Removing rows from R data frame
I have the following data frame:
> str(df)
'data.frame': 3149 obs. of 9 variables:
$ mkod : int 5029 5035 5036 5042 5048 5050 5065 5071 5072 5075 ...
$ mad : Factor w/ 65 levels "Akgün Kasetçilik ",..: 58 29 59 40 56 11 33 34 19 20 ...
$ yad : Factor w/ 44 levels "BAKUGAN","BARBIE",..: 1 1 1 1 1 1 1 1 1 1 ...
$ donem: int 201101 201101 201101 201101 201101 201101 201101 201101 201101 201101 ...
$ sayi : int 201101 201101 201101 201101 201101 201101 201101 201101 201101 201101 ...
$ plan : int 2 2 3 2 2 2 7 3 2 7 ...
$ sevk : int 2 2 3 2 2 2 6 3 2 7 ...
$ iade : int 0 0 3 1 2 2 6 2 2 3 ...
$ satis: int 2 2 0 1 0 0 0 1 0 4 ...
I want to remove 21 specific rows from this data frame.
> a <- df[df$plan==0 & df$sevk==0,]
> nrow(a)
[1] 21
So when I remove those 21 rows, I will have a new data frame with 3149 - 21 = 3128 rows. I found the following solution:
> b <- df[df$plan!=0 | df$sevk!=0,]
> nrow(b)
[1] 3128
My above solution uses a modified logical expression (!= instead of == and | instead of &). Other than modifying the original logical expression, how can I obtain the new data frame without those 21 rows? I need something like that:
> df[-a,] #does not work
EDIT (especially for the downvoters, I hope they understand why I need an alternative solution): I asked for a different solution because I'm writing a long code, and there are various variable assignments (like a's in my example) in various parts of my code. So, when I need to remove rows in advancing parts of my code, I don't want to go back and try to write the inverse of the logical expressions inside a-like expressions. That's why df[-a,] is more usable for me.
Answer
Just negate your logical subscript:
a <- df[!(df$plan==0 & df$sevk==0),]