Deleting rows based on identity variable

Is this what you want:

d1[which(id != 4),]

id x    y
 [1,]  1 1 0.00
 [2,]  1 2 0.10
 [3,]  1 3 0.50
 [4,]  1 4 0.40
 [5,]  1 5 0.20
 [6,]  2 1 0.00
 [7,]  2 2 0.10
 [8,]  2 3 0.50
 [9,]  2 4 0.40
[10,]  2 5 0.12
[11,]  3 1 0.00
[12,]  3 2 0.10
[13,]  3 3 0.50
[14,]  3 4 0.55
[15,]  3 5 0.20
[16,]  5 1 0.00
[17,]  5 2 0.10
[18,]  5 3 0.60
[19,]  5 4 0.40
[20,]  5 5 0.10

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of gina patel
Sent: Thursday, April 02, 2009 4:30 PM
To: R-help at r-project.org
Subject: [R] Deleting rows based on identity variable


I have created this data frame to help illustrate my problem.

id<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
x<-rep((seq(1:5)),5)
y<-c(0, 0.1, 0.5, 0.4, 0.2, 0, 0.1, 0.5, 0.4, 0.12, 0, 0.1, 0.5, 0.55, 0.2, 0, 0.1, 0.5, 0.3, 0.2, 0, 0.1, 0.6, 0.4, 0.1)
d1<-cbind(id,x,y)

I would like to delete all rows where id=4, however, when I tried the command

 d2=d1[-c(id==4),]

and looked at d2 no data was removed.

Thanks in advance, I appreciate any help.




      


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On Thu, Apr 2, 2009 at 3:37 PM, Rowe, Brian Lee Yung (Portfolio

Is this what you want:

d1[which(id != 4),]

Or just

d1[id != 4, ]

Hadley

http://had.co.nz/

I have created this data frame to help illustrate my problem.

id<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
x<-rep((seq(1:5)),5)
y<-c(0, 0.1, 0.5, 0.4, 0.2, 0, 0.1, 0.5, 0.4, 0.12, 0, 0.1, 0.5,  
0.55, 0.2, 0, 0.1, 0.5, 0.3, 0.2, 0, 0.1, 0.6, 0.4, 0.1)
d1<-cbind(id,x,y)

I would like to delete all rows where id=4, however, when I tried  
the command

 d2=d1[-c(id==4),]

and looked at d2 no data was removed.

Thanks in advance, I appreciate any help.

Others have told you how to do what you want.  It would be useful
for you to understand why what you ***did*** didn't work.

Note that ``id==4'' yields a *logical* vector with entries TRUE and  
FALSE.

When an arithmetic operator is applied, as in -(id==4) [NOTE: THE ``c 
()'' wrapper
is TOTALLY UNNECESSARY HERE.] the vector is coerced to a numeric  
vector of
-1's (for TRUE) and 0's (for FALSE).  Thus you are asking d2 to be d1 
[i,] where
i is a vector of -1's and 0's.  The -1's say to throw away the ``1- 
th'' (first)
row of d1; the 0's say to pick out the 0-th row of d1 (and there is  
no 0-th row
so this doesn't do anything).  Thus you are in effect asking for d1 
[2:25,].

And that's what you get.  Look carefully --- it is not correct to say  
that when
you looked at d2 *no* data were removed; d2 is equal to d1 with its  
first row
removed.  I.e. your operation didn't remove what you wanted, but it  
removed
*something*.

	cheers,

		Rolf Turner

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