timings_consecutive_nans.py

from itertools import groupby
import math

import numpy as np
import numba as nb


def Divakar(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()

def Divakar2(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()
    
@nb.njit
def mine(arr):
    max_ = 0
    current = 0
    idx = 0
    while idx < arr.size:
        while idx < arr.size and math.isnan(arr[idx]):
            current += 1
            idx += 1
        if current > max_:
            max_ = current
        current = 0
        idx += 1
    return max_

def Tagc_special(sequence):
    # This doesn't work with numpy arrays
    return max((sum(1 for _ in group) for key, group in groupby(sequence) if key is nan), default=0)

def Tagc_other(sequence):
    # This doesn't work with numpy arrays
    return max((sum(1 for _ in group) for key, group in groupby(np.isnan(sequence)) if key), default=0)

def pltrdy(arr):
    count, max_count = 0, 0
    for e in arr:
        if np.isnan(e):
            count += 1
            max_count = max(max_count, count)
        else:
            count = 0

    return max_count

def Kasramvd(arr):
    arr = np.isnan(arr)
    return max(i.size for i in np.split(arr, np.where(np.diff(arr) != 0)[0]+1) if i.any())

from numpy import nan
arr = np.random.rand(10000)

arr[np.random.choice(range(len(arr)),size=1000,replace=0)] = np.nan

mine(arr)  # numba needs one call to compile the function, so do a warmup

%timeit mine(arr)
%timeit Divakar2(arr)
%timeit Divakar(arr)
%timeit Tagc_special(arr)
%timeit Tagc_other(arr)
%timeit Kasramvd(arr)
%timeit pltrdy(arr)