用指数估计填空

Question

我正在尝试用显示指数增长的数字填充 NA 值。下面是我正在尝试做的数据示例。


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)
  ))

我已经使用了 %>% arrange(month, X2009H1N1) %>% group_by(X2009H1N1, type ) %>% mutate(aprox_MidLevelRange = zoo::na.approx(MidLevelRange, na.rm = FALSE))，但结果对我来说并不是指数级的。谢谢

Answer 1

确定你的结果不是指数的，你正在使用函数 na.approx() 来使用线性插值来估算值。您正在使用的 zoo 包提供使用 na.spline() 函数使用三次样条插值进行插值，但此函数也不会产生指数曲线。

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 = zoo::na.spline(MidLevelRange))

三次样条插值的问题是您的最低值将被插值为负值，这取决于您是否正在寻找这种行为：

# A tibble: 8 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          -18568160.
3 0-17 years Cases 2009-06-12            NA       NA          -25223342.
4 0-17 years Cases 2009-07-12            NA       NA          -22929832.
5 0-17 years Cases 2009-08-12            NA       NA          -14651914.
6 0-17 years Cases 2009-09-12            NA       NA           -3353875.
7 0-17 years Cases 2009-10-12       8000000  5000000            8000000.

Answer 2

看看 imputeTS 包。它为时间序列提供了大量的插补函数。查看此 paper 以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。

此处为您提供的示例：

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

所以只是比较插值函数，我想这可能是最好的选择。

自己绘制不同的插值选项，看看差异。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")

来自 imputeTS 的线性/样条选项与 zoo::approx()/ zoo::spline() 相同。 stine 不存在于动物园中。

用指数估计填空

fill in blanks with exponential estimates

r

exponential

imputets