Loop function for comparing the columns - r

Question

I have a very large data set including 400 string and numeric variables. I want to compare each two consequiative columns 3&4, 5&6, etc. I am going to compare the third variable (.x) with fourth (.y) , fifth with sixth one, seventh one with eightth one and so on in the following way: if (.y) is NA then we replace the NA with the value of corresponding row from (.x) . For example if number .y is NA we replace NA with the corresponding value from number .x which would be 5. Again, if day.y is NA we replace NA in day.y with the corresponding value from day.x which would be 3. How can I write a loope function to do that?
A<-c(1,2,3,4,5,6,7,NA,NA,5,5,6)
B<-c(3,4,5,6,1,2,7,6,7,NA,NA,6)
number.x<-c(1,2,3,4,5,6,7,NA,NA,5,5,6)
number.y<-c(3,4,5,6,1,2,7,6,7,NA,NA,6)
day.x<-c(1,3,4,5,6,7,8,1,NA,3,5,3)
day.y<-c(4,5,6,7,8,7,8,1,2,3,5,NA)
school.x<-c("a","b","b","c","n","f","h","NA","F","G","z","h")
school.y<-c("a","b","b","c","m","g","h","NA","NA","G","H","T")
city.x<- c(1,2,3,7,5,8,7,5,6,7,5,1)
city.y<- c(1,2,3,5,5,7,7,NA,NA,3,4,5)
df<-data.frame(A,B,number.x,number.y,day.x,day.y,school.x,school.y,city.x,city.y)

Answer 1

This is a hacked approach to your question and it requires that every two columns are going to be compared against one another.
library(dplyr)
start_group <- seq(1, length(df), by = 2)
df2 <- data.frame(id = 1:nrow(df))
for(i in start_group){
i <- i
j <- i + 1
dnames <- df[, c(i, j)] %>%
names
df_ <- data.frame(col1 = df[, i],
col2 = df[, j]) %>%
mutate(col1 = ifelse(is.na(col1), col2 %>% paste, col1 %>% paste)) %>%
mutate(col2 = ifelse(is.na(col2), col1 %>% paste, col2 %>% paste))
names(df_) <- dnames
df2 <- cbind(df2, df_)
}
df2[, -1]
number.x number.y day.x day.y school.x school.y city.x city.y
1 1 3 1 4 a a 1 1
2 2 4 3 5 b b 2 2
3 3 5 4 6 b b 3 3
4 4 6 5 7 c c 7 5
5 5 1 6 8 n m 5 5
6 6 2 7 7 f g 8 7
7 7 7 8 8 h h 7 7
8 6 6 1 1 NA NA 5 5
9 7 7 2 2 F F 6 6
10 5 5 3 3 G G 7 3
11 5 5 5 5 z H 5 4
12 6 6 3 3 h T 1 5

Answer 2

Consider the following base R solution. Essentially, it loops through a distinct list of column stem names (number, day, school, class) and replaces NA values in .x columns with corresponding NA values in .y columns and vice versa. NOTE: Schools column require conversion from factor to character and one of its rows has NA in both .x and .y columns
# CONVERT TO CHARACTER (NOTE: NA VALUE BECOME "NA" STRINGS)
df[,c('school.x', 'school.y')] <-
sapply(df[,c('school.x', 'school.y')], as.character)
# SET UP FINAL DF
finaldf <- df
# OBTAIN UNIQUE LIST OF COLUMNS STEM (W/O x AND y SUFFIXES)
distinctcols <- unique(gsub("[.][x]|[.][y]", "", names(df)[49:ncol(df)]))
# LOOP THROUGH COLUMN STEM REPLACING NA VALUES
for (col in distinctcols) {
# REPLACE NA .x COLUMN VALUES
finaldf[is.na(finaldf[paste0(col,'.x')])|finaldf[paste0(col,'.x')]=="NA",
paste0(col,'.x')] <-
finaldf[is.na(finaldf[paste0(col,'.x')])|finaldf[paste0(col,'.x')]=="NA",
paste0(col,'.y')]
# REPLACE NA .y COLUMN VALUES
finaldf[is.na(finaldf[paste0(col,'.y')])|finaldf[paste0(col,'.y')]=="NA",
paste0(col,'.y')] <-
finaldf[is.na(finaldf[paste0(col,'.y')])|finaldf[paste0(col,'.y')]=="NA",
paste0(col,'.x')]
}
OUTPUT
number.x number.y day.x day.y school.x school.y city.x city.y
1 1 3 1 4 a a 1 1
2 2 4 3 5 b b 2 2
3 3 5 4 6 b b 3 3
4 4 6 5 7 c c 7 5
5 5 1 6 8 n m 5 5
6 6 2 7 7 f g 8 7
7 7 7 8 8 h h 7 7
8 6 6 1 1 NA NA 5 5
9 7 7 2 2 F F 6 6
10 5 5 3 3 G G 7 3
11 5 5 5 5 z H 5 4
12 6 6 3 3 h T 1 5