如何用R中的时间顺序列表填充NA？

Question

Say I have a chronological list of household incomes and a dataframe of Town IDs with household incomes but there are some NAs I want to fill in.

HouseholdIncome_list <- c(10000, 20000, 30000,40000,50000, 60000, 70000) 
                                                 
Town_ID <- c("A", "A", "A", "A", "B", "B", "B", "B", "B")
HouseholdIncome <- c(10000, 40000, 50000, NA, 20000, 40000, NA, NA, 60000)

df <- data.frame(Town_ID, HouseholdIncome)

  Town_ID HouseholdIncome
1       A           10000
2       A           40000
3       A           50000
4       A              NA
5       B           20000
6       B           40000
7       B              NA
8       B              NA
9       B           60000

How do fill in the NAs in the dataframe so that the missing values are the ones in the list. So it looks like the df below

  Town_ID HouseholdIncome
1       A           10000
2       A           40000
3       A           50000
4       A           60000
5       B           20000
6       B           40000
7       B           50000
8       B           50000
9       B           60000

I have spent time searching for some sort of na fill option but cannot find one that realtes to a given list

Answer 1

This is a terrible solution, but it get's your job done.

library(tidyr)
library(dplyr)

df %>% 
  group_by(grp = cumsum(!is.na(HouseholdIncome))) %>% 
  rowwise() %>%
  mutate(Income = ifelse(length(which(HouseholdIncome_list == HouseholdIncome)) > 0,
    HouseholdIncome_list[which(HouseholdIncome_list == HouseholdIncome) + 1],
    NA_real_)) %>% 
  ungroup() %>% 
  fill(Income) %>% 
  mutate(HouseholdIncome = ifelse(is.na(HouseholdIncome), Income, HouseholdIncome)) %>% 
  select(Town_ID, HouseholdIncome)

returns

# A tibble: 9 x 2
  Town_ID HouseholdIncome
  <chr>             <dbl>
1 A                 10000
2 A                 40000
3 A                 50000
4 A                 60000
5 B                 20000
6 B                 40000
7 B                 50000
8 B                 50000
9 B                 60000

If your first item is NA this won't work.

Answer 2

here is another approach based on joins that will also impute the frist value of a group in case that is missing:

library(tidyverse)

rdf <- data.frame(HouseholdIncome_list = c(10000, 20000, 30000,40000,50000, 60000, 70000)) %>%
    dplyr::mutate(rn = as.double(dplyr::row_number()))
                                             
df <- data.frame(Town_ID = c("A", "A", "A", "A", "B", "B", "B", "B", "B"),
                 HouseholdIncome = c(10000, 40000, 50000, NA, 20000, 40000, NA, NA, 60000))

df %>%
    dplyr::left_join(rdf, by = c("HouseholdIncome" = "HouseholdIncome_list")) %>%
    dplyr::group_by(Town_ID) %>%
    tidyr::fill(rn, .direction = "down") %>%
    tidyr::fill(rn, .direction = "up") %>%
    dplyr::mutate(rn2 = dplyr::row_number()) %>%
    dplyr::ungroup() %>% 
    dplyr::mutate(rn = case_when(is.na(HouseholdIncome) & rn2 == 1 & rn == min(rdf$rn) ~ rn,
                                 is.na(HouseholdIncome) & rn2 == 1 ~ rn - 1,
                                 is.na(HouseholdIncome) & rn < max(rdf$rn) ~ rn + 1,
                                 TRUE ~ rn)) %>%
    dplyr::left_join(rdf, by = "rn") %>%
    select(Town_ID, HouseholdIncome = HouseholdIncome_list)

# A tibble: 9 x 2
  Town_ID HouseholdIncome
  <chr>             <dbl>
1 A                 10000
2 A                 40000
3 A                 50000
4 A                 60000
5 B                 20000
6 B                 40000
7 B                 50000
8 B                 50000
9 B                 60000

Answer 3

I would "cheat" a bit using tidyverse . Clearly, the Household Income is in 10.000 intervals, and we can therefore utilise this,

df %>% mutate(
        is_na = as.numeric(is.na(HouseholdIncome)) * 10000
) %>% fill(
        HouseholdIncome, .direction = "down"
) %>% mutate(
        HouseholdIncome =(HouseholdIncome + is_na),
        is_na = NULL
)

First we check for NA , here is_na = 1 * 10000 if TRUE , and then we use fill to carry the last values forward.

In the end we sum our cheater variable is_na and HouseholdIncome to get the next HouseholdIncome interval.

The output is the following,

  Town_ID HouseholdIncome
1       A           10000
2       A           40000
3       A           50000
4       A           60000
5       B           20000
6       B           40000
7       B           50000
8       B           50000
9       B           60000

Answer 4

A possible base R option

transform(
    df,
    HouseholdIncome = ave(
        HouseholdIncome,
        Town_ID,
        FUN = function(x) replace(x, is.na(x), x[min(which(is.na(x))) - 1] + 1e4)
    )
)

gives

  Town_ID HouseholdIncome
1       A           10000
2       A           40000
3       A           50000
4       A           60000
5       B           20000
6       B           40000
7       B           50000
8       B           50000
9       B           60000