):    mask = np.concatenate(([False],np.isnan(a),[False]))    if ~mask.any():        return 0    else:        idx = np.nonzero(mask[1:] != mask[:-1])[0]        return (idx[1::2] - idx[::2]).max()

基准测试

@pltrdy's comment

In [77]: a = np.random.rand(10000)In [78]: a[np.random.choice(range(len(a)),size=1000,replace=0)] = np.nanIn [79]: %timeit contiguous_NaN(a) #@pltrdy's solution100 loops, best of 3: 15.8 ms per loopIn [80]: %timeit max_repeatedNaNs(a)10000 loops, best of 3: 103 µs per loopIn [81]: %timeit max_repeatedNaNs_v2(a)10000 loops, best of 3: 86.4 µs per loop