Python，Pandas：获得列表中第一个None位置的更好方法，它给出了最大的连续无计数

Question

I have lists that contain None like the following lists.

l1 = [None, 1, None, None, 2, None, None]
l2 = [None, 1, 1, None, None, None, 2, None, None]

I want to get the first None position in this list which gives the maximum consecutive None count.

get_start_None_pos(l1) # should return 2
get_start_None_pos(l2) # should return 3

My current approach with Pandas which works fine but it too slow when I have so many lists to deal with.

def get_start_None_pos(l: list) -> int:
    s = pd.Series(l)
    s = s.isna()
    s = s.cumsum() - s.cumsum().where(~s).ffill().fillna(0)
    return int(s.idxmax() - s.max() + 1)

I would like to know, is there any better way to solve something like this?

Answer 1

Here's one with NumPy -

def maxconsecNone_start(l):
    a = np.isnan(np.asarray(l, dtype=np.float64))
    a1 = np.r_[False,a,False]
    idx = np.flatnonzero(a1[:-1] != a1[1:])
    return idx[2*(idx[1::2]-idx[::2]).argmax()]

Sample runs -

In [49]: l1
Out[49]: [None, 1, None, None, 2, None, None]

In [50]: l2
Out[50]: [None, 1, 1, None, None, None, 2, None, None]

In [51]: maxconsecNone_start(l1)
Out[51]: 2

In [52]: maxconsecNone_start(l2)
Out[52]: 3

Answer 2

itertools.groupby

l=[list(y) for x,y in itertools.groupby(l2)]
x=max([(x,y)for x , y in enumerate(l) if all(v is None for v in y)], key = lambda x: len(x[1]))
sum(list(map(len,l[:x[0]])))
Out[465]: 3