Python 熊猫:标记连续值
我有一个熊猫系列的表格Python 熊猫:标记连续值,python,pandas,Python,Pandas,我有一个熊猫系列的表格[0,1,0,1,1,1,0,0,1,1,0,1,0,0,1]。 0: indicates economic increase. 1: indicates economic decline. 经济衰退的信号是连续两次下降(1) 经济衰退结束的信号是连续两次增长(0) 在上面的数据集中,我有两次衰退,从指数3开始,从指数5结束,从指数8开始,从指数11结束 我对如何对待熊猫感到困惑。我想确定衰退开始和结束的指数。任何协助都将不胜感激 下面是我在解决方案方面的python尝试
[0,1,0,1,1,1,0,0,1,1,0,1,0,0,1]。
0: indicates economic increase.
1: indicates economic decline.
经济衰退的信号是连续两次下降(1)
经济衰退结束的信号是连续两次增长(0)
在上面的数据集中,我有两次衰退,从指数3开始,从指数5结束,从指数8开始,从指数11结束
我对如何对待熊猫感到困惑。我想确定衰退开始和结束的指数。任何协助都将不胜感激
下面是我在解决方案方面的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?
Leon您可以使用:
使用rolling(2)
我减去.5
,因此当衰退开始时滚动
和为1
,当衰退停止时为-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)
获取正值或负值时的真值
使用
shift
类似的方法,但将结果写入单个布尔列:
# 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
运行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))
((x == 1) & (x_next == 0) & (x_next2 == 0))
也就是说,运行开始时的值为1,前一个值不是1,下一个值为1。类似地,运行结束时满足该条件
x_prev = x.shift(1)
x_next = x.shift(-1)
((x_prev != 1) & (x == 1) & (x_next == 1))
((x == 1) & (x_next == 0) & (x_next2 == 0))
因为运行结束时的值为1,接下来的两个值为0。
我们可以使用np.flatnonzero
找到这些条件为真的索引:
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']])
屈服
start end
0 3 5
1 8 11
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]
也许只是用0来
.fillna
使它整洁。Hi@piRSquared。我怎样才能修改你们在衰退开始和衰退结束两栏中所做的工作。如果衰退开始是真的,那么衰退开始是假的。当然,如果衰退结束是真的,那么衰退结束是假的。
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]