Python 对列中的数字进行迭代，得到n次重复开始的第一个行号

pandas approach

假设本例的阈值为4(而不是280):

df = pd.DataFrame({'differences': [0,0,0,0,0,1,2,0,3,0,0,0,0,0,0,4,0,5]})

    differences
0             0  # 0: first stretch of >4
1             0
2             0
3             0
4             0
5             1
6             2
7             0
8             3
9             0  # 9: second stretch of >4
10            0
11            0
12            0
13            0
14            0
15            4
16            0
17            5

您可以执行groupby.size来过滤groupby.first:

thresh = 4

m = df['differences'].eq(0)
group = (~m).cumsum().to_numpy()

g = df.reset_index()[m].groupby(group[m])
g.size()

out = g['index'].first()[g.size()>thresh].to_numpy()

输出:array([ 0, 9])

loop approach

lst = [0, 0, 0, 0, 0, 1, 2, 0, 3, 0, 0, 0, 0, 0, 0, 4, 0, 5, 0, 0, 0, 0, 0]
thresh = 4

start = -1
zeros = False
count = 0
out = []
for i, v in enumerate(lst+[-1]):
    if v==0:
        if not zeros:
            count = 0
            start = i
            zeros = True
        count += 1
        continue
    if count > thresh:
        if zeros:
            out.append(start)
    zeros = False

out
# [0, 9, 18]

itertools.groupby approach:

from itertools import groupby

lst = [0, 0, 0, 0, 0, 1, 2, 0, 3, 0, 0, 0, 0, 0, 0, 4, 0, 5, 0, 0, 0, 0, 0]
thesh  = 4

out = [x[0][0] for k,g in groupby(enumerate(lst), key=lambda x: x[1]==0)
       if k and len(x:=list(g))>thresh]
# [0, 9, 18]