[R] Need help with lm function on MAC OS X. R version - 3.2.0

samarvir singh samarvir1996 at gmail.com
Thu May 14 14:07:32 CEST 2015


I Have a data frame named BSE and CP is my independent variable
and here;s the error I get if I try to run an lm function
any idea whats wrong

P.S - all my data is in numeric except company which is a factor. I have
2700 row and 450 variable

P.P.S -  I have no missing data, I have 0 in Empty field.

> BSE_Reg <- lm(CP ~.-company, data = bse)
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
  contrasts can be applied only to factors with 2 or more levels

> BSE_Reg <- lm(CP ~., data = bse)
Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
  contrasts can be applied only to factors with 2 or more levels
Error in as.character(tools:::httpdPort) :
  cannot coerce type 'closure' to vector of type 'character'
Error in as.character(tools:::httpdPort) :
  cannot coerce type 'closure' to vector of type 'character'
Error in as.character(tools:::httpdPort) :
  cannot coerce type 'closure' to vector of type 'character'
Error in as.character(tools:::httpdPort) :
  cannot coerce type 'closure' to vector of type 'character'





> R.Version()
$platform
[1] "x86_64-apple-darwin13.4.0"

$arch
[1] "x86_64"

$os
[1] "darwin13.4.0"

$system
[1] "x86_64, darwin13.4.0"

$status
[1] ""

$major
[1] "3"

$minor
[1] "2.0"

$year
[1] "2015"

$month
[1] "04"

$day
[1] "16"

$`svn rev`
[1] "68180"

$language
[1] "R"

$version.string
[1] "R version 3.2.0 (2015-04-16)"

$nickname
[1] "Full of Ingredients"

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