熊猫:标志连续值

Question

我有一个熊猫系列的形式 [0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1].

0: indicates economic increase.
1: indicates economic decline.

衰退是由连续两次下降(1)发出信号.

经济衰退的结束是连续两次增加(0).

在上面的数据集中,我有两次经济衰退,从指数3开始,在指数5结束,在指数11结束时从指数8结束.

对于如何用熊猫来解决这个问题我感到很遗憾.我想确定经济衰退开始和结束的指数.任何援助将不胜感激.

这是我对soln的python尝试.

np_decline =  np.array([0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1])
recession_start_flag = 0
recession_end_flag = 0
recession_start = []
recession_end = []

for i in range(len(np_decline) - 1):
    if recession_start_flag == 0 and np_decline[i] == 1 and np_decline[i + 1] == 1:
        recession_start.append(i)
        recession_start_flag = 1
    if recession_start_flag == 1 and np_decline[i] == 0 and np_decline[i + 1] == 0:
        recession_end.append(i - 1)
        recession_start_flag = 0

print(recession_start)
print(recession_end)

这是一种更加以熊猫为中心的方法吗？莱昂

Answer 1

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

Answer 2

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总和就是1衰退开始和结束的-1时间。

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

因为两者1和-1评估到True我可以掩盖滚动信号只是开始和停止和ffill。使用来获取真值的正负值gt(0)。

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

Answer 3

使用相似的想法shift，但是将结果写为单个布尔列：

# 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

Answer 4

开始运行1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，上一个值不为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))

由于运行结束时的值为1，接下来的两个值为0。我们可以使用以下条件找到满足这些条件的索引np.flatnonzero：

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']])

产量

   start  end
0      3    5
1      8   11