I have a R data frame df below
a b c
1 6 NA
2 NA 4
3 7 NA
NA 8 1
4 9 10
NA NA 7
5 10 8
I want to remove the row which has NA in BOTH a & b
My desired output will be
a b c
1 6 NA
2 NA 4
3 7 NA
NA 8 1
4 9 10
5 10 8
I tried something like this below
df1<-df[(is.na(df$a)==FALSE & is.na(df$b)==FALSE),]
but this removes all the NAs (performs an OR function). I need to do AND operation here.
How do i do it ?
using rowSums
df[!rowSums(is.na(df))==2,]
better one by saving a character[1]
df[rowSums(is.na(df))!=2,]
output:
a b
1 1 6
2 2 NA
3 3 7
4 NA 8
5 4 9
7 5 10
can be generalized using ncol
df[!rowSums(is.na(df))==ncol(df),]
[1] credits: alistaire
We can use rowSums on a logical matrix (is.na(df1)) and convert that to a logical vector (rowSums(...) < ncol(df1)) to subset the rows.
df1[rowSums(is.na(df1)) < ncol(df1),]
Or another option is Reduce with lapply
df1[!Reduce(`&`, lapply(df1, is.na)),]
Another approach
df[!apply(is.na(df),1,all),]
# a b
#1 1 6
#2 2 NA
#3 3 7
#4 NA 8
#5 4 9
#7 5 10
Data
df <- structure(list(a = c(1L, 2L, 3L, NA, 4L, NA, 5L), b = c(6L, NA,
7L, 8L, 9L, NA, 10L)), .Names = c("a", "b"), class = "data.frame", row.names = c(NA,
-7L))
this will also work:
df[apply(df, 1, function(x) sum(is.na(x)) != ncol(df)),]
a b
1 1 6
2 2 NA
3 3 7
4 NA 8
5 4 9
7 5 10
My thought is basically the same with other replies.
Considering any dataset with a specific row having all NAs, the sum of !is.na(ROW) will always be zero. So you just have to take out that row.
So you can just do:
df1 = df[-which(rowSums(!is.na(df))==0),]
