Question

我有时间序列数据。但数据不连续。（2005-03-02 02:08:00缺失。）

我需要一个新的C列C(i)=A(i)+B(i)+average，其中我的平均值是B到不连续点(02:08:00)的平均值。

average=Data.loc['2005-03-02 02:05:30':'2005-03-02 02:07:30',['B']].mean(axis=0)  
After discontinuity we have to again recalculate average till next discontinuity  
average=Data.loc['2005-03-02 02:08:30':'2005-03-02 02:11:00',['B']].mean(axis=0)

输入

Date,A,B  
2005-03-02 02:05:30,1,3   
2005-03-02 02:06:00,2,4   
2005-03-02 02:06:30,3,5  
2005-03-02 02:07:00,4,6  
2005-03-02 02:07:30,5,7  
2005-03-02 02:08:30,7,9  
2005-03-02 02:09:00,7,9  
2005-03-02 02:09:30,7,9  
2005-03-02 02:10:00,8,12  
2005-03-02 02:10:30,9,13  
2005-03-02 02:11:00,10,14

输出

Date,A,B,C  
2005-03-02 02:05:30,1,3,9  
2005-03-02 02:06:00,2,4,11  
2005-03-02 02:06:30,3,5,13  
2005-03-02 02:07:00,4,6,15  
2005-03-02 02:07:30,5,7,17  
2005-03-02 02:08:30,7,9,28  
2005-03-02 02:09:00,7,9,28  
2005-03-02 02:09:30,7,9,28  
2005-03-02 02:10:00,8,12,32  
2005-03-02 02:10:30,9,13,34  
2005-03-02 02:11:00,10,14,36

如何在索引中找出不连续性？

如何使用pandas完成所有内容？

Answer 1

我建议使用：

#if unique values in index use reindex 
df = Data.reindex(pd.date_range(Data.index.min(), Data.index.max(), freq='30S'))
#if non unique values in index
#df = df.resample('30s').mean()

#get mask for NaNs rows
mask = df.isnull().all(axis=1)
#get sum of all columns
s1 = df.sum(axis=1)
#if need sum only A, B columns
#s1 = df[['A', 'B']].sum(axis=1)
#create column for grouping
df['C'] = mask.cumsum()
#filter out NaNs rows
df = df[~mask]
#transform mean and add sum
df['C'] = df.groupby('C')['B'].transform('mean') + s1
print (df)
                        A     B     C
2005-03-02 02:05:30   1.0   3.0   9.0
2005-03-02 02:06:00   2.0   4.0  11.0
2005-03-02 02:06:30   3.0   5.0  13.0
2005-03-02 02:07:00   4.0   6.0  15.0
2005-03-02 02:07:30   5.0   7.0  17.0
2005-03-02 02:08:30   7.0   9.0  27.0
2005-03-02 02:09:00   7.0   9.0  27.0
2005-03-02 02:09:30   7.0   9.0  27.0
2005-03-02 02:10:00   8.0  12.0  31.0
2005-03-02 02:10:30   9.0  13.0  33.0
2005-03-02 02:11:00  10.0  14.0  35.0

另一个解决方案，谢谢@iDrwish的建议：

首先获得索引的差异（diff）（尚未实现，因此首先按to_series将索引转换为系列），与30 s Timedelta进行比较并按cumsum创建组

上次使用transform mean并添加列数：

g = Data.index.to_series().diff().gt(pd.Timedelta(30, unit='s')).cumsum()
Data['C'] = Data.groupby(g)['B'].transform('mean') + Data.sum(axis=1)
#if need specify columns
#Data['C'] = Data.groupby(g)['B'].transform('mean') + Data['A'] + Data['B']
print (Data)
                      A   B   C
Date                           
2005-03-02 02:05:30   1   3   9
2005-03-02 02:06:00   2   4  11
2005-03-02 02:06:30   3   5  13
2005-03-02 02:07:00   4   6  15
2005-03-02 02:07:30   5   7  17
2005-03-02 02:08:30   7   9  27
2005-03-02 02:09:00   7   9  27
2005-03-02 02:09:30   7   9  27
2005-03-02 02:10:00   8  12  31
2005-03-02 02:10:30   9  13  33
2005-03-02 02:11:00  10  14  35

Answer 2

第1步：读取数据框

import pandas as pd
from io import StringIO

y = '''Date,A,B
2005-03-02 02:05:30,1,3   
2005-03-02 02:06:00,2,4   
2005-03-02 02:06:30,3,5  
2005-03-02 02:07:00,4,6  
2005-03-02 02:07:30,5,7  
2005-03-02 02:08:30,7,9  
2005-03-02 02:09:00,7,9  
2005-03-02 02:09:30,7,9  
2005-03-02 02:10:00,8,12  
2005-03-02 02:10:30,9,13  
2005-03-02 02:11:00,10,14'''

df = pd.read_csv(StringIO(y), index_col='Date')

第2步：转换为日期时间索引

df.index = pd.to_datetime(df.index)

第2步：持续30秒重新取样

new = df.resample('30s').mean()

输出：

                        A   B  
Date                           
2005-03-02 02:05:30   1.0   3.0
2005-03-02 02:06:00   2.0   4.0
2005-03-02 02:06:30   3.0   5.0
2005-03-02 02:07:00   4.0   6.0
2005-03-02 02:07:30   5.0   7.0
2005-03-02 02:08:00   NaN   NaN
2005-03-02 02:08:30   7.0   9.0
2005-03-02 02:09:00   7.0   9.0
2005-03-02 02:09:30   7.0   9.0
2005-03-02 02:10:00   8.0  12.0
2005-03-02 02:10:30   9.0  13.0
2005-03-02 02:11:00  10.0  14.0

步骤3：按NaN行拆分数据框并获取组ID

new["group_no"] = new.T.isnull().all().cumsum()

输出：

                        A   B    group_no
Date                                     
2005-03-02 02:05:30   1.0   3.0         0
2005-03-02 02:06:00   2.0   4.0         0
2005-03-02 02:06:30   3.0   5.0         0
2005-03-02 02:07:00   4.0   6.0         0
2005-03-02 02:07:30   5.0   7.0         0
2005-03-02 02:08:00   NaN   NaN         1
2005-03-02 02:08:30   7.0   9.0         1
2005-03-02 02:09:00   7.0   9.0         1
2005-03-02 02:09:30   7.0   9.0         1
2005-03-02 02:10:00   8.0  12.0         1
2005-03-02 02:10:30   9.0  13.0         1
2005-03-02 02:11:00  10.0  14.0         1

第4步：获取每个group_no的平均值

new['Bmean'] = new.groupby('group_no').transform('mean').B

输出：

A B group_no Bmean Date 2005-03-02 02:05:30 1.0 3.0 0 5.0 2005-03-02 02:06:00 2.0 4.0 0 5.0 2005-03-02 02:06:30 3.0 5.0 0 5.0 2005-03-02 02:07:00 4.0 6.0 0 5.0 2005-03-02 02:07:30 5.0 7.0 0 5.0 2005-03-02 02:08:00 NaN NaN 1 11.0 2005-03-02 02:08:30 7.0 9.0 1 11.0 2005-03-02 02:09:00 7.0 9.0 1 11.0 2005-03-02 02:09:30 7.0 9.0 1 11.0 2005-03-02 02:10:00 8.0 12.0 1 11.0 2005-03-02 02:10:30 9.0 13.0 1 11.0 2005-03-02 02:11:00 10.0 14.0 1 11.0

第5步：应用必要的转化并删除额外的列

new['C'] = new['A'] + new['B'] + new['Bmean'] new.drop(['group_no', 'Bmean'], axis=1, inplace=True)

输出：

A B C Date 2005-03-02 02:05:30 1.0 3.0 9.0 2005-03-02 02:06:00 2.0 4.0 11.0 2005-03-02 02:06:30 3.0 5.0 13.0 2005-03-02 02:07:00 4.0 6.0 15.0 2005-03-02 02:07:30 5.0 7.0 17.0 2005-03-02 02:08:00 NaN NaN NaN 2005-03-02 02:08:30 7.0 9.0 27.0 2005-03-02 02:09:00 7.0 9.0 27.0 2005-03-02 02:09:30 7.0 9.0 27.0 2005-03-02 02:10:00 8.0 12.0 31.0 2005-03-02 02:10:30 9.0 13.0 33.0 2005-03-02 02:11:00 10.0 14.0 35.0

Answer 3

如果一个点被描述为P（v，t）。 A =（3,1）和B =（10,5）。

因此任何C（v，t）= A（v）+（B（v）-A（v））*（（C（t）-A（t））:( B（t）-A（吨））。

A(v,1) = 3 
C(v,2) = 3 + (10-3) * ((2-1):(5-1)) = 4,75
C(v,3) = 3 + (10-3) * ((3-1):(5-1)) = 6,5
C(v,4) = 3 + (10-3) * ((4-1):(5-1)) = 8,25
B(v,5) = 10

如何找出不连续的日期时间指数？如何取平均连续指数？

3 个答案: