Question

我有一个[0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0 , 0 , 1].

形式的熊猫系列

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

经济衰退是连续两次下降（1）。

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。

我对如何用熊猫来解决这个问题感到迷茫。我想确定经济衰退开始和结束的指数。任何帮助将不胜感激。

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，因此当经济衰退开始时rolling总和为1，而当它停止时为-1。

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

由于1和-1都评估为True，因此我可以屏蔽滚动信号以启动和停止ffill。通过gt(0)获取正面或负面的真值。

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

Answer 3

使用Guid guid = Guid.NewGuid(); byte[] guidBytes = guid.ToByteArray(); // Is the result (uniqueId) as unique as guid.ToString()? string uniqueId = string.Join(string.Empty, guidBytes);的类似想法，但将结果写为单个布尔列：

shift

结果输出：

# 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)

Answer 4

1的运行开始满足条件

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.同样，运行结束时满足条件

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

因为运行结束时的值是1，接下来的两个值是0。我们可以使用np.flatnonzero找到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']])

产量

   start  end
0      3    5
1      8   11

Answer 5

您可以使用scipy.signal.find_peaks解决此问题。

    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]

熊猫：标记连续值

5 个答案: