William, David, and Peter,
Thank you all so much for your help on this. Though I had read the help files
on 'subset' and '[', I had not been able to discern from that text what the
problem was. I could not have solved it without your help.
The help file on 'subset' mentions For ordinary
On 2012-08-28 07:44, Mauricio Cornejo wrote:
William, David, and Peter,
Thank you all so much for your help on this. Though I had read the help files
on 'subset' and '[', I had not been able to discern from that text what the
problem was. I could not have solved it without your help.
The
...@ucalgary.ca
Cc: William Dunlap wdun...@tibco.com; David Winsemius
dwinsem...@comcast.net; peter dalgaard pda...@gmail.com;
r-help@r-project.org r-help@r-project.org
Sent: Tuesday, August 28, 2012 2:08 PM
Subject: Re: [R] Inexplicably different results using subset vs bracket
notation
ehl...@ucalgary.ca
Cc: William Dunlap wdun...@tibco.com; David Winsemius dwinsem...@comcast.net
; peter dalgaard pda...@gmail.com; r-help@r-project.org r-help@r-project.org
Sent: Tuesday, August 28, 2012 2:08 PM
Subject: Re: [R] Inexplicably different results using subset vs
bracket notation
Hi,
Would anyone have any idea as to why I would obtain completely different
results when subsetting using the subset function vs bracket notation?
I have a data frame with 65 variables and 4382 rows. When I use execute the
following subset command I get the correct results (125 rows)
] Inexplicably different results using subset vs bracket notation
on logical
variable
Hi,
Would anyone have any idea as to why I would obtain completely different
results when
subsetting using the subset function vs bracket notation?
I have a data frame with 65 variables and 4382 rows. When I use
On Aug 27, 2012, at 5:08 PM, Mauricio Cornejo wrote:
Hi,
Would anyone have any idea as to why I would obtain completely
different results when subsetting using the subset function vs
bracket notation?
I have a data frame with 65 variables and 4382 rows. When I use
execute the
On Aug 28, 2012, at 05:11 , David Winsemius wrote:
That's exactly it. If a logical index returns NA, its row is included in the
output of [ extraction. You can correct what I consider a failing and
others consider a feature with:
df[df$Renewal==TRUE !is.na(df$Renewal), 1:2]
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