Question

我正在尝试使用python的pandas库确定两个时间序列重叠的时间百分比。数据是非同步的，因此每个数据点的时间不对齐。这是一个例子：

时间序列1

2016-10-05 11:50:02.000734    0.50
2016-10-05 11:50:03.000033    0.25
2016-10-05 11:50:10.000479    0.50
2016-10-05 11:50:15.000234    0.25
2016-10-05 11:50:37.000199    0.50
2016-10-05 11:50:49.000401    0.50
2016-10-05 11:50:51.000362    0.25
2016-10-05 11:50:53.000424    0.75
2016-10-05 11:50:53.000982    0.25
2016-10-05 11:50:58.000606    0.75

时间序列2

2016-10-05 11:50:07.000537    0.50
2016-10-05 11:50:11.000994    0.50
2016-10-05 11:50:19.000181    0.50
2016-10-05 11:50:35.000578    0.50
2016-10-05 11:50:46.000761    0.50
2016-10-05 11:50:49.000295    0.75
2016-10-05 11:50:51.000835    0.75
2016-10-05 11:50:55.000792    0.25
2016-10-05 11:50:55.000904    0.75
2016-10-05 11:50:57.000444    0.75

假设系列保持其值直到下一次更改，确定它们具有相同值的时间百分比的最有效方法是什么？

示例

让我们计算这些系列重叠的时间，从11：50：07.000537开始，到2016-10-05 11：50：57.000444 0.75结束，因为我们有这个时期的两个系列的数据。有重叠的时间：

11：50：10.000479 - 11：50：15.000234（两者的值均为0.5） 4.999755秒
11：50：37.000199 - 11：50：49.000295（两者的值均为0.5） 12.000096秒
11：50：53.000424 - 11：50：53.000982（两者的值均为0.75） 0.000558秒
11：50：55.000792 - 11：50：55.000904（两者的值均为0.25） 0.000112秒

结果（4.999755 + 12.000096 + 0.000558 + 0.000112）/ 49.999907 = 34％

其中一个问题是我的实际时间序列有更多的数据，例如1000 - 10000次观测，我需要运行更多对。我考虑过向前填充一个系列，然后简单地比较行并将总匹配数除以总行数，但我不认为这会非常有效。

Answer 1

<强> 设置
创建2个时间序列

from StringIO import StringIO
import pandas as pd


txt1 = """2016-10-05 11:50:02.000734    0.50
2016-10-05 11:50:03.000033    0.25
2016-10-05 11:50:10.000479    0.50
2016-10-05 11:50:15.000234    0.25
2016-10-05 11:50:37.000199    0.50
2016-10-05 11:50:49.000401    0.50
2016-10-05 11:50:51.000362    0.25
2016-10-05 11:50:53.000424    0.75
2016-10-05 11:50:53.000982    0.25
2016-10-05 11:50:58.000606    0.75"""

s1 = pd.read_csv(StringIO(txt1), sep='\s{2,}', engine='python',
                 parse_dates=[0], index_col=0, header=None,
                 squeeze=True).rename('s1').rename_axis(None)

txt2 = """2016-10-05 11:50:07.000537    0.50
2016-10-05 11:50:11.000994    0.50
2016-10-05 11:50:19.000181    0.50
2016-10-05 11:50:35.000578    0.50
2016-10-05 11:50:46.000761    0.50
2016-10-05 11:50:49.000295    0.75
2016-10-05 11:50:51.000835    0.75
2016-10-05 11:50:55.000792    0.25
2016-10-05 11:50:55.000904    0.75
2016-10-05 11:50:57.000444    0.75"""

s2 = pd.read_csv(StringIO(txt2), sep='\s{2,}', engine='python',
                 parse_dates=[0], index_col=0, header=None,
                 squeeze=True).rename('s2').rename_axis(None)

<强> TL; DR

df = pd.concat([s1, s2], axis=1).ffill().dropna()
overlap = df.index.to_series().diff().shift(-1) \
            .fillna(0).groupby(df.s1.eq(df.s2)).sum()
overlap.div(overlap.sum())

False    0.666657
True     0.333343
Name: duration, dtype: float64

<强> 解释

构建基础pd.DataFrame df

使用pd.concat来对齐索引
使用ffill让值向前传播
使用dropna在另一个系列开始之前删除一个系列的值

df = pd.concat([s1, s2], axis=1).ffill().dropna()
df

计算'duration'
从当前时间戳到下一个

df['duration'] = df.index.to_series().diff().shift(-1).fillna(0)
df

计算重叠

df.s1.eq(df.s2)给出了s1与s2重叠的布尔系列
在布尔系列之上使用groupby汇总True和False

overlap = df.groupby(df.s1.eq(df.s2)).duration.sum()
overlap

False   00:00:33.999548
True    00:00:17.000521
Name: duration, dtype: timedelta64[ns]

具有相同值的时间百分比

overlap.div(overlap.sum())

False    0.666657
True     0.333343
Name: duration, dtype: float64

Answer 2

很酷的问题。我粗暴地用大熊猫或者numpy强迫这个，但是我得到了你的答案（感谢你的解决）。我还没有测试过任何其他东西。我也不知道它有多快，因为它只通过每个数据帧一次，但不进行任何矢量化。

import pandas as pd
#############################################################################
#Preparing the dataframes
times_1 = ["2016-10-05 11:50:02.000734","2016-10-05 11:50:03.000033",
           "2016-10-05 11:50:10.000479","2016-10-05 11:50:15.000234",
           "2016-10-05 11:50:37.000199","2016-10-05 11:50:49.000401",
           "2016-10-05 11:50:51.000362","2016-10-05 11:50:53.000424",
           "2016-10-05 11:50:53.000982","2016-10-05 11:50:58.000606"]
times_1 = [pd.Timestamp(t) for t in times_1]
vals_1 = [0.50,0.25,0.50,0.25,0.50,0.50,0.25,0.75,0.25,0.75]

times_2 = ["2016-10-05 11:50:07.000537","2016-10-05 11:50:11.000994",
           "2016-10-05 11:50:19.000181","2016-10-05 11:50:35.000578",
           "2016-10-05 11:50:46.000761","2016-10-05 11:50:49.000295",
           "2016-10-05 11:50:51.000835","2016-10-05 11:50:55.000792",
           "2016-10-05 11:50:55.000904","2016-10-05 11:50:57.000444"]
times_2 = [pd.Timestamp(t) for t in times_2]
vals_2 = [0.50,0.50,0.50,0.50,0.50,0.75,0.75,0.25,0.75,0.75]

data_1 = pd.DataFrame({"time":times_1,"vals":vals_1})
data_2 = pd.DataFrame({"time":times_2,"vals":vals_2})
#############################################################################

shared_time = 0      #Keep running tally of shared time
t1_ind = 0           #Pointer to row in data_1 dataframe
t2_ind = 0           #Pointer to row in data_2 dataframe

#Loop through both dataframes once, incrementing either the t1 or t2 index
#Stop one before the end of both since do +1 indexing in loop
while t1_ind < len(data_1.time)-1 and t2_ind < len(data_2.time)-1:
    #Get val1 and val2
    val1,val2 = data_1.vals[t1_ind], data_2.vals[t2_ind]

    #Get the start and stop of the current time window
    t1_start,t1_stop = data_1.time[t1_ind], data_1.time[t1_ind+1]
    t2_start,t2_stop = data_2.time[t2_ind], data_2.time[t2_ind+1]

    #If the start of time window 2 is in time window 1
    if val1 == val2 and (t1_start <= t2_start <= t1_stop):
        shared_time += (min(t1_stop,t2_stop)-t2_start).total_seconds()
        t1_ind += 1
    #If the start of time window 1 is in time window 2
    elif val1 == val2 and t2_start <= t1_start <= t2_stop:
        shared_time += (min(t1_stop,t2_stop)-t1_start).total_seconds()
        t2_ind += 1
    #If there is no time window overlap and time window 2 is larger
    elif t1_start < t2_start:
        t1_ind += 1
    #If there is no time window overlap and time window 1 is larger
    else:
        t2_ind += 1

#How I calculated the maximum possible shared time (not pretty)
shared_start = max(data_1.time[0],data_2.time[0])
shared_stop = min(data_1.time.iloc[-1],data_2.time.iloc[-1])
max_possible_shared = (shared_stop-shared_start).total_seconds()

#Print output
print "Shared time:",shared_time
print "Total possible shared:",max_possible_shared
print "Percent shared:",shared_time*100/max_possible_shared,"%"

输出：

Shared time: 17.000521
Total possible shared: 49.999907
Percent shared: 34.0011052421 %

确定Python中重叠时间序列的最有效方法

2 个答案: