CODEWITHSUNDEEP
rdataframelapply
user63230
user63230

Reputation: 4636

R extract coefficients from regression models add string and create dataframe using lapply and sprintf

Lets say I have multiple models (two survival and logistic here for convenience) and I just want to look at sex estimates.

library(survival)
data(colon)
sn <- Surv(colon$time, colon$status)
fit <- coxph(sn ~ sex + perfor + age, data = colon)
fit1 <- coxph(sn ~ sex + perfor + surg + rx , data = colon)
fit2 <- glm(factor(status) ~ sex + age, data=colon, family=binomial(link = "logit")) 
fit3 <- glm(factor(status) ~ sex + age + nodes, data=colon, family=binomial(link = "logit"))

I want the following dataframe (df) as my final output with model name and estimate. I also want a modified version of df, df2, so that there is a grouping effect where logistic and survival models are in different columns. The fact that they are ordered (2 survival followed by 2 logistic) makes this easier. Is there a way to extend this to a more general form and pre-define the layout of the dataset, lets say we had five models of survival/logistic/lme so we need 5 rows x 3 column type dataset. 

> df
  model                    estimate
1   fit 0.97 (95 % CI 0.85 to 1.10)
2  fit1 0.94 (95 % CI 0.83 to 1.07)
3  fit2 0.97 (95 % CI 0.81 to 1.17)
4  fit3 0.98 (95 % CI 0.81 to 1.18)

and also

> df2
  model_survival                sur_estimate model_logistic           logistic_estimate
1            fit 0.97 (95 % CI 0.85 to 1.10)           fit2 0.97 (95 % CI 0.81 to 1.17)
2           fit1 0.94 (95 % CI 0.83 to 1.07)           fit3 0.98 (95 % CI 0.81 to 1.18)

My attempt so far: I have used lapply which I think is better than a for loop and have pretty much worked it out but I would like the section that is outside lapply, to be inside so it is more automated if I had more models etc. See below.

mylist<-list(fit,fit1,fit2,fit3)
results <- list()
results <- lapply(mylist, function(x) {
  sprintf("%.2f (95 %% CI %.2f to %.2f)",     
          exp(coef(x))["sex"], 
          exp(confint(x)[,1])["sex"], 
          exp(confint(x)[,2])["sex"])
})          
results <- do.call(rbind.data.frame, results)

I can make results to look like df by doing the following but I would like this inside lapply so I don't need to get names again, just use them from mylist etc. but results$model<-names(mylist) doesn't work.

colnames(results)[1]<-"estimate"
results$model<-c("fit","fit1","fit2","fit3")

To get df2 I could do some long to wide conversion but can I pre-define the layout and column names inside lapply (I know I will probably need two separate lapply - one for df and df2). Thanks.

Upvotes: 1

Views: 253

Answers (1)

akrun
akrun

Reputation: 886938

We can use map with stack

library(tidyverse)
out <- mget(ls(pattern = "fit\\d*")) %>% 
        map(~sprintf("%.2f (95 %% CI %.2f to %.2f)",     
           exp(coef(.x))["sex"], 
           exp(confint(.x)[,1])["sex"], 
           exp(confint(.x)[,2])["sex"])) %>%
        stack %>%
        select(model = ind, estimate = values)
out
#  model                    estimate
#1   fit 0.97 (95 % CI 0.85 to 1.10)
#2  fit1 0.94 (95 % CI 0.83 to 1.07)
#3  fit2 0.97 (95 % CI 0.81 to 1.17)
#4  fit3 0.98 (95 % CI 0.81 to 1.18)

From the 'out', we can get the second output

library(data.table)#using dcast as it can take multiple value.vars
out %>%
   group_by(group = rep(c("model_survival", "model_logistic"), each = 2)) %>%
   mutate(rn = row_number()) %>%
   as.data.table %>%
   dcast(., rn ~ group, value.var = c('model', 'estimate')) %>% 
   select(-rn)
# model_model_logistic model_model_survival     estimate_model_logistic     estimate_model_survival
#1:                 fit2                  fit 0.97 (95 % CI 0.81 to 1.17) 0.97 (95 % CI 0.85 to 1.10)
#2:                 fit3                 fit1 0.98 (95 % CI 0.81 to 1.18) 0.94 (95 % CI 0.83 to 1.07)

Upvotes: 1

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