CODEWITHSUNDEEP
rdplyrmodelformulaeval
zhiwei li
zhiwei li

Reputation: 1711

Can not evaluate an object in r

I have a model.

library(dlnm)
library(splines)
data(chicagoNMMAPS)

cb1.pm <- crossbasis(chicagoNMMAPS$pm10, lag=15, argvar=list(fun="lin"),
                     arglag=list(fun="poly",degree=4))


model1 <- glm(death ~ cb1.pm + ns(time, 7*14) + dow,
              family=quasipoisson(), chicagoNMMAPS)

pred1.pm1 <- crosspred(cb1.pm, model1, bylag=0.2, cumul=TRUE)

I want to use a character vector to call the object cb1.pm so I tried this.

# use character vector

i = 'pm'

cb1.var = paste0('cb1.', i) %>% get()

model2 <- glm(death ~ cb1.var + ns(time, 7*14) + dow,
              family=quasipoisson(), chicagoNMMAPS)


pred1.pm2 <- crosspred(cb1.pm, model2, bylag=0.2, cumul=TRUE) # Error occur here 

identical(cb1.pm, cb1.var) # TRUE

The result identical(cb1.pm, cb1.var) is TURE, but why can't these two variables replace each other?

Any help will be highly appreciated!

Upvotes: 0

Views: 55

Answers (1)

Allan Cameron
Allan Cameron

Reputation: 173793

The problem is that crosspred uses non-standard evaluation to extract the name of the first argument from your call and tries to match it up to the variables in model2. It does this using grep. Since you have used cb1.var as the name in your model, the crosspred function doesn't find any coefficients containing the string "cb1.pm" and throws an error.

The way round this is to ensure that cb1.var remains as a symbol and is substituted for cb1.pm before the model is built. This is also more efficient, since you don't need to copy the whole cb1.pm data:

i = 'pm'

cb1.var = paste0('cb1.', i)

form <- as.formula(paste("death ~", cb1.var, "+  ns(time, 7*14) + dow"))

model2 <- glm(form, family = quasipoisson(), chicagoNMMAPS)

pred1.pm2 <- crosspred(cb1.pm, model2, bylag = 0.2, cumul=TRUE)

We no longer get a warning, and we can confirm we have the correct model:

identical(pred1.pm1, pred1.pm2)
#> [1] TRUE

Upvotes: 1

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