一串1的开始满足条件

x_prev = x.shift(1)
x_next = x.shift(-1)
((x_prev != 1) & (x == 1) & (x_next == 1))

也就是说，运行开始时的值为1，前一个值不为1，下一个值为1。类似地，运行结束时满足条件

((x == 1) & (x_next == 0) & (x_next2 == 0))

因为一轮结束时的值为1，接下来的两个值value为0。我们可以使用np.flatnonzero查找满足这些条件的索引：

import numpy as np
import pandas as pd

x = pd.Series([0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1])
x_prev = x.shift(1)
x_next = x.shift(-1)
x_next2 = x.shift(-2)
df = pd.DataFrame(
    dict(start = np.flatnonzero((x_prev != 1) & (x == 1) & (x_next == 1)),
         end = np.flatnonzero((x == 1) & (x_next == 0) & (x_next2 == 0))))
print(df[['start', 'end']])

收益率

start  end
0      3    5
1      8   11

您可以使用shift：

df = pd.DataFrame([0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1], columns=['signal'])
df_prev = df.shift(1)['signal']
df_next = df.shift(-1)['signal']
df_next2 = df.shift(-2)['signal']
df.loc[(df_prev != 1) & (df['signal'] == 1) & (df_next == 1), 'start'] = 1
df.loc[(df['signal'] != 0) & (df_next == 0) & (df_next2 == 0), 'end'] = 1
df.fillna(0, inplace=True)
df = df.astype(int)

    signal  start  end
0        0      0    0
1        1      0    0
2        0      0    0
3        1      1    0
4        1      0    0
5        1      0    1
6        0      0    0
7        0      0    0
8        1      1    0
9        1      0    0
10       0      0    0
11       1      0    1
12       0      0    0
13       0      0    0
14       1      0    0

使用shift的想法类似，但将结果写成单个布尔列：

# Boolean indexers for recession start and stops.
rec_start = (df['signal'] == 1) & (df['signal'].shift(-1) == 1)
rec_end = (df['signal'] == 0) & (df['signal'].shift(-1) == 0)

# Mark the recession start/stops as True/False.
df.loc[rec_start, 'recession'] = True
df.loc[rec_end, 'recession'] = False

# Forward fill the recession column with the last known Boolean.
# Fill any NaN's as False (i.e. locations before the first start/stop).
df['recession'] = df['recession'].ffill().fillna(False)

结果输出：

signal recession
0        0     False
1        1     False
2        0     False
3        1      True
4        1      True
5        1      True
6        0     False
7        0     False
8        1      True
9        1      True
10       0      True
11       1      True
12       0     False
13       0     False
14       1     False

使用rolling(2)

s = pd.Series([0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1])

我减去.5，所以rolling的和在衰退开始时是1，在衰退停止时是-1。

s2 = s.sub(.5).rolling(2).sum()

由于1和-1的计算结果均为True，因此我可以将滚动信号屏蔽为仅开始和停止以及ffill。使用gt(0)获取它们何时为正或负的真值。

pd.concat([s, s2.mask(~s2.astype(bool)).ffill().gt(0)], axis=1, keys=['signal', 'isRec'])

您可以使用scipy.signal.find_peaks来解决这个问题。

from scipy.signal import find_peaks

    np_decline =  np.array([0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1])
    peaks = find_peaks(np_decline,width=2)
    recession_start_loc =  peaks[1]['left_bases'][0]

def function2(dd:pd.DataFrame):
    if dd.iat[0,1]>=2:
        if dd.query("col1==0").pipe(len)==1:
            return (dd.index.min(),dd.index.max()+1)
        else:
            dd1=dd.query("col1==1")
            return (dd1.index.min(),dd1.index.max())

col2=df1.col1.diff().eq(1).cumsum()
df1.groupby(col2).apply(lambda dd:dd.assign(col3=dd.col1.cumprod().sum()))\
    .groupby('col3',sort=False).apply(function2).dropna()

输出：

col3
3     (3, 5)
2    (8, 11)
dtype: object