How to apply list to a function which give data frame as output

Question

I have applied a single argument function which outputs a dataframe.

library(tidyverse) 
my_function <-function(x) {mtcars %>%
                         filter_(x) %>%
                         group_by(cyl) %>%
                         summarise(mean(disp), mean(drat)) %>%
                         mutate(group=x)}

When I feed a single argument into this function, it gives the output as expected i.e. a dataframe

   my_function('mpg>15')

   cyl      mean(disp)   mean(drat)     group
    4       105           4.07          mpg>15
    6       183           3.59          mpg>15
    8       105           3.20          mpg>15

Now, how do I apply such a function to a list of arguments that gives output as one dataframe which combines all the reults over the list.

For example I would like to apply my_function to a list

c('mpg>15', 'drat>4.2')

and, as the result, to obtain a single dataframe:

cyl      mean(disp)   mean(drat)     group
 4       105           4.07          mpg>15
 6       183           3.59          mpg>15
 8       105           3.20          mpg>15
 4       89            4.53          drat>4.2
 8       351           4.22          drat>4.2

How to do that ?

Answer 1

If you use  tidyverse, you can use something like this:

c("mpg>15", "am==1") %>% map(myfun) %>% bind_rows

But, we can also shorten this by using map_df, which returns a data frame:

c("mpg>15", "am==1") %>% map_df(my_function)

A mixed option, three equivalent ways, using lapply:

lapply(c("mpg>15", "am==1"), my_function) %>% bind_rows 
c("mpg>15", "am==1") %>% lapply(my_function) %>% bind_rows
bind_rows(lapply(c("mpg>15", "am==1"), my_function))

You can also mix base and tidyverse:

c("mpg>15", "am==1") %>% lapply(my_function) %>% do.call(rbind, .)

Another option is as follows:

do.call(rbind, lapply(c("mpg>15", "am==1"), my_function))