Question

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 1

You can use the rownames to specify a “complementary” dataframe. Its easier if they are numerical rownames:

df[-as.numeric(rownames(a)),]

But more generally you can use:

df[setdiff(rownames(df),rownames(a)),]