I have 3 columns
2 are the same length
1 is of a lesser length
here are the columns:
column1 <- letters[1:10]
column2 <- letters[1:15]
column3 <- letters[1:15]
I want all 3 columns to be joined together but have the missing 5 values in column1 to be NA?
What can i do to achieve this? a tibble?
You can change length of a vector
column1 <- letters[1:10]
column2 <- letters[1:15]
length(column1) <- length(column2)
Now
> column1
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" NA NA NA NA NA
We can wrap it in function
cbind_dif <- function(x = list()){
# Find max length
max_length <- max(unlist(lapply(x, length)))
# Set length of each vector as
res <- lapply(x, function(x){
length(x) <- max_length
return(x)
})
return(as.data.frame(res))
}
# Example usage:
> cbind_dif(list(column1 = column1, column2 = column2))
column1 column2
1 a a
2 b b
3 c c
4 d d
5 e e
6 f f
7 g g
8 h h
9 i i
10 j j
11 <NA> k
12 <NA> l
13 <NA> m
14 <NA> n
15 <NA> o
n <- max(length(column1), length(column2), length(column3))
data.frame(column1[1:n],column2[1:n],column3[1:n])
column1.1.n. column2.1.n. column3.1.n.
1 a a a
2 b b b
3 c c c
4 d d d
5 e e e
6 f f f
7 g g g
8 h h h
9 i i i
10 j j j
11 <NA> k k
12 <NA> l l
13 <NA> m m
14 <NA> n n
15 <NA> o o
Using cbind.fill from rowr package you can do it easily.
library(rowr)
new<- cbind.fill(column1,column2,column3)
I hope this helps
column1 <- letters[1:10]
column2 <- letters[1:15]
column3 <- letters[1:15]
tibble(a = c(column1, rep(NA, length(column2) - length(column1))), b = column2, c = column3)
# A tibble: 15 × 3
a b c
<chr> <chr> <chr>
1 a a a
2 b b b
3 c c c
4 d d d
5 e e e
6 f f f
7 g g g
8 h h h
9 i i i
10 j j j
11 NA k k
12 NA l l
13 NA m m
14 NA n n
15 NA o o
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.