在r中处理时间序列中的缺失值

我处理的是时间序列数据，我需要有连续的时间戳，但在如下所示的捕获过程中，很少有数据时间戳点丢失

ID Time_Stamp                     A           B                 C
1  02/02/2018 07:45:00           123          567               434     
2   02/02/2018 07:45:01    
.....                  ...

5   02/02/2018 07:46:00         mean(A1:A5)
5.1  02/02/2018 07:46:01        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.2  02/02/2018 07:46:02        mean(A1:A5)  mean(B1:B5)         mean(C1:C5) 
5.3  02/02/2018 07:46:03        mean(A1:A5)  mean(B1:B5)         mean(C1:C5) 
5.4  02/02/2018 07:46:04        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.5  02/02/2018 07:46:05        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.6  02/02/2018 07:46:06        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.7  02/02/2018 07:46:07        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.8  02/02/2018 07:46:08        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.9  02/02/2018 07:46:09        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
6   02/02/2018 07:46:10         112         2323            2323
6.1 02/02/2018 07:46:11         mean(A1:A15) mean(B1:B15)       mean(C1:C15)

DF

ID Time_Stamp             A           B                 C
1  02/02/2018 07:45:00   123          567               434     
2   02/02/2018 07:45:01    
.....                  ...

5   02/02/2018 07:46:00   
6   02/02/2018 07:46:10    112          2323            2323

如上面的示例df所示，时间戳在

第5行

之前是连续的，但在

5行

和

6行

之间错过了

10秒

的捕获数据。我的数据框大约有60000行，手动识别缺少的值是一件乏味的事情。因此，我正在寻找使用R自动化处理缺失值的过程

我的结果数据框如下所示

ID Time_Stamp                     A           B                 C
1  02/02/2018 07:45:00           123          567               434     
2   02/02/2018 07:45:01    
.....                  ...

5   02/02/2018 07:46:00         mean(A1:A5)
5.1  02/02/2018 07:46:01        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.2  02/02/2018 07:46:02        mean(A1:A5)  mean(B1:B5)         mean(C1:C5) 
5.3  02/02/2018 07:46:03        mean(A1:A5)  mean(B1:B5)         mean(C1:C5) 
5.4  02/02/2018 07:46:04        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.5  02/02/2018 07:46:05        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.6  02/02/2018 07:46:06        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.7  02/02/2018 07:46:07        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.8  02/02/2018 07:46:08        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
5.9  02/02/2018 07:46:09        mean(A1:A5)  mean(B1:B5)         mean(C1:C5)
6   02/02/2018 07:46:10         112         2323            2323
6.1 02/02/2018 07:46:11         mean(A1:A15) mean(B1:B15)       mean(C1:C15)

甚至可以是该时间间隔内前几行的平均值

 6.1 02/02/2018 07:46:11         mean(A14:A17) mean(B14:B17)      mean(C14:C17)

即缺失，但缺失时间值除外

我已经完成了以下代码，以获得整个列的平均值

library(dplyr)
library(tidyr)

df %>%
  complete(Time_Stamp = seq(min(Time_Stamp), max(Time_Stamp), by = "sec")) %>%
  mutate_at(vars(A:C), ~replace(., is.na(.), mean(., na.rm = TRUE))) %>%
  mutate(ID = row_number())

它给出列中所有行的所有平均值的输出

就像下面的一样，它工作得很完美，但我需要修改。我怎么能做到呢。

---

请帮助

这里有一个结合了

tidyverse

和base R的方法来实现这个结果。我们首先创建一个新列，每个列的累积平均值。然后，我们

完成

缺失的观察值，并用其他列中的相应平均值替换

NA

s

library(tidyverse)
cols <- c("A", "B", "C")

df1 <- df %>%
        mutate_at(cols, list(mean = ~cummean(.))) %>%
        complete(Time_Stamp = seq(min(Time_Stamp), max(Time_Stamp), by = "sec")) %>%
        fill(ends_with("mean")) %>%
        mutate(ID = row_number())

mean_cols <- grep("_mean$", names(df1))
df1[cols] <- Map(function(x, y) ifelse(is.na(x), y, x), df1[cols], df1[mean_cols])

df1[names(df)]

#     ID Time_Stamp              A     B     C
#   <int> <dttm>              <dbl> <dbl> <dbl>
# 1     1 2018-02-02 07:45:00  123   567   434 
# 2     2 2018-02-02 07:45:01  234   100   110 
# 3     3 2018-02-02 07:45:02  234   100   110 
# 4     4 2018-02-02 07:45:03  197   256.  218 
# 5     5 2018-02-02 07:45:04  197   256.  218 
# 6     6 2018-02-02 07:45:05  197   256.  218 
# 7     7 2018-02-02 07:45:06  197   256.  218 
# 8     8 2018-02-02 07:45:07  197   256.  218 
# 9     9 2018-02-02 07:45:08  197   256.  218 
#10    10 2018-02-02 07:45:09  197   256.  218 
#11    11 2018-02-02 07:45:10  112  2323  2323 
#12    12 2018-02-02 07:45:11  176.  772.  744.
#13    13 2018-02-02 07:45:12  176.  772.  744.
#14    14 2018-02-02 07:45:13  176.  772.  744.
#15    15 2018-02-02 07:45:14  176.  772.  744.
#16    16 2018-02-02 07:45:15  100    23    12 

数据

df <- structure(list(ID = c(1, 2, 3, 4, 5), Time_Stamp = structure(c(1517557500, 
1517557501, 1517557502, 1517557510, 1517557515), class = c("POSIXct", 
"POSIXt"), tzone = "UTC"), A = c(123, 234, 234, 112, 100), B = c(567, 
100, 100, 2323, 23), C = c(434, 110, 110, 2323, 12)), row.names = c(NA, 
-5L), class = "data.frame")

---

有一个非常直观的软件包正是为了这个目的而设计的，叫做“padr”。 我想你会发现它满足了你的需要：

---

我建议查看

tidyr:：fill

或可能使用

tsibble

软件包中的时间填充功能。请为我们提供可复制的数据集，以测试我们提出的解决方案on@RonakShah这对我有用。很抱歉迟了答复。

df
#  ID          Time_Stamp   A    B    C
#1  1 2018-02-02 07:45:00 123  567  434
#2  2 2018-02-02 07:45:01 234  100  110
#3  3 2018-02-02 07:45:02 234  100  110
#4  4 2018-02-02 07:45:10 112 2323 2323
#5  5 2018-02-02 07:45:15 100   23   12