CODEWITHSUNDEEP
rdplyrlapplyglmpredict
Stata_user
Stata_user

Reputation: 564

Predicting from list using lapply

I am trying to estimate a group of models using dplyr and lapply. I estimate a probit regression, where results are stored in a list. Then I would like to use predict function to predict values on a new dataset. My model runs, but I get zero values as results. What am I doing wrong?

# Code from the original question
library(dplyr)

year <- rep(2014:2015, length.out=10000)
group <- sample(c(0,1,2,3,4,5,6), replace=TRUE, size=10000)
value <- sample(10000, replace=T)
female <- sample(c(0,1), replace=TRUE, size=10000)
smoker <- sample(c(0,1), replace=TRUE, size=10000)
dta <- data.frame(year=year, group=group, value=value, female=female, smoker=smoker)

# cut the dataset into list
table_list <- dta %>%
  group_by(year, group) %>%
  group_split()

# fit model per subgroup
model_list <- lapply(table_list, function(x) glm(smoker ~ female, data=x,
                                                 family=binomial(link="probit")))

# create new dataset where female =1
dat_new <- data.frame(dta[, c("smoker", "year", "group")], female=1) 

# cut into list
pred_list <- dat_new %>%
  group_by(year, group) %>%
  group_split()

# do prediction
pred2 <- Map(function(x, y) predict.glm(x, type = "response", newdata = y), 
             model_list, pred_list)

I get zero results predicted. Why?

Upvotes: 1

Views: 663

Answers (1)

Ronak Shah
Ronak Shah

Reputation: 388862

You should lapply over model_list instead.

pred1 <- lapply(model_list, function(x) predict.glm(x, type = "response"))

Or if you want to pass data use Map.

pred2 <- Map(function(x, y) predict.glm(x, type = "response", newdata = y), 
          model_list, pred_list)

Upvotes: 1

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