Reputation: 751
I have the following data.frame.
a <- c(rep("A", 3), rep("B", 3), rep("C",2), "D")
b <- c(NA,1,2,4,1,NA,2,NA,NA)
c <- c(1,1,2,4,1,1,2,2,2)
d <- c(1,2,3,4,5,6,7,8,9)
df <-data.frame(a,b,c,d)
a b c d
1 A NA 1 1
2 A 1 1 2
3 A 2 2 3
4 B 4 4 4
5 B 1 1 5
6 B NA 1 6
7 C 2 2 7
8 C NA 2 8
9 D NA 2 9
I want to remove duplicate rows (based on column A & C) so that the row with values in column B are kept. In this example, rows 1, 6, and 8 are removed.
Upvotes: 0
Views: 223
Reputation: 11
First create two datasets, one with duplicates in column a and one without duplicate in column a using the below function :
x = df[df$a %in% names(which(table(df$a) > 1)), ]
x1 = df[df$a %in% names(which(table(df$a) ==1)), ]
Now use na.omit function on data set x to delete the rows with NA and then rbind x and x1 to the final data set.
rbind(na.omit(x),x1)
Answer:
a b c d
2 A 1 1 2
3 A 2 2 3
4 B 4 4 4
5 B 1 1 5
7 C 2 2 7
9 D NA 2 9
Upvotes: 1
Reputation: 2725
You can use dplyr
to do this.
df %>% distinct(a, c, .keep_all = TRUE)
a b c d
1 A NA 1 1
2 A 2 2 3
3 B 4 4 4
4 B 1 1 5
5 C 2 2 7
6 D NA 2 9
There are other options in dplyr
, check this question for details: Remove duplicated rows using dplyr
Upvotes: 0
Reputation: 887971
One way to do this is to order
by 'a', 'b' and the the logical vector based on 'b' so that all 'NA' elements will be last for each group of 'a', and 'b'. Then, apply the duplicated
and keep only the non-duplicate elements
df1 <- df[order(df$a, df$b, is.na(df$b)),]
df2 <- df1[!duplicated(df1[c('a', 'c')]),]
df2
# a b c d
#2 A 1 1 2
#3 A 2 2 3
#5 B 1 1 5
#4 B 4 4 4
#7 C 2 2 7
#9 D NA 2 9
setdiff(seq_len(nrow(df)), row.names(df2) )
#[1] 1 6 8
Upvotes: 1