Removing rows from R data frame

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.

-1 You have a solution contained within the question. There is no problem to solve (as the question is currently worded). — Richie Cotton
– Richie Cotton, Commented Oct 27, 2011 at 13:10
@RichieCotton: My solution uses a modified (different) logical expression which ends up with the result I need; but what I want to see is how to remove specific rows from a data frame. I included my solution in my question because I didn't want to see it in the answers. — Mehper C. Palavuzlar
– Mehper C. Palavuzlar, Commented Oct 27, 2011 at 13:16
I've added a few lines to my question to explain what I want to know. — Mehper C. Palavuzlar
– Mehper C. Palavuzlar, Commented Oct 27, 2011 at 13:22
I think there is confusion over why you want something like df[-a,], when df[df$plan!=0 | df$sevk!=0,] seems to be the correct approach. Could you comment why, in the bigger picture, something like df[-a,] is preferable? Perhaps, in the bigger picture, there is an approach which avoids this problem. — jthetzel
– jthetzel, Commented Oct 27, 2011 at 21:50
It's 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. — Mehper C. Palavuzlar
– Mehper C. Palavuzlar, Commented Oct 28, 2011 at 6:53

Joshua Ulrich · Accepted Answer · 2011-10-27 15:03:08Z

15

Just negate your logical subscript:

a <- df[!(df$plan==0 & df$sevk==0),]

answered Oct 27, 2011 at 15:03

Joshua Ulrich

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Comments

James · Accepted Answer · 2011-10-27 12:10:14Z

12

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)),]

edited Oct 27, 2011 at 12:10

answered Oct 27, 2011 at 12:04

James

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1 Comment

Carl Witthoft Over a year ago

Of course this assumes you have rownames, which the OP did in this case, but it's not a general solution

jthetzel · Accepted Answer · 2011-10-27 12:50:38Z

9

Are you looking for subset()?

dat <- airquality
dat.sub <- subset(dat, Temp > 80 & Month < 10)

dim(dat)
dim(dat.sub)

Applied to your example:

df.sub <- subset(df, plan != 0 & sevk != 0)

answered Oct 27, 2011 at 12:50

jthetzel

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2 Comments

Mehper C. Palavuzlar Over a year ago

This is the same as my solution: df[df$plan!=0 | df$sevk!=0,] which selects a subset; but thanks anyway.

Mehper C. Palavuzlar Over a year ago

BTW, the & operator must be | (OR) operator in subset(df, plan != 0 & sevk != 0).

tim riffe · Accepted Answer · 2011-10-27 13:10:47Z

2

You're almost there. 'a' needs to be a vector of indices:

    df <- data.frame(plan=runif(10),sevk=runif(10))
    a <- c(df$plan<.1 | df$sevk < .1) # some logical thing
    df[-a,]

or, with your data:

    a <- c(df$plan==0 & df$sevk==0)
    df[-a,]

edited Oct 27, 2011 at 13:10

answered Oct 27, 2011 at 11:48

tim riffe

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2 Comments

Mehper C. Palavuzlar Over a year ago

I tried the last two lines of your code with my data, but it gives the wrong result (3148 rows instead of 3128). (BTW, b[-a,] should be df[-a,] I guess)

tim riffe Over a year ago

sorry about the slop- it works with my self-contained little example above, so I guess whatever is going on with your data is over my head

Carl Witthoft · Accepted Answer · 2011-10-27 12:08:36Z

0

I don't see why you object to your solution, but here's another way.

which( df[df$plan==0 & df$sevk==0,], arr.ind=TRUE) ->killlist 
newdf <- df[-c(killlist[1,])]

answered Oct 27, 2011 at 12:08

Carl Witthoft

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