我正在处理时间序列数据，我需要有连续的时间戳，但是在捕获数据时几乎没有丢失数据时间戳记点，如下所示，
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，时间戳记一直持续到row 5，但是在10 seconds和5th之间丢失了6th row的捕获数据。我的数据框是关于60000 rows的信息，因此手动识别丢失的值很乏味。因此，我一直在寻找使用R
我的结果数据框如下，
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中前几行的平均值。
6.1 02/02/2018 07:46:11 mean(A14:A17) mean(B14:B17) mean(C14:C17)
即缺少时间值。
我已经完成以下代码来获取整列的均值。
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均值的输出。
就像跟随this code一样，效果很好，但我需要进行此修改。怎么做 请帮助
答案 0 :(得分：1)
这里是tidyverse和基本R方法的组合，以实现结果。我们首先创建一个新列，并为每列累积平均值。然后，我们complete缺少的观察值，并用其他列中的相应方法替换NA。
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
如果您需要每个NA值的均值，它将变得更加简单
df %>%
complete(Time_Stamp = seq(min(Time_Stamp), max(Time_Stamp), by = "sec")) %>%
mutate_at(cols, ~ifelse(is.na(.), cummean(na.omit(.)), .)) %>%
mutate(ID = row_number())
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")
看起来像
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
答案 1 :(得分：1)
为此专门制作了一个非常直观的程序包，称为“ padr”。 我认为您会发现它可以满足您的需求： cran padr vignette