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Defining partial list of variables

12 messages · Jeff Newmiller, Eric Berger, Steven Yen +2 more

#
I constantly define variable lists from a data frame (e.g., to define a 
regression equation). Line 3 below does just that. Placing each variable 
name in quotation marks is too much work especially for a long list so I 
do that with line 4. Is there an easier way to accomplish this----to 
define a list of variable names containing "a","c","e"? Thank you!

 > data<-as.data.frame(matrix(1:30,nrow=6))
 > colnames(data)<-c("a","b","c","d","e"); data

 ? a? b? c? d? e
1 1? 7 13 19 25
2 2? 8 14 20 26
3 3? 9 15 21 27
4 4 10 16 22 28
5 5 11 17 23 29
6 6 12 18 24 30
 > x1<-c("a","c","e"); x1 # line 3
[1] "a" "c" "e"
 > x2<-colnames(subset(data,select=c(a,c,e))); x2 # line 4

[1] "a" "c" "e"
#
see below

Steven Yen wrote/hat geschrieben on/am 05.01.2021 08:14:
What about:
x3 <- names(data)[c(1,3,5)]
x3
[1] "a" "c" "e"

If I have to compile longer vectors of variable names I do it as follows:
First I use:
dput(names(data))
resulting in a vector of names.
c("a", "b", "c", "d", "e")
Then I edit the output by hand, e.g.
x4 <- c("a", "b", "c", "d", "e")
x4 <- c("a", "c", "e")
This is especially useful with long names, where I could easily make
typing errors.

regards,
Heinz
#
IMO if you want to hardcode a formula then simply hardcode a formula. If you want 20 formulas, write 20 formulas. Is that really so bad?

If you want to have an abbreviated way to specify sets of variables without conforming to R syntax then put them into data files and read them in using a format of your choice.

But using NSE to avoid using quotes for entering what amounts to in-script data is abuse of the language justified by laziness... the amount of work you put yourself and anyone else who reads your code through is excessive relative to the benefit gained.

NSE has its strengths... but as a method of creating data objects it sucks. Note that even the tidyverse (now) requires you to use quotes when you are not directly referring to something that already exists. And if you were... you might as well be creating a formula.
On January 4, 2021 11:14:54 PM PST, Steven Yen <styen at ntu.edu.tw> wrote:

  
    
#
Thank you, Jeff. IMO, we are all here to make R work better to suit our 
various needs. All I am asking is an easier way to define variable list 
zx, differently from the way z0 , x0, and treat are defined.

 > zx<-colnames(subset(mydata,select=c(
+ age,exercise,income,white,black,hispanic,base,somcol,grad,employed,
+???? unable,homeowner,married,divorced,widowed)))
 > z0<-c("fruit","highblood")
 > x0<-c("vgood","poor")
 > treat<-"depression"
 > eq1 <-my.formula(y="depression",x=zx,z0)
 > eq2 <-my.formula(y="bmi",?????? x=zx,x0)
 > eq2t<-my.formula(y="bmi",?????? x=zx,treat)
 > eqs<-list(eq1,eq2); eqs
[[1]]
depression ~ age + exercise + income + white + black + hispanic +
 ??? base + somcol + grad + employed + unable + homeowner + married +
 ??? divorced + widowed + fruit + highblood

[[2]]
bmi ~ age + exercise + income + white + black + hispanic + base +
 ??? somcol + grad + employed + unable + homeowner + married +
 ??? divorced + widowed + vgood + poor

 > eqt<-list(eq1,eq2t); eqt
[[1]]
depression ~ age + exercise + income + white + black + hispanic +
 ??? base + somcol + grad + employed + unable + homeowner + married +
 ??? divorced + widowed + fruit + highblood

[[2]]
bmi ~ age + exercise + income + white + black + hispanic + base +
 ??? somcol + grad + employed + unable + homeowner + married +
 ??? divorced + widowed + depression
On 2021/1/5 ?? 04:18, Jeff Newmiller wrote:
#
zx<-strsplit("age,exercise,income,white,black,hispanic,base,somcol,grad,employed,unable,homeowner,married,divorced,widowed",",")
On Tue, Jan 5, 2021 at 11:01 AM Steven Yen <styen at ntu.edu.tw> wrote:

            

  
  
#
Here we go! BUT, it works great for a continuous line. With line 
break(s), I got the nuisance "\n" inserted.

 > x<-strsplit("hhsize,urban,male,gov,nongov,married",","); x
[[1]]
[1] "hhsize"? "urban"?? "male"??? "gov"???? "nongov"? "married"

 > x<-strsplit("hhsize,urban,male,
+???????????? gov,nongov,married",","); x
[[1]]
[1] "hhsize"??????????? "urban"???????????? "male" "\n??????????? gov"
[5] "nongov"??????????? "married"
On 2021/1/5 ?? 05:34, Eric Berger wrote:

  
  
#
If your column names have no spaces the following should work

 x<-strsplit(gsub("[\n ]","",
             "hhsize,urban,male,
+             gov,nongov,married"),","); x
On Tue, Jan 5, 2021 at 11:47 AM Steven Yen <styen at ntu.edu.tw> wrote:

            

  
  
#
Thanks Eric. Perhaps I should know when to stop. The approach produces a 
slightly different variable list (note the [[1]]). Consequently, I was 
not able to use xx in defining my regression formula.

 > x<-colnames(subset(mydata,select=c(

+??? hhsize,urban,male,
+??? age3045,age4659,age60, # age1529
+??? highsc,tert,?????????? # primary
+??? gov,nongov,??????????? # unemp
+??? married))); x
 ?[1] "hhsize"? "urban"?? "male"??? "age3045" "age4659" "age60" 
"highsc"? "tert"
 ?[9] "gov"???? "nongov"? "married"
 > xx<-strsplit(gsub("[\n ]","",
+??? "hhsize,urban,male,
+???? age3045,age4659,age60,
+???? highsc,tert,
+???? gov,nongov,
+???? married"
+ ),","); xx
[[1]]
 ?[1] "hhsize"? "urban"?? "male"??? "age3045" "age4659" "age60" 
"highsc"? "tert"
 ?[9] "gov"???? "nongov"? "married"

 > eq1<-my.formula(y="cig",x=x); eq1
cig ~ hhsize + urban + male + age3045 + age4659 + age60 + highsc +
 ??? tert + gov + nongov + married
 > eq2<-my.formula(y="cig",x=xx); eq2
cig ~ c("hhsize", "urban", "male", "age3045", "age4659", "age60",
 ??? "highsc", "tert", "gov", "nongov", "married")
On 2021/1/5 ?? 06:01, Eric Berger wrote:

  
  
#
wrap it in unlist

xx <- unlist(strsplit( .... ))
On Tue, Jan 5, 2021 at 12:59 PM Steven Yen <styen at ntu.edu.tw> wrote:

            

  
  
#
Thanks Eric. Yes, "unlist" makes a difference. Below, I am doing not 
regression but summary to keep the example simple.

 > set.seed(123)
 > data<-matrix(runif(1:25),nrow=5)
 > colnames(data)<-c("x1","x2","x3","x4","x5"); data
 ??????????? x1??????? x2??????? x3???????? x4??????? x5
[1,] 0.2875775 0.0455565 0.9568333 0.89982497 0.8895393
[2,] 0.7883051 0.5281055 0.4533342 0.24608773 0.6928034
[3,] 0.4089769 0.8924190 0.6775706 0.04205953 0.6405068
[4,] 0.8830174 0.5514350 0.5726334 0.32792072 0.9942698
[5,] 0.9404673 0.4566147 0.1029247 0.95450365 0.6557058
 > j<-strsplit(gsub("[\n ]","","x1,x3,x5"),",")
 > j<-unlist(j); j
[1] "x1" "x3" "x5"
 > summary(data[,j])
 ?????? x1?????????????? x3?????????????? x5
 ?Min.?? :0.2876?? Min.?? :0.1029?? Min.?? :0.6405
 ?1st Qu.:0.4090?? 1st Qu.:0.4533?? 1st Qu.:0.6557
 ?Median :0.7883?? Median :0.5726?? Median :0.6928
 ?Mean?? :0.6617?? Mean?? :0.5527?? Mean?? :0.7746
 ?3rd Qu.:0.8830?? 3rd Qu.:0.6776?? 3rd Qu.:0.8895
 ?Max.?? :0.9405?? Max.?? :0.9568?? Max.?? :0.9943
On 2021/1/5 ?? 07:08, Eric Berger wrote:

  
  
#
What about the Cs()-function in Hmisc?
library(Hmisc)
Cs(a,b,c)
[1] "a" "b" "c"

Steven Yen wrote/hat geschrieben on/am 05.01.2021 13:29:
#
I may not I properly understand the context of this discussion, and, in
particular what the my.formula() function does. But if I do, the following,
from ?formula, seems relevant and would indicate that the discussion is
unnecessary:

"There are two special interpretations of . in a formula. The usual one is
in the context of a data argument of model fitting functions and means ?all
columns not otherwise in the formula?:"

This means you can fit different models just by indexing the columns -- by
number --  you wish to use in a data argument, viz:

y <- runif(100)
dat <- data.frame(matrix(runif(500), ncol = 5))
names(dat) <- letters[1:5]
head(dat)

## Use columns 1,3, and 5 only
mdl1 <- lm(y ~ ., data = dat[,c(1,3,5)])

## Result:
 summary(mdl1)

Call:
lm(formula = y ~ ., data = dat[, c(1, 3, 5)])

Residuals:
     Min       1Q   Median       3Q      Max
-0.52334 -0.27494  0.01245  0.28637  0.51998

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.51461    0.08236   6.248 1.14e-08 ***
a            0.01516    0.10928   0.139    0.890
c            0.03517    0.10399   0.338    0.736
e           -0.09437    0.10967  -0.861    0.392
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.299 on 96 degrees of freedom
Multiple R-squared:  0.008256, Adjusted R-squared:  -0.02274
F-statistic: 0.2664 on 3 and 96 DF,  p-value: 0.8495


If I have misunderstood and this is unhelpful, just ignore without comment.
You don't need to waste time explaining it to me.

Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue, Jan 5, 2021 at 4:49 AM Heinz Tuechler <tuechler at gmx.at> wrote: