This is the output of dput(your data)
structure(list(Ca = c(NA, NA, 24.4, NA, 21.4, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 28, 32, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, 34.7, NA, 42.5, NA, 26, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.6, 21.4, NA, 48.3,
63.5
There is only one row with a complete set of observations; I think lm() is
throwing out the rest.
Rich Shepard wrote:
>
> On Wed, 9 Nov 2011, John C Frain wrote:
>
>> As far as I know if there is an NA in any variable in an observation the
>> default is to drop the entire observation. Thus the
On 09-Nov-11 19:39:54, Rich Shepard wrote:
> On Wed, 9 Nov 2011, John C Frain wrote:
>
>> As far as I know if there is an NA in any variable in an
>> observation the default is to drop the entire observation.
>> Thus there are no observations in your calculation
>
> John,
>
> Hadn't realized th
On Wed, 9 Nov 2011, John C Frain wrote:
As far as I know if there is an NA in any variable in an observation the
default is to drop the entire observation. Thus there are no observations
in your calculation
John,
Hadn't realized that. I know there are NA's in other data frames that
yield mo
On Wed, 9 Nov 2011, Marc Schwartz wrote:
# 'DF' is the result of copying your data above from the
# clipboard on OSX
DF <- read.table(pipe("pbpaste"), header = TRUE)
Marc,
Oh? I don't do Apple so there's no OSX here.
After removing incomplete records (any records with NA values) which is
As far as I know if there is an NA in any variable in an observation
the default is to drop the entire observation. Thus there are no
observations in your calculation
Best Regards
John
On 9 November 2011 19:17, Rich Shepard wrote:
> On Wed, 9 Nov 2011, Daniel Nordlund wrote:
>
>> I would guess
On Nov 9, 2011, at 2:17 PM, Rich Shepard wrote:
On Wed, 9 Nov 2011, Daniel Nordlund wrote:
I would guess that there is something problematic with the how the
data
frame is structured relative to what lm() is expecting.
Dan,
I was not comfortable with my explanation, but the formula (and
On Nov 9, 2011, at 1:17 PM, Rich Shepard wrote:
> On Wed, 9 Nov 2011, Daniel Nordlund wrote:
>
>> I would guess that there is something problematic with the how the data
>> frame is structured relative to what lm() is expecting.
>
> Dan,
>
> I was not comfortable with my explanation, but the
On Wed, 9 Nov 2011, Daniel Nordlund wrote:
I would guess that there is something problematic with the how the data
frame is structured relative to what lm() is expecting.
Dan,
I was not comfortable with my explanation, but the formula (and data
frame) was equivalent to those of the other 8
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Rich Shepard
> Sent: Wednesday, November 09, 2011 9:42 AM
> To: r-help@r-project.org
> Subject: Re: [R] Interpreting Multiple Linear Regression Summary
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Rich Shepard
> Sent: Wednesday, November 09, 2011 9:05 AM
> To: r-help@r-project.org
> Subject: [R] Interpreting Multiple Linear Regression Summary
>
&g
On Wed, 9 Nov 2011, David Winsemius wrote:
I don't see a data= argument specified, so you are telling lm() that your
workspace has individual vectors by those names in the formula. That is
not what is implied by hte rest of your message.
David,
That's because I attached the data frame befor
On Nov 9, 2011, at 12:04 PM, Rich Shepard wrote:
I would appreciate pointers on what I should read to understand this
output:
summary(lm(TDS ~ Cond + Ca + Cl + Mg + Na + SO4))
I don't see a data= argument specified, so you are telling lm() that
your workspace has individual vectors by tho
Please see ?dput
use dput(your data) and paste the output into a reply, thanks.
This way we know what you are working with.
Rich Shepard wrote:
>
> I would appreciate pointers on what I should read to understand this
> output:
>
> summary(lm(TDS ~ Cond + Ca + Cl + Mg + Na + SO4))
>
> Ca
I would appreciate pointers on what I should read to understand this
output:
summary(lm(TDS ~ Cond + Ca + Cl + Mg + Na + SO4))
Call:
lm(formula = TDS ~ Cond + Ca + Cl + Mg + Na + SO4)
Residuals:
ALL 1 residuals are 0: no residual degrees of freedom!
Coefficients: (6 not defined because of s
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