示例数据
df <- data.frame(id=c("A","A","A","A","B","B","B","B"),year=c(2014,2014,2015,2015),month=c(1,2),marketcap=c(4,6,2,6,23,2,5,34),return=c(NA,0.23,0.2,0.1,0.4,0.9,NA,0.6))
df1
id year month marketcap return
1: A 2014 1 4 NA
2: A 2014 2 6 0.23
3: A 2015 1 2 0.20
4: A 2015 2 6 0.10
5: B 2014 1 23 0.40
6: B 2014 2 2 0.90
7: B 2015 1 5 NA
8: B 2015 2 34 0.60
所需数据
desired_df <- data.frame(id=c("A","A","A","A","B","B","B","B"),year=c(2014,2014,2015,2015),month=c(1,2),marketcap=c(4,6,2,6,23,2,5,34),return=c(0.23,0.23,0.2,0.1,0.4,0.9,0.75,0.6))
desired_df
id year month marketcap return
1 A 2014 1 4 0.23
2 A 2014 2 6 0.23
3 A 2015 1 2 0.20
4 A 2015 2 6 0.10
5 B 2014 1 23 0.40
6 B 2014 2 2 0.90
7 B 2015 1 5 0.75
8 B 2015 2 34 0.60
我想通过id替换NA
值与时间序列中的相邻值来插值返回。假设只有两个月:一年1,2。
(B,2015,1)的第二个NA
被0.75 =(0.9 + 0.6)/ 2取代
由于没有以前的数据,因此(A,2014,1)的第一个NA
被替换为0.23。
data.table解决方案是首选的
更新: 使用如下代码结构(适用于样本)
df[,returnInterpolate:=na.approx(return,rule=2), by=id]
我遇到了错误: 约(x [!na],y [!na],xout,...)中的错误: 需要至少两个非NA值才能进行插值
我想可能有一些id没有非NA值要插值。 。有什么建议?
答案 0 :(得分:5)
library(data.table)
df <- data.frame(id=c("A","A","A","A","B","B","B","B"),
year=c(2014,2014,2015,2015),
month=c(1,2),
marketcap=c(4,6,2,6,23,2,5,34),
return=c(NA,0.23,0.2,0.1,0.4,0.9,NA,0.6))
setDT(df)
library(zoo)
df[, returnInterpol := na.approx(return, rule = 2), by = id]
# id year month marketcap return returnInterpol
#1: A 2014 1 4 NA 0.23
#2: A 2014 2 6 0.23 0.23
#3: A 2015 1 2 0.20 0.20
#4: A 2015 2 6 0.10 0.10
#5: B 2014 1 23 0.40 0.40
#6: B 2014 2 2 0.90 0.90
#7: B 2015 1 5 NA 0.75
#8: B 2015 2 34 0.60 0.60
修改强>
如果您的群组只有NA
个值,或者只有一个非NA
,那么您可以这样做:
df <- data.frame(id=c("A","A","A","A","B","B","B","B","C","C","C","C"),
year=c(2014,2014,2015,2015),
month=c(1,2),
marketcap=c(4,6,2,6,23,2,5,34, 1:4),
return=c(NA,0.23,0.2,0.1,0.4,0.9,NA,0.6,NA,NA,0.3,NA))
setDT(df)
df[, returnInterpol := switch(as.character(sum(!is.na(return))),
"0" = return,
"1" = {na.omit(return)},
na.approx(return, rule = 2)), by = id]
# id year month marketcap return returnInterpol
# 1: A 2014 1 4 NA 0.23
# 2: A 2014 2 6 0.23 0.23
# 3: A 2015 1 2 0.20 0.20
# 4: A 2015 2 6 0.10 0.10
# 5: B 2014 1 23 0.40 0.40
# 6: B 2014 2 2 0.90 0.90
# 7: B 2015 1 5 NA 0.75
# 8: B 2015 2 34 0.60 0.60
# 9: C 2014 1 1 NA 0.30
# 10: C 2014 2 2 NA 0.30
# 11: C 2015 1 3 0.30 0.30
# 12: C 2015 2 4 NA 0.30
答案 1 :(得分:0)
简单的imputeTS
解决方案,无需关心ID,将是:
library("imputeTS")
na.interpolate(df)
由于插补应该根据ID进行,因此稍微复杂一点-因为按ID过滤时,似乎常常剩下的值不足。我会采用罗兰(Roland)发布的解决方案,并在可能的情况下使用imputeTS::na.interpolation()
,在其他情况下,也许可以使用imputeTS::na.mean()
的总体均值或对总体边界imputeTS::na.random()
的随机猜测。 >
在这种情况下,超越单变量时间序列插值/插值法可能也是一个好主意。还有许多其他变量可以帮助估计缺失值(如果存在相关性)。像AMELIA
这样的软件包可以在这里提供帮助。