If it's not already clear, perhaps this slight variation might help
explain what's happening:
b0<-1
b1<-1
x<-1
str2expr<-function(y){eval(parse(text=y))} ## Note change from "x" to "y" here
test1<-"b0+b1*sqrt(x)"
str2expr(test1)
Cheers,
Bert
On Sun, Oct 31, 2010 at 1:58 PM, Duncan Murdoch
wrot
On 31/10/2010 4:47 PM, Wu Gong wrote:
Hi Duncan:
I'm curious about the environment setting. ?eval says:
"If envir is not specified, then the default is parent.frame() (the
environment where the call to eval was made). "
So what's the difference between set envir=parent.frame() or not?
If yo
Hi Duncan:
I'm curious about the environment setting. ?eval says:
"If envir is not specified, then the default is parent.frame() (the
environment where the call to eval was made). "
So what's the difference between set envir=parent.frame() or not?
Thank you.
Wu
-
A R learner.
--
View t
On Oct 31, 2010, at 3:22 PM, Yilong Zhang wrote:
Dear all:
when I use parse() there is some problems. Below is an example:
b0<-1
b1<-1
x<-1
str2expr<-function(x){eval(parse(text=x))}
test1<-"b0+b1*sqrt(x)"
test2<-"b0+b1"
str2expr(test1)
str2expr(test2)
I don't think the scoping rules are up
On 31/10/2010 3:22 PM, Yilong Zhang wrote:
Dear all:
when I use parse() there is some problems. Below is an example:
b0<-1
b1<-1
x<-1
str2expr<-function(x){eval(parse(text=x))}
test1<-"b0+b1*sqrt(x)"
test2<-"b0+b1"
str2expr(test1)
str2expr(test2)
it can work well for test2 but not for test1.
Dear all:
when I use parse() there is some problems. Below is an example:
b0<-1
b1<-1
x<-1
str2expr<-function(x){eval(parse(text=x))}
test1<-"b0+b1*sqrt(x)"
test2<-"b0+b1"
str2expr(test1)
str2expr(test2)
it can work well for test2 but not for test1.
Could you tell me how to fix this problem or
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