Dear all,
According to the post I was trying:
factorA = c(2,2,3,3,4,4,3,4,2,2)
levels(factor - c(lv1,lv2,lv3) )
But this returns NULL and doesn't change factor names.
Actually, my factor is included in a data.frame, so I also tried:
levels(df$factorA)[levels(df$factorA)==2] - lv1 Also
Dear all,
According to the post I was trying:
factorA = c(2,2,3,3,4,4,3,4,2,2)
levels(factorA - c(lv1,lv2,lv3) )
But this returns NULL and doesn't change factor names.
Actually, my factor is included in a data.frame, so I also tried:
levels(df$factorA)[levels(df$factorA)==2] - lv1 Also
On Mar 25, 2011, at 8:30 AM, agent dunham wrote:
Dear all,
According to the post I was trying:
factorA = c(2,2,3,3,4,4,3,4,2,2)
levels(factorA - c(lv1,lv2,lv3) )
Well, this is wrong. Try:
levels(factorA) - c(lv1,lv2,lv3)
factorA
[1] 2 2 3 3 4 4 3 4 2 2
attr(,levels)
[1] lv1 lv2 lv3
-21-09 5:02 PM
To: r-help@r-project.org
Subject: [R] factor with numeric names
Hi all,
I have a pretty basic question about categorical variables but I
can't
seem to be able to find answer so I am hoping someone here can
help. I
found that if the factor names are all in numbers, fitting
Hi all,
I have a pretty basic question about categorical variables but I can't
seem to be able to find answer so I am hoping someone here can help. I
found that if the factor names are all in numbers, fitting the model
in lm would return labels that are not very recognizable.
# Example:
this helps,
John
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Saiwing Yeung
Sent: March-21-09 5:02 PM
To: r-help@r-project.org
Subject: [R] factor with numeric names
Hi all,
I have a pretty basic question about categorical
Hi Saiwing,
If all you are asking is how to rename a factor vector, the easiest way
would be to use:
levels(Loblolly$Seed) - c( a vector of level names you would like to use
for the factor - separated by commas)
If you are asking how to make your output look better, I am not sure I have
an idea
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