我正在尝试用显示指数增长的数字填充 NA 值。以下是我正在尝试执行的操作的数据示例。
library(tidyverse)
expand.grid(X2009H1N1 = "0-17 years",
type = "Cases",
month = seq(as.Date("2009-04-12") , to = as.Date("2010-03-12"), by = "month")) %>%
bind_cols( data.frame(
MidLevelRange = c(0,NA,NA,NA,NA,NA,8000000,16000000,18000000,19000000,19000000,19000000),
lowEst = c(0,NA,NA,NA,NA,NA,5000000,12000000,12000000,13000000,14000000,14000000)
))
我使用了%>%排列(月,X2009H1N1)%>%
group_by(X2009H1N1, 类型) %>%
mutate(aprox_MidLevelRange = Zoo::na.approx(MidLevelRange, na.rm = FALSE))
但结果对我来说看起来并不是指数级的。谢谢
最佳答案
看看 imputeTS 包。 它为时间序列提供了大量的插补函数。看看这个paper全面了解所有提供的选项
在您的情况下,使用 Stineman 插值( imputeTS::na_interpolation(x, option ="stine"
) 可能是一个合适的选择。
这里是您提供的示例:
x <- expand.grid(
X2009H1N1 = "0-17 years",
type = "Cases",
month = seq(as.Date("2009-04-12"),
to = as.Date("2010-03-12"),
by = "month"
)
) %>%
bind_cols(data.frame(
MidLevelRange = c(0, NA, NA, NA, NA, NA, 8000000, 16000000, 18000000, 19000000, 19000000, 19000000),
lowEst = c(0, NA, NA, NA, NA, NA, 5000000, 12000000, 12000000, 13000000, 14000000, 14000000)
))
x %>%
arrange(month, X2009H1N1) %>%
group_by(X2009H1N1, type) %>%
mutate(aprox_MidLevelRange = imputeTS::na_interpolation(MidLevelRange, option = "stine"))
这给你:
# A tibble: 12 x 6
# Groups: X2009H1N1, type [1]
X2009H1N1 type month MidLevelRange lowEst aprox_MidLevelRange
<fct> <fct> <date> <dbl> <dbl> <dbl>
1 0-17 years Cases 2009-04-12 0 0 0
2 0-17 years Cases 2009-05-12 NA NA 593718.
3 0-17 years Cases 2009-06-12 NA NA 1335612.
4 0-17 years Cases 2009-07-12 NA NA 2289061.
5 0-17 years Cases 2009-08-12 NA NA 3559604.
6 0-17 years Cases 2009-09-12 NA NA 5336975.
7 0-17 years Cases 2009-10-12 8000000 5000000 8000000
8 0-17 years Cases 2009-11-12 16000000 12000000 16000000
9 0-17 years Cases 2009-12-12 18000000 12000000 18000000
10 0-17 years Cases 2010-01-12 19000000 13000000 19000000
11 0-17 years Cases 2010-02-12 19000000 14000000 19000000
12 0-17 years Cases 2010-03-12 19000000 14000000 19000000
因此,仅比较插值函数,我想这可能是最好的选择。
只需绘制不同的插值选项,即可看到差异。 一般来说,这是插值选项:
imputeTS::na_interpolation(x, option ="linear")
imputeTS::na_interpolation(x, option ="spline")
imputeTS::na_interpolation(x, option ="stine")
imputeTS 中的线性/样条选项与zoo::approx()/zoo::spline() 相同。动物园里不存在斯汀。
关于r - 用指数估计填空,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60816527/