Hi John,
Thanks for the comment... but that appears to mean that SPSS has a big problem.
I have always been told that to include an interaction term in a regression,
the only way is to do the multiplication by hand. But then it seems to be
impossible to stop SPSS from re-standardizing the vari
Well, one correction -- the 'standardized coefficients' that SPSS shows are
based on standardizing all variables separately (so x1, x2, and x1*x2 are all
standardized). So with respect to that, the criticism certainly stands.
-Original Message-
From: Viechtbauer Wolfgang (SP)
Sent: Frid
Totally agree that standardizing the interaction term is nonsense. But in all
fairness, SPSS doesn't do that. In fact, the 'REGRESSION' command in SPSS
doesn't compute any interaction terms -- one has to first compute them 'by
hand' and then add them to the model like any other variable. So some
Regarding the anonymous-function-in-a-pipeline point one can already
do this which does use brackets but even so it involves fewer
characters than the example shown. Here { . * 2 } is basically a
lambda whose argument is dot. Would this be sufficient?
library(magrittr)
1.5 %>% { . * 2 }
##
Dear Nick,
On 2017-05-05, 9:40 AM, "R-devel on behalf of Nick Brown"
wrote:
>>I conjecture that something in the vicinity of
>> res <- lm(DEPRESSION ~ scale(ZMEAN_PA) + scale(ZDIVERSITY_PA) +
>>scale(ZMEAN_PA * ZDIVERSITY_PA), data=dat)
>>summary(res)
>> would reproduce the SPSS Beta values.
>
On Fri, May 5, 2017 at 1:00 PM, Antonin Klima wrote:
> Dear Sir or Madam,
>
> I am in 2nd year of my PhD in bioinformatics, after taking my Master’s in
> computer science, and have been using R heavily during my PhD. As such, I
> have put together a list of certain features in R that, in my opin
Dear Sir or Madam,
I am in 2nd year of my PhD in bioinformatics, after taking my Master’s in
computer science, and have been using R heavily during my PhD. As such, I have
put together a list of certain features in R that, in my opinion, would be
beneficial to add, or could be improved. The fir
>I conjecture that something in the vicinity of
> res <- lm(DEPRESSION ~ scale(ZMEAN_PA) + scale(ZDIVERSITY_PA) +
> scale(ZMEAN_PA * ZDIVERSITY_PA), data=dat)
>summary(res)
> would reproduce the SPSS Beta values.
Yes, that works. Thanks!
- Original Message -
From: "peter dalgaard"
Thanks, I was getting to try this, but got side tracked by actual work...
Your analysis reproduces the SPSS unscaled estimates. It still remains to
figure out how Nick got
>
coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1))
(Intercept) ZMEAN_PA
Thanks for the report, handled in configure in 72661 (R-devel).
I'll also port to R-patched.
Best
Tomas
On 05/04/2017 03:49 PM, Tomas Kalibera wrote:
>
> There is no way to control this at runtime.
> We will probably have to add a configure test.
>
> Best,
> Tomas
>
> On 05/04/2017 03:23 PM, Kasp
I had no problems running regression models in SPSS and R that yielded the same
results for these data.
The difference you are observing is from fitting different models. In R, you
fitted:
res <- lm(DEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=dat)
summary(res)
The interaction term is the produ
Greetings R-devel group. When dealing with Inf dates, as.POSIXct seems to
return Inf, but is printed as NA:
> x1 <- as.POSIXct(Inf, origin = '1970-01-01')
> print(x1)
[1] NA
> is.na(x1)
[1] FALSE
> is.infinite(x1)
[1] TRUE
>
POSIXlt at least evaluates and prints the result consistently:
> x1 <
Hi,
Here is (I hope) all the relevant output from R.
> mean(s1$ZDEPRESSION, na.rm=T) [1] -1.041546e-16 > mean(s1$ZDIVERSITY_PA,
> na.rm=T) [1] -9.660583e-16 > mean(s1$ZMEAN_PA, na.rm=T) [1] -5.430282e-15 >
> lm.ridge(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)$coef ZMEAN_PA
>Z
I asked you before, but in case you missed it: Are you looking at the right
place in SPSS output?
The UNstandardized coefficients should be comparable to R, i.e. the "B" column,
not "Beta".
-pd
> On 5 May 2017, at 01:58 , Nick Brown wrote:
>
> Hi Simon,
>
> Yes, if I uses coefficients() I
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