Question

我的SPI时间序列的长度为324，范围为-3到+3。我想获取连续3个或更多时间步长低于阈值-1的地方的索引

我对本网站和其他地方进行了彻底搜索，但未取得重大成功，例如Check if there are 3 consecutive values in an array which are above some threshold却不能完全满足我的要求

>0: option
>2: option
   value: "6"
   text: "apple"
   spellcheck: true
   textContent: "apple"
>3: option
   value: "2"
   text: "test"
   spellcheck: true
   textContent: "test"
>4: option
...

这就是我想做的事，我们将不胜感激，谢谢

Answer 1

这是一个有注释的分步食谱。

a = [-3,4,5,-1,-2,-5,1,4,6,9,-3,-3,-1,-2,4,1,4]
th = -1
a = np.array(a)

# create mask of events; find indices where mask switches
intervals = np.where(np.diff(a<=th, prepend=0, append=0))[0].reshape(-1,2)

# discard short stretches
intervals = intervals[np.subtract(*intervals.T) <= -3]

intervals
# array([[ 3,  6],
#        [10, 14]])

# get corresponding data
stretches = np.split(a, intervals.reshape(-1))[1::2]

stretches
# [array([-1, -2, -5]), array([-3, -3, -1, -2])]

# count events
-np.subtract(*intervals.T)
# array([3, 4])

# sum events
np.add.reduceat(a, intervals.reshape(-1))[::2]
# array([-8, -9])

Answer 2

自从您标记了熊猫：

s = pd.Series([-3,4,5,-1,-2,-5,1,4,6,9,-3,-3,-1,-2,4,1,4])

# thresholding
a = (s<1)

# blocks
b = (a!=a.shift()).cumsum()

# groupby
df = s[a].groupby(b).agg([list,'size','sum'])
df = df[df.size>=3]

输出

           list      size   sum
3       [-1, -2, -5]    3   -8
5   [-3, -3, -1, -2]    4   -9

Answer 3

使用np.logical_and.reduce + shift，检查连续的行是否低于阈值。然后使用groupby来获取您需要的所有聚合：

import numpy as np
import pandas as pd

def get_grps(s, thresh=-1, Nmin=3):
    """
    Nmin : int > 0
        Min number of consecutive values below threshold.
    """
    m = np.logical_and.reduce([s.shift(-i).le(thresh) for i in range(Nmin)])
    if Nmin > 1:
        m = pd.Series(m, index=s.index).replace({False: np.NaN}).ffill(limit=Nmin-1).fillna(False)
    else:
        m = pd.Series(m, index=s.index)

    # Form consecutive groups
    gps = m.ne(m.shift(1)).cumsum().where(m)

    # Return None if no groups, else the aggregations
    if gps.isnull().all():
        return None
    else:
        return s.groupby(gps).agg([list, sum, 'size']).reset_index(drop=True)

get_grps(pd.Series(a))
#               list  sum  size
#0      [-1, -2, -5]   -8     3
#1  [-3, -3, -1, -2]   -9     4

get_grps(pd.Series(a), thresh=-1, Nmin=1)
#               list  sum  size
#0              [-3]   -3     1
#1      [-1, -2, -5]   -8     3
#2  [-3, -3, -1, -2]   -9     4

get_grps(pd.Series(a), thresh=-100, Nmin=1)
#None

连续事件低于阈值

3 个答案: