[R] Problem with LM
rsherry8
r@herry8 @ending from comc@@t@net
Wed Dec 19 03:14:32 CET 2018
Richard,
It is now working.
Thank you very much.
Bob
On 12/18/2018 7:10 PM, Richard M. Heiberger wrote:
> ## This example, with your variable names, works correctly.
>
> z2 <- data.frame(y=1:5, x=c(1,5,2,3,5), x2=c(1,5,2,3,5)^2)
> z2
> class(z2)
> length(z2)
> dim(z2)
>
> lm(y ~ x + x2, data=z2)
>
> ## note that that variable names y, x, x2 are column names of the
> ## data.frame z2
>
> ## please review the definitions and examples of data.frame in ?data.frame
> ## also the argument requirements for lm in ?lm
>
> On Tue, Dec 18, 2018 at 6:32 PM rsherry8 <rsherry8 using comcast.net> wrote:
>> The values read into z2 came from a CSV file. Please consider this R
>> session:
>>
>> > length(x2)
>> [1] 1632
>> > length(x)
>> [1] 1632
>> > length(z2)
>> [1] 1632
>> > head(z2)
>> [1] 28914.0 28960.5 28994.5 29083.0 29083.0 29083.0
>> > tail(z2)
>> [1] 32729.65 32751.85 32386.05 32379.75 32379.15 31977.15
>> > lm ( y ~ x2 + x, z2 )
>> Error in eval(predvars, data, env) :
>> numeric 'envir' arg not of length one
>> > lm ( y ~ x2 + x, as.data.frme(z2) )
>> Error in as.data.frme(z2) : could not find function "as.data.frme"
>> > lm ( y ~ x2 + x, as.data.frame(z2) )
>> Error in eval(predvars, data, env) :
>> numeric 'envir' arg not of length one
>> lm(formula = y ~ x2 + x, data = as.data.frame(z2))
>>
>> Coefficients:
>> (Intercept) x2 x
>> -1.475e-09 1.000e+00 6.044e-13
>>
>> > min(z2)
>> [1] 24420
>> > max(z2)
>> [1] 35524.85
>> > class(z2)
>> [1] "numeric"
>> >
>>
>> where x is set to x = seq(1:1632)
>> and x2 is set to x^2
>>
>> I am looking for an interpolating polynomial of the form:
>> Ax^2 + Bx + C
>> I do not think the results I got make sense. I believe that I have a
>> data type error. I do not understand why
>> I need to convert z2 to a data frame if it is already numeric.
>>
>> Thanks,
>> Bob
>>
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