rcorty - 2 months ago 4

R Question

I have something along the lines of

`y ~ x + z`

And I would like to transform it to

`y ~ x_part1 + x_part2 + z`

More generally, I would like to have a function that takes a formula and returns that formula with all terms that match "^x$" replaced by "x_part1" and "x_part2". Here's my current solution, but it just feels so kludgey...

`my.formula <- fruit ~ apple + banana`

var.to.replace <- 'apple'

my.terms <- labels(terms(my.formula))

new.terms <- paste0('(',

paste0(var.to.replace,

c('_part1', '_part2'),

collapse = '+'),

')')

new.formula <- reformulate(termlabels = gsub(pattern = var.to.replace,

replacement = new.terms,

x = my.terms),

response = my.formula[[2]])

An additional caveat is that the input formula may be specified with interactions.

`y ~ b*x + z`

should output one of these (equivalent) formulae

`y ~ b*(x_part1 + x_part2) + z`

y ~ b + (x_part1 + x_part2) + b:(x_part1 + x_part2) + z

y ~ b + x_part1 + x_part2 + b:x_part1 + b:x_part2 + z

MrFlick has advocated the use of

substitute(y ~ b*x + z, list(x=quote(x_part1 + x_part2)))

but when I have stored the formula I want to modify in a variable, as in

`my.formula <- fruit ~ x + banana`

This approach seems to require a little more massaging:

`substitute(my.formula, list(x=quote(apple_part1 + apple_part2)))`

# my.formula

Answer

You can use the `substitute`

function for this

```
substitute(y ~ b*x + z, list(x=quote(x_part1 + x_part2)))
# y ~ b * (x_part1 + x_part2) + z
```

Here we use the named list to tell R to replace the variable `x`

with the expression `x_part1 + x_part2`

Source (Stackoverflow)

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