Question

我希望在两个分类变量上聚合时间序列数据集时构建移动平均线。虽然我已经看过其他一些教程，但它们似乎都没有捕捉到我想要实现的具体任务。

我的原始数据集（df）为每个人（id）提供了一系列日期范围为0-180（Days）的行。个人可以是两个数据子集之一（Group）的成员。

然后我汇总这个数据框以获得两组的每日均值。

library(plyr)
summary <- ddply(df, .(Group,Days), summarise,
                      DV = mean(variable), resp=length(unique(Id)))

然而，下一步是在两个组内构建移动平均值。在下面的示例数据框中，我使用前5天构建了一个5天的平均值。

Group Days DV 5DayMA exceeded 0 2859 exceeded 1 2948 exceeded 2 4412 exceeded 3 5074 exceeded 4 5098 4078 exceeded 5 5147 4536 exceeded 6 4459 4838 exceeded 7 4730 4902 exceeded 8 4643 4815 exceeded 9 4698 4735 exceeded 10 4818 4670 exceeded 11 4521 4682 othergroup 0 2859 othergroup 1 2948 othergroup 2 4412 othergroup 3 5074 othergroup 4 5098 4078 othergroup 5 5147 4536 othergroup 6 4459 4838 othergroup 7 4730 4902 othergroup 8 4643 4815 othergroup 9 4698 4735 othergroup 10 4818 4670 othergroup 11 4521 4682

关于如何做到这一点的任何想法？

Answer 1

您可以尝试zoo::rollmean

df <- structure(list(Group = c("exceeded", "exceeded", "exceeded", 
"exceeded", "exceeded", "exceeded", "exceeded", "exceeded", "exceeded", 
"exceeded", "exceeded", "exceeded", "othergroup", "othergroup", 
"othergroup", "othergroup", "othergroup", "othergroup", "othergroup", 
"othergroup", "othergroup", "othergroup", "othergroup", "othergroup"
), Days = c(0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 
0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L), DV = c(2859L, 
2948L, 4412L, 5074L, 5098L, 5147L, 4459L, 4730L, 4643L, 4698L, 
4818L, 4521L, 2859L, 2948L, 4412L, 5074L, 5098L, 5147L, 4459L, 
4730L, 4643L, 4698L, 4818L, 4521L), X5DayMA = c(NA, NA, NA, NA, 
4078L, 4536L, 4838L, 4902L, 4815L, 4735L, 4670L, 4682L, NA, NA, 
NA, NA, 4078L, 4536L, 4838L, 4902L, 4815L, 4735L, 4670L, 4682L
)), .Names = c("Group", "Days", "DV", "X5DayMA"), class = "data.frame", row.names = c(NA, 
-24L))

head(df)
     Group Days   DV X5DayMA
1 exceeded    0 2859      NA
2 exceeded    1 2948      NA
3 exceeded    2 4412      NA
4 exceeded    3 5074      NA
5 exceeded    4 5098    4078
6 exceeded    5 5147    4536

library(plyr)
library(zoo)
ddply(
  df, "Group",
  transform,
  5daymean = rollmean(DV, 5, align="right", na.pad=TRUE ))

        Group Days   DV X5DayMA 5daymean
1    exceeded    0 2859      NA       NA
2    exceeded    1 2948      NA       NA
3    exceeded    2 4412      NA       NA
4    exceeded    3 5074      NA       NA
5    exceeded    4 5098    4078   4078.2
6    exceeded    5 5147    4536   4535.8
7    exceeded    6 4459    4838   4838.0
8    exceeded    7 4730    4902   4901.6
9    exceeded    8 4643    4815   4815.4
10   exceeded    9 4698    4735   4735.4
11   exceeded   10 4818    4670   4669.6
12   exceeded   11 4521    4682   4682.0
13 othergroup    0 2859      NA       NA
14 othergroup    1 2948      NA       NA
15 othergroup    2 4412      NA       NA
16 othergroup    3 5074      NA       NA
17 othergroup    4 5098    4078   4078.2
18 othergroup    5 5147    4536   4535.8
19 othergroup    6 4459    4838   4838.0
20 othergroup    7 4730    4902   4901.6
21 othergroup    8 4643    4815   4815.4
22 othergroup    9 4698    4735   4735.4
23 othergroup   10 4818    4670   4669.6
24 othergroup   11 4521    4682   4682.0

使用dplyr

甚至更快

library(dplyr)
df %.%
  dplyr:::group_by(Group) %.%
  dplyr:::mutate('5daymean' = rollmean(DV, 5, align="right", na.pad=TRUE ))

或超快data.table

library(data.table)
dft <- data.table(df)
dft[ , `:=` ('5daymean' = rollmean(DV, 5, align="right", na.pad=TRUE )) , by=Group ]

Answer 2

ave和filter：

with(df, ave(DV, Group, FUN=function(x) filter(x,rep(1/5,5),sides=1)))
# [1]     NA     NA     NA     NA 4078.2 4535.8 4838.0 4901.6 4815.4 4735.4
#[11] 4669.6 4682.0     NA     NA     NA     NA 4078.2 4535.8 4838.0 4901.6
#[21] 4815.4 4735.4 4669.6 4682.0

在R中构造分类变量的移动平均值

2 个答案: