在 dplyr group/variable 对上跳过 na_interpolation，在 R 中具有完整的 NA

Question

我有一个如下所示的数据框：

   Country Year acnt_class     wages
3      AZE 2010         NA        NA
4      AZE 2011  0.4206776        NA
5      AZE 2012         NA        NA
6      AZE 2013         NA        NA
7      AZE 2014  0.7735889 0.4273174
8      AZE 2015         NA        NA
9      AZE 2016         NA        NA
10     AZE 2017  0.5108674 0.4335978
11     AZE 2018         NA        NA
15     BDI 2010         NA        NA
16     BDI 2011  0.3140646        NA
17     BDI 2012         NA        NA
18     BDI 2013         NA        NA
19     BDI 2014  0.1224175        NA
20     BDI 2015         NA        NA
21     BDI 2016         NA        NA
22     BDI 2017         NA        NA
23     BDI 2018         NA        NA
27     BEL 2010         NA        NA
28     BEL 2011  0.9576057        NA
29     BEL 2012         NA        NA
30     BEL 2013         NA        NA
31     BEL 2014  1.0083120 0.9623492
32     BEL 2015         NA        NA
33     BEL 2016         NA        NA
34     BEL 2017  1.0036910 0.9499486
35     BEL 2018         NA        NA

我正在尝试运行此函数使用 stine 插值法跨两个变量列 "acnt_class" 和 "wages":

按组填充缺失的 NA

DF <- DF %>% 
  group_by(Country) %>% 
  mutate_at(.vars = c("acnt_class", "wages"), 
            .funs = ~na_interpolation(., option = "stine"))

只要我运行它在每组至少有两个观察值的列上工作，但是，在这里，我运行进入这个错误：

Error in na_interpolation(., option = "stine") : 
  Input data needs at least 2 non-NA data point for applying na_interpolation

由于组 "BDI" 具有变量 "wages" 的完整 NA。

理想情况下，我正在寻找一个修改后的函数，它将 "skip" group/variable 与完整的 NAs/1 观察配对并保持原样。解决方案？谢谢！

Answer 1

找到解决方案：

仅用于插值：

library(TSimpute)
library(dplyr)
library(zoo)

DF <- DF %>% 
  group_by(Country) %>% 
  mutate_at(vars(acnt_class, wages), funs(if(sum(!is.na(.))<2) {.} else{replace(na_interpolation(., option = "stine"), is.na(na.approx(., na.rm=FALSE)), NA)}))

Answer 2

TiberiusGracchus2020 提供的答案效果很好。如果它对任何人都有帮助，我已将该代码片段转换为一个带有大量注释的函数，以便更清楚地了解每个阶段发生的事情。

# Modify imputeTS::na_interpolate function
#   (1) doesn't break on all NA vectors
#   (2) won't impute leading and lagging NAs

na_interpolation2 <- function(x, option = "linear") {
  library(TSimpute)
  library(dplyr)

  total_not_missing <- sum(!is.na(x))
  
  # check there is sufficient data for na_interpolation 
  if(total_not_missing < 2) {x} 

    else

    # replace takes an input vector, a T/F vector & replacement value
    {replace(
        # input vector is interpolated data
        # this will impute leading/lagging NAs which we don't want 
        imputeTS::na_interpolation(x, option = option), 

        # create T/F vector for NAs,  
        is.na(na.approx(x, na.rm = FALSE)), 

        # replace TRUE with NA in input vector  
        NA) 
      }
}

# example data
data1 <- c(NA, NA, NA, NA, NA) 
data2 <- c(NA, NA, 1, NA, 3, NA)

na_interpolation(data1)
# Error in na_interpolation(data1) : Input data needs at 
# least 2 non-NA data point for applying na_interpolation

na_interpolation(data2)
# [1] 1 1 1 2 3 3

na_interpolation2(data1)
# [1] NA NA NA NA NA

na_interpolation2(data2)
# [1] NA NA  1  2  3 NA

在 dplyr group/variable 对上跳过 na_interpolation，在 R 中具有完整的 NA

Skip na_interpolation on dplyr group/variable pairs with full NAs in R

interpolation

r

na

dplyr