R replace_na values conditionally by column with multiple conditions - r

Question

My question is similar to other replace_na posts but I can't find the right combination of answers.
I have a dataframe with inflation rates for all countries over 8 years (wide format - countries as rows and years as columns).
Some countries have NAs for all 8 years (columns 3:10), and in that case I want to replace all NAs with the column mean
library(tidyverse)
sample %>%
mutate_if((rowSums(is.na[,3:10]))!=8, replace_na = colMeans(na.rm=T))
This is close but something is wrong.
Other countries only have NAs in some columns, in which case I want to replace NA with the previous year's value
library(zoo)
sample %>%
mutate_if((rowSums(is.na[,3:10]))!=8, replace_na = colMeans(na.rm=T)),
is.na[,4:10], na.locf(fromLast = TRUE)))
Tried using na.locf from the zoo package but can't get it right with the other conditions
The final condition is that, if the NA is in the first year (2007), I want to replace it with the 2007 column mean instead of the next year (2008 was the financial crisis so all the inflation rates went nuts).
mutate_if((rowSums(is.na[,3:10]))!=8, replace_na = colMeans(na.rm=T)),
is.na[,4:10], na.locf(fromLast = TRUE)),
is.na("2007"), replace = colMeans("2007", na.rm = TRUE))
But this is full of errors and I'm stuck trying to link all these conditions together - pretty new to ifelse statements. I'm trying to find a dplyr solution as that's the syntax I'm most familiar with, but maybe it's easier in base R or data.table
running R 3.6.1
sample <- structure(list(`Country Name` = c("Aruba", "Afghanistan", "Angola",
"Albania", "Andorra", "Arab World", "United Arab Emirates", "Argentina",
"Armenia", "American Samoa", "Antigua and Barbuda", "Australia"
), `Country Code` = c("ABW", "AFG", "AGO", "ALB", "AND", "ARB",
"ARE", "ARG", "ARM", "ASM", "ATG", "AUS"), `2007` = c(5.39162036843645,
8.68057078513406, 12.2514974459487, 2.93268248162318, NA, 4.74356585295154,
NA, NA, 4.40736089644519, NA, 1.41605259409743, 2.32761128891476
), `2008` = c(8.95722105296535, 26.4186641547444, 12.4758291326398,
3.36313757366391, NA, 11.2706652380848, 12.2504202448139, NA,
8.94995335353386, NA, 5.33380639820232, 4.35029854990047), `2009` = c(-2.13630037272305,
-6.81116108898995, 13.7302839288409, 2.23139683475865, NA, 2.92089711805365,
1.55980098148558, NA, 3.40676682683799, NA, -0.550159995508869,
1.77111716621252), `2010` = c(2.07773902027782, 2.1785375238942,
14.4696564932574, 3.61538461538463, NA, 3.91106195534027, 0.879216764156813,
NA, 8.17636138473956, NA, 3.3700254022015, 2.91834002677376),
`2011` = c(4.31633194082721, 11.8041858089129, 13.4824679218511,
3.44283593170005, NA, 4.75316388885632, 0.877346595685083,
NA, 7.6500080785929, NA, 3.45674967234599, 3.30385015608744
), `2012` = c(0.627927921638161, 6.44121280934118, 10.2779049218839,
2.03642235579081, NA, 4.61184432206646, 0.662268900269082,
NA, 2.55802007757907, NA, 3.37688044338879, 1.76278015613193
), `2013` = c(-2.37226328015073, 7.38577178397857, 8.77781429332619,
1.92544399507649, NA, 3.23423783752364, 1.10111836375706,
NA, 5.78966778544654, NA, 1.05949782356168, 2.44988864142539
), `2014` = c(0.421637771012246, 4.67399603536339, 7.28038730361125,
1.61304235314414, NA, 2.77261158414198, 2.34626865671643,
NA, 2.98130868933673, NA, 1.08944157435363, 2.48792270531403
)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-12L))

Answer 1

First compute a logical vector, all.na having one component per row which is TRUE if that row's numeric data is all NAs and FALSE otherwise. Then use na.aggregate to fill in all-NA rows. Also use na.aggregate on 2007. Then convert to long form and apply na.locf0 by country and convert back to wide form.
library(dplyr)
library(tidyr)
library(zoo)
all.na <- sample %>%
select_if(is.numeric) %>%
{ rowSums(is.na(.)) == ncol(.) }
sample %>%
mutate_at(-(1:3), ~ if_else(all.na, na.aggregate(.x), .x)) %>%
mutate(`2007` = na.aggregate(`2007`)) %>%
gather(key, value, -`Country Name`, -`Country Code`) %>%
group_by(`Country Name`, `Country Code`) %>%
mutate(value = na.locf0(value)) %>%
ungroup %>%
spread(key, value)
or using only zoo:
library(zoo)
all.na <- apply(is.na(sample[grep("^2", names(sample))]), 1, all)
ix <- -(1:3)
sample.out <- sample
Fill <- function(x) ifelse(all.na, na.aggregate(x), x)
sample.out[ix] <- lapply(sample[ix], Fill)
sample.out$"2007" <- na.aggregate(sample.out$"2007")
sample.out[ix] <- t(apply(sample.out[ix], 1, na.locf0))

Answer 2

Tried to do the same by only using the tidyverse (dplyr and tidyr), this is what I got:
Created the Replacement according to condition 1. Column Means.
# Getting the Column Means to Replace according to Condition 1 and 3.
replacement <- sample %>%
select_if(is.numeric) %>%
summarize_all( ~ mean(., na.rm = TRUE)) %>%
#Transformed to List since it is a requirement for tidyr::replace_na()
as.list()
Then I created everything into just one pipeline.
sample %>%
pivot_longer(`2007`:`2014`, names_to = "year", values_to = "int_rate") %>%
group_by(`Country Name`) %>%
summarize(na_num = is.na(int_rate) %>% sum) %>%
#Joining the number of NAs na_num as a new column
left_join(sample, by = "Country Name") %>%
#Replacing 2007 missing as a first value. Condition 3.
mutate(`2007` = if_else(between(na_num, 1, 7) &
is.na(`2007`), replacement[[1]] , `2007`)) %>%
#Making dataset wider
pivot_longer(`2007`:`2014`, names_to = "year", values_to = "int_rate") %>%
group_by(`Country Name`) %>%
#Using fill to impute NAs with the previous one. Condition 2.
fill(int_rate) %>%
pivot_wider(names_from = year, values_from = int_rate) %>%
#Replacing Values when all values are missing. Condition 1.
replace_na(replace = replacement)