I have built a matrix whose names are those of a regressor subset that i want to insert in a regression model formula in R. For example:
data$age is the response variable
X is the design matrix whose column names are, for example, data$education and data$wage.
The problem is that the column names of X are not fixed (i.e. i don't know which are them in advance), so i tried to code this:
best_model <- lm(data$age ~ paste(colnames(x[, GA@solution == 1]), sep = "+"))
But it doesn't work.
Rather than writing formula by yourself, using pipe(%>%) and dplyr::select() appropriately might be helpful. (Here, change your matrix to data frame.)
library(tidyverse)
mpg
#> # A tibble: 234 x 11
#> manufacturer model displ year cyl trans drv cty hwy fl class
#> <chr> <chr> <dbl> <int> <int> <chr> <chr> <int> <int> <chr> <chr>
#> 1 audi a4 1.8 1999 4 auto… f 18 29 p comp…
#> 2 audi a4 1.8 1999 4 manu… f 21 29 p comp…
#> 3 audi a4 2 2008 4 manu… f 20 31 p comp…
#> 4 audi a4 2 2008 4 auto… f 21 30 p comp…
#> 5 audi a4 2.8 1999 6 auto… f 16 26 p comp…
#> 6 audi a4 2.8 1999 6 manu… f 18 26 p comp…
#> 7 audi a4 3.1 2008 6 auto… f 18 27 p comp…
#> 8 audi a4 q… 1.8 1999 4 manu… 4 18 26 p comp…
#> 9 audi a4 q… 1.8 1999 4 auto… 4 16 25 p comp…
#> 10 audi a4 q… 2 2008 4 manu… 4 20 28 p comp…
#> # ... with 224 more rows
dplyr::select() subsets column.
mpg %>%
select(hwy, manufacturer, displ, cyl, cty) %>% # subsetting
lm(hwy ~ ., data = .)
#>
#> Call:
#> lm(formula = hwy ~ ., data = .)
#>
#> Coefficients:
#> (Intercept) manufacturerchevrolet manufacturerdodge
#> 2.65526 -1.08632 -2.55442
#> manufacturerford manufacturerhonda manufacturerhyundai
#> -2.29897 -2.98863 -0.94980
#> manufacturerjeep manufacturerland rover manufacturerlincoln
#> -3.36654 -1.87179 -1.10739
#> manufacturermercury manufacturernissan manufacturerpontiac
#> -2.64828 -2.44447 0.75427
#> manufacturersubaru manufacturertoyota manufacturervolkswagen
#> -3.04204 -2.73963 -1.62987
#> displ cyl cty
#> -0.03763 0.06134 1.33805
Denote that -col.name exclude that column. %>% enables formula to use . notation.
Lots of data sets group their columns using underscore.
nycflights13::flights
#> # A tibble: 336,776 x 19
#> year month day dep_time sched_dep_time dep_delay arr_time
#> <int> <int> <int> <int> <int> <dbl> <int>
#> 1 2013 1 1 517 515 2 830
#> 2 2013 1 1 533 529 4 850
#> 3 2013 1 1 542 540 2 923
#> 4 2013 1 1 544 545 -1 1004
#> 5 2013 1 1 554 600 -6 812
#> 6 2013 1 1 554 558 -4 740
#> 7 2013 1 1 555 600 -5 913
#> 8 2013 1 1 557 600 -3 709
#> 9 2013 1 1 557 600 -3 838
#> 10 2013 1 1 558 600 -2 753
#> # ... with 336,766 more rows, and 12 more variables: sched_arr_time <int>,
#> # arr_delay <dbl>, carrier <chr>, flight <int>, tailnum <chr>,
#> # origin <chr>, dest <chr>, air_time <dbl>, distance <dbl>, hour <dbl>,
#> # minute <dbl>, time_hour <dttm>
For instance, both dep_delay and arr_delay are about delay time. Select helpers such as starts_with(), ends_with(), and contains() can handle this kind of columns.
nycflights13::flights %>%
select(starts_with("sched"),
ends_with("delay"),
distance)
#> # A tibble: 336,776 x 5
#> sched_dep_time sched_arr_time dep_delay arr_delay distance
#> <int> <int> <dbl> <dbl> <dbl>
#> 1 515 819 2 11 1400
#> 2 529 830 4 20 1416
#> 3 540 850 2 33 1089
#> 4 545 1022 -1 -18 1576
#> 5 600 837 -6 -25 762
#> 6 558 728 -4 12 719
#> 7 600 854 -5 19 1065
#> 8 600 723 -3 -14 229
#> 9 600 846 -3 -8 944
#> 10 600 745 -2 8 733
#> # ... with 336,766 more rows
After that, just %>% lm().
nycflights13::flights %>%
select(starts_with("sched"),
ends_with("delay"),
distance) %>%
lm(dep_delay ~ ., data = .)
#>
#> Call:
#> lm(formula = dep_delay ~ ., data = .)
#>
#> Coefficients:
#> (Intercept) sched_dep_time sched_arr_time arr_delay
#> -0.151408 0.002737 0.000951 0.816684
#> distance
#> 0.001859
