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Drop column from a data frame

4 messages · John Sorkin, jim holtman, Phil Spector +1 more

#
I am trying to drop a column of a data frame. The code below attempts to drop a numeric column (which does not work but gives no error or warning) and a factor column (which does not work but gives an error).
I would appreciate someone telling me why my code does not work, and suggesting code that will work.
Thanks,
John

rm(dfxyz,dfxz,dfxy)

# create the data frame.
dfxyz <- data.frame(x=1:10,y=11:20,z=factor(c(rep(0,5),rep(1,5))))
dfxyz

names(dfxyz)

# try to drop y column
# does not work, does not produce error message
dfxz <- dfxyz[,-(dfxyz$y)]
dfxz

# try to drop z column
# does not work, produces error message:
# In Ops.factor(df$z) : - not meaningful for factors
dfxy <- dfxyz[,-dfxyz$z]
dfxy



John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
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#
assign NULL to the column:
x  y z
1   1 11 0
2   2 12 0
3   3 13 0
4   4 14 0
5   5 15 0
6   6 16 1
7   7 17 1
8   8 18 1
9   9 19 1
10 10 20 1
x z
1   1 0
2   2 0
3   3 0
4   4 0
5   5 0
6   6 1
7   7 1
8   8 1
9   9 1
10 10 1
On Sun, Dec 26, 2010 at 8:22 PM, John Sorkin
<jsorkin at grecc.umaryland.edu> wrote:
#
John -
    You can use a syntax similar to what you've tried with
the select= argument of the subset function:
x z
1   1 0
2   2 0
   . . .
x  y
1   1 11
2   2 12
   . . .


 					- Phil Spector
 					 Statistical Computing Facility
 					 Department of Statistics
 					 UC Berkeley
 					 spector at stat.berkeley.edu
On Sun, 26 Dec 2010, John Sorkin wrote:

            
#
On Dec 26, 2010, at 8:22 PM, John Sorkin wrote:

            
You are misusing the syntax of the "[" operation. When using negative  
indices you can only use numeric or logical values :

?"["

Character indices always need to be "positive".

dfxyz[ , -2]  # works
dfxyz[ , c(T,F,T)] # works

 > dfxyz[ , -"y"]
Error in -"y" : invalid argument to unary operator

This next mechanism also works and us especially useful on dataframes  
with lots of columns:

dfxyz[ , -grep("y", names(dfxyz))]

But you need to be careful to make sure you know which columns will  
match and its good practice to test the grepping expression first:
 > grep("y", names(dfxyz))
[1] 2

If you only wanted to remove "y" and not "y2" you would need to add  
qualifiers to the pattern.
Well, dfxyz$y does evaluate to a numeric vector with values 11:20 and  
there were no columns in that range. So it behaved as documented. You  
asked for the dataframe without some non-existent (numbered) columns  
and it obliged.
Right, you cannot subtract (or negate) factors.

As Phil suggests, subset()-ting is often safer.