Fill missing values with new data R-Python - r

Question

I have two dataset x and y
> x
a index b
1 1 1 5
2 NA 2 6
3 2 3 NA
4 NA 4 9
> y
index a
1 2 100
2 4 101
>
I would like to fill the missing values of x with the values contained in y.
I have tried to use the merge function but the result is not what I want.
> merge(x,y, by = 'index', all=T)
index a.x b a.y
1 1 1 5 NA
2 2 NA 6 100
3 3 2 7 NA
4 4 NA 9 101
In the real problem there are additional limitations:
1 - y does not fill all the missing values
2 - x and y have in common more variables (so not only a and index)
EDIT : More realistic example
> x
a index b c
1 1 1 5 NA
2 NA 2 6 NA
3 2 3 NA 5
4 NA 4 9 NA
5 NA 5 10 6
> y
index a c
1 2 100 4
2 4 101 NA
>
The solution would be accepted both in python or R

Answer 1

I used your merge idea and did the following using dplyr. I am sure there will be better ways of doing this task.
index <- 1:5
a <- c(1, NA, 2, NA, NA)
b <- c(5,6,NA,9,10)
c <- c(NA,NA,5,NA,6)
ana <- data.frame(index, a,b,c, stringsAsFactors=F)
index <- c(2,4)
a <- c(100, 101)
c <- c(4, NA)
bob <- data.frame(index, a,c, stringsAsFactors=F)
> ana
index a b c
1 1 1 5 NA
2 2 NA 6 NA
3 3 2 NA 5
4 4 NA 9 NA
5 5 NA 10 6
> bob
index a c
1 2 100 4
2 4 101 NA
ana %>%
merge(., bob, by = "index", all = TRUE) %>%
mutate(a.x = ifelse(a.x %in% NA, a.y, a.x)) %>%
mutate(c.x = ifelse(c.x %in% NA, c.y, c.x))
index a.x b c.x a.y c.y
1 1 1 5 NA NA NA
2 2 100 6 4 100 4
3 3 2 NA 5 NA NA
4 4 101 9 NA 101 NA
5 5 NA 10 6 NA NA
I overwrote a.x (ana$$a) using a.y (bob$a) using mutate. I did a similar thing for c.x (ana$c). If you remove a.y and c.y in the end, that will be the outcome you expect, I think.

Answer 2

Try:
xa = x[,c(1,2)]
m1 = merge(y,xa,all=T)
m1 = m1[!duplicated(m1$index),]
m1$b = x$b[match(m1$index, x$index)]
m1$c = x$c[match(m1$index, x$index)]
m1
index a b c
1 1 1 5 NA
2 2 100 6 NA
4 3 2 NA 5
5 4 101 9 NA
7 5 NA 10 6
or, if there many other columns like b and c:
xa = x[,c(1,2)]
m1 = merge(y,xa,all=T)
m1 = m1[!duplicated(m1$index),]
for(nn in names(x)[3:4]) m1[,nn] = x[,nn][match(m1$index, x$index)]
m1
index a b c
1 1 1 5 NA
2 2 100 6 NA
4 3 2 NA 5
5 4 101 9 NA
7 5 NA 10 6

Answer 3

If there are multiple columns to replace, you could try converting from wide to long form as shown in the first two methods and replace in one step
m1 <- merge(x,y, by="index", all=TRUE)
m1L <- reshape(m1, idvar="index", varying=grep("\\.", colnames(m1)), direction="long", sep=".")
row.names(m1L) <- 1:nrow(m1L)
lst1 <- split(m1L, m1L$time)
indx <- is.na(lst1[[1]][,4:5])
lst1[[1]][,4:5][indx] <- lst1[[2]][,4:5][indx]
res <- lst1[[1]][,c(4,1,2,5)]
res
# a index b c
#1 1 1 5 NA
#2 100 2 6 4
#3 2 3 NA 5
#4 101 4 9 NA
#5 NA 5 10 6
Or you could use dplyr with tidyr
library(dplyr)
library(tidyr)
z <- left_join(x, y, by="index") %>%
gather(Var, Val, matches("\\.")) %>%
separate(Var, c("Var1", "Var2"))
indx1 <- which(is.na(z$Val) & z$Var2=="x")
z$Val[indx1] <- z$Val[indx1+nrow(z)/2]
z %>%
spread(Var1, Val) %>%
filter(Var2=="x") %>%
select(-Var2)
# index b a c
#1 1 5 1 NA
#2 2 6 100 4
#3 3 NA 2 5
#4 4 9 101 NA
#5 5 10 NA 6
Or split the columns by matching names before the . and use lapply to replace the NA's.
indx <- grep("\\.", colnames(m1),value=TRUE)
res <- cbind(m1[!names(m1) %in% indx],
sapply(split(indx, gsub("\\..*", "", indx)), function(x) {
x1 <- m1[x]
indx1 <- is.na(x1[,1])
x1[,1][indx1] <- x1[,2][indx1]
x1[,1]} ))
res
# index b a c
#1 1 5 1 NA
#2 2 6 100 4
#3 3 NA 2 5
#4 4 9 101 NA
#5 5 10 NA 6